100 results found for "no-real-roots" in Class 10.
कथन: \(x^2+3x+7=0\) के वास्तविक मूल नहीं हैं। कारण: (D<0) होने पर वास्तविक मूल नहीं होते। सही विकल्प चुनिए।
Assertion: \(x^2+3x+7=0\) has no real roots. Reason: When (D<0), real roots do not exist. Choose the correct option.
#quadratic-equations
#assertion-reason
#no-real-roots
A कथन और कारण दोनों सही हैं / Both assertion and reason are correct
B कथन सही है, कारण गलत है / Assertion is correct, reason is wrong
C कथन गलत है, कारण सही है / Assertion is wrong, reason is correct
D कथन और कारण दोनों गलत हैं / Both assertion and reason are wrong
Explanation opens after your attempt
Correct Answer
A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct
Step 1
Concept
Here (D=32 -4(1)(7)=-19). Since (D<0), the assertion is correct.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct. Here (D=32 -4(1)(7)=-19). Since (D<0), the assertion is correct.
Step 3
Exam Tip
यहाँ (D=32 -4(1)(7)=-19) है। (D<0) होने से कथन सही है।
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\(x^2+12x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+12x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(0<\lambda<36\)
B \(\lambda=36\)
C \(\lambda>36\)
D \(\lambda<0\)
Explanation opens after your attempt
Correct Answer
A. \(0<\lambda<36\)
Step 1
Concept
For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).
Step 2
Why this answer is correct
The correct answer is A. \(0<\lambda<36\). For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-12) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(144-4\lambda>0\), इसलिए \(0<\lambda<36\)।
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\(x^2+10x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+10x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(\lambda<0\)
B \(0<\lambda<25\)
C \(\lambda=25\)
D \(\lambda>25\)
Explanation opens after your attempt
Correct Answer
B. \(0<\lambda<25\)
Step 1
Concept
For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).
Step 2
Why this answer is correct
The correct answer is B. \(0<\lambda<25\). For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-10) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(100-4\lambda>0\), इसलिए \(0<\lambda<25\)।
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\(x^2+2x+\lambda=0\) की जड़ें वास्तविक और भिन्न हों तथा दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+2x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(0<\lambda<1\)
B \(\lambda>1\)
C \(\lambda<0\)
D \(\lambda=1\)
Explanation opens after your attempt
Correct Answer
A. \(0<\lambda<1\)
Step 1
Concept
For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).
Step 2
Why this answer is correct
The correct answer is A. \(0<\lambda<1\). For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-2) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(4-4\lambda>0\), इसलिए \(0<\lambda<1\)।
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यदि \(kx^2-12x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हैं, तो (k) पर सही शर्त क्या है?
If \(kx^2-12x+k=0\) has real reciprocal roots, what is the correct condition on (k)?
#quadratic-roots
#reciprocal-roots
#real-roots
A \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\)
B (k=0)
C \(k^2>36\)
D (k=12) केवल / (k=12) only
Explanation opens after your attempt
Correct Answer
A. \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\)
Step 1
Concept
The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(144-4k^2\ge0\), hence \(k^2\le36\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\). The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(144-4k^2\ge0\), hence \(k^2\le36\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(144-4k^2\ge0\), अतः \(k^2\le36\)।
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यदि \(kx^2-10x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हैं, तो (k) पर सही शर्त कौन-सी है?
If \(kx^2-10x+k=0\) has real reciprocal roots, which condition on (k) is correct?
#quadratic-roots
#reciprocal-roots
#real-roots
A (k=0)
B \(k^2>25\)
C \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\)
D (k=10) केवल / (k=10) only
Explanation opens after your attempt
Correct Answer
C. \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\)
Step 1
Concept
The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(100-4k^2\ge0\), hence \(k^2\le25\).
Step 2
Why this answer is correct
The correct answer is C. \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\). The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(100-4k^2\ge0\), hence \(k^2\le25\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(100-4k^2\ge0\), अतः \(k^2\le25\)।
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यदि \(kx^2-8x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हैं, तो (k) पर सही शर्त क्या है?
If the roots of \(kx^2-8x+k=0\) are real and reciprocal, what is the correct condition on (k)?
#quadratic-roots
#reciprocal-roots
#real-roots
A \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\)
B (k=0)
C \(k^2>16\)
D (k=8) केवल / (k=8) only
Explanation opens after your attempt
Correct Answer
A. \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\)
Step 1
Concept
For reciprocal roots, \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(64-4k^2\ge0\), hence \(k^2\le16\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\). For reciprocal roots, \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(64-4k^2\ge0\), hence \(k^2\le16\).
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(64-4k^2\ge0\), अतः \(k^2\le16\)।
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\(x^2-4x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हों, तो (k) का मान क्या है?
If the roots of \(x^2-4x+k=0\) are real and reciprocal, what is (k)?
#quadratic-roots
#reciprocal-roots
#real-roots
A (1)
B (2)
C (4)
D (-1)
Explanation opens after your attempt
Step 1
Concept
For reciprocal roots, \(\alpha\beta=1\). Here \(\alpha\beta=k\), so (k=1), and (D=12>0) confirms real roots.
Step 2
Why this answer is correct
The correct answer is A. (1). For reciprocal roots, \(\alpha\beta=1\). Here \(\alpha\beta=k\), so (k=1), and (D=12>0) confirms real roots.
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए \(\alpha\beta=1\) होता है। यहाँ \(\alpha\beta=k\), इसलिए (k=1), और (D=12>0) से जड़ें वास्तविक भी हैं।
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यदि \(x^2-6x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=26\), तो जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-6x+c=0\) and \(\alpha^2+\beta^2=26\), what are the roots?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (5) / (1) and (5)
B (2) और (4) / (2) and (4)
C (3) और (3) / (3) and (3)
D (0) और (6) / (0) and (6)
Explanation opens after your attempt
Correct Answer
A. (1) और (5) / (1) and (5)
Step 1
Concept
Here \(\alpha+\beta=6\) and \(\alpha^2+\beta^2=26\). From \(36-2\alpha\beta=26\), \(\alpha\beta=5\), so the roots are (1) and (5).
Step 2
Why this answer is correct
The correct answer is A. (1) और (5) / (1) and (5). Here \(\alpha+\beta=6\) and \(\alpha^2+\beta^2=26\). From \(36-2\alpha\beta=26\), \(\alpha\beta=5\), so the roots are (1) and (5).
Step 3
Exam Tip
\(\alpha+\beta=6\) और \(\alpha^2+\beta^2=26\) है। \(36-2\alpha\beta=26\) से \(\alpha\beta=5\), इसलिए जड़ें (1) और (5) हैं।
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यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(2\alpha+3\) और \(2\beta+3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), which equation has roots \(2\alpha+3\) and \(2\beta+3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-20x+99=0\)
B \(x^2-14x+99=0\)
C \(x^2-20x+91=0\)
D \(x^2+20x+99=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+99=0\)
Step 1
Concept
The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+99=0\). The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 3
Exam Tip
मूल जड़ें (3) और (4) हैं। नई जड़ें (9) और (11) हैं, इसलिए समीकरण \(x^2-20x+99=0\) है।
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यदि \(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha+\beta\) और \(\alpha\beta\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are roots of \(x^2-5x+6=0\), which equation has roots \(\alpha+\beta\) and \(\alpha\beta\)?
#quadratic-roots
#new-equation
#sum-product-roots
A \(x^2-11x+30=0\)
B \(x^2+11x+30=0\)
C \(x^2-5x+6=0\)
D \(x^2-30x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+30=0\)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=6\) हैं। नई जड़ें (5) और (6) हैं, इसलिए समीकरण \(x^2-11x+30=0\) है।
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यदि \(x^2-5x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=17\), तो जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-5x+c=0\) and \(\alpha^2+\beta^2=17\), what are the roots?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (4) / (1) and (4)
B (2) और (3) / (2) and (3)
C (0) और (5) / (0) and (5)
D (-1) और (6) / (-1) and (6)
Explanation opens after your attempt
Correct Answer
A. (1) और (4) / (1) and (4)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 2
Why this answer is correct
The correct answer is A. (1) और (4) / (1) and (4). Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha^2+\beta^2=17\) है। \(25-2\alpha\beta=17\) से \(\alpha\beta=4\), इसलिए जड़ें (1) और (4) हैं।
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यदि \(x^2-9x+14=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-3\) और \(\beta-3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-9x+14=0\), which equation has roots \(\alpha-3\) and \(\beta-3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-3x-4=0\)
B \(x^2+3x-4=0\)
C \(x^2-3x+4=0\)
D \(x^2-9x+14=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-4=0\)
Step 1
Concept
The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-4=0\). The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (7) हैं। नई जड़ें (-1) और (4) होंगी, इसलिए समीकरण \(x^2-3x-4=0\) है।
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यदि \(x^2-4x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=10\), तो समीकरण की जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-4x+c=0\) and \(\alpha^2+\beta^2=10\), what are the roots of the equation?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (3) / (1) and (3)
B (2) और (2) / (2) and (2)
C (0) और (4) / (0) and (4)
D (-1) और (5) / (-1) and (5)
Explanation opens after your attempt
Correct Answer
A. (1) और (3) / (1) and (3)
Step 1
Concept
Here \(\alpha+\beta=4\) and \(\alpha^2+\beta^2=10\). From \(16-2\alpha\beta=10\), \(\alpha\beta=3\), so the roots are (1) and (3).
Step 2
Why this answer is correct
The correct answer is A. (1) और (3) / (1) and (3). Here \(\alpha+\beta=4\) and \(\alpha^2+\beta^2=10\). From \(16-2\alpha\beta=10\), \(\alpha\beta=3\), so the roots are (1) and (3).
Step 3
Exam Tip
\(\alpha+\beta=4\) और \(\alpha^2+\beta^2=10\) है। \(16-2\alpha\beta=10\) से \(\alpha\beta=3\), इसलिए जड़ें (1) और (3) हैं।
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यदि \(x^2-6x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(3\alpha-2\) और \(3\beta-2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-6x+5=0\), which equation has roots \(3\alpha-2\) and \(3\beta-2\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-14x+13=0\)
B \(x^2-18x+45=0\)
C \(x^2-14x+25=0\)
D \(x^2+14x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+13=0\)
Step 1
Concept
The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+13=0\). The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 3
Exam Tip
मूल जड़ें (1) और (5) हैं, इसलिए नई जड़ें (1) और (13) हैं। उनका समीकरण \(x^2-14x+13=0\) है।
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यदि \(x^2+px+q=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha+1,\beta+1\), \(x^2-5x+6=0\) की जड़ें हैं, तो (p,q) क्या होंगे?
If \(\alpha,\beta\) are the roots of \(x^2+px+q=0\) and \(\alpha+1,\beta+1\) are the roots of \(x^2-5x+6=0\), what are (p,q)?
#quadratic-roots
#transformed-roots
#parameter
A (p=-3,\ q=2)
B (p=3,\ q=2)
C (p=-2,\ q=3)
D (p=2,\ q=-3)
Explanation opens after your attempt
Correct Answer
A. (p=-3,\ q=2)
Step 1
Concept
The sum of new roots is \(\alpha+\beta+2=5\), so (p=-3). From product (q-p+1=6), we get (q=2).
Step 2
Why this answer is correct
The correct answer is A. (p=-3,\ q=2). The sum of new roots is \(\alpha+\beta+2=5\), so (p=-3). From product (q-p+1=6), we get (q=2).
Step 3
Exam Tip
नई जड़ों का योग \(\alpha+\beta+2=5\) है, इसलिए (p=-3)। गुणनफल (q-p+1=6) से (q=2) मिलता है।
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यदि \(x^2-3x-2=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2,\beta^2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-3x-2=0\), which equation has roots \(\alpha^2,\beta^2\)?
#quadratic-roots
#squared-roots
#new-equation
A \(x^2-13x+4=0\)
B \(x^2+13x+4=0\)
C \(x^2-9x+4=0\)
D \(x^2-13x-4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+4=0\)
Step 1
Concept
Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+4=0\). Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 3
Exam Tip
\(\alpha+\beta=3\) और \(\alpha\beta=-2\) है। इसलिए \(\alpha^2+\beta^2=13\) और \(\alpha^2\beta^2=4\), अतः समीकरण \(x^2-13x+4=0\) है।
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\(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं। \(\alpha+1,\beta+1\) जड़ों वाला समीकरण कौन-सा है?
The roots of \(x^2-5x+6=0\) are \(\alpha,\beta\). Which equation has roots \(\alpha+1,\beta+1\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-7x+12=0\)
B \(x^2-5x+12=0\)
C \(x^2-6x+8=0\)
D \(x^2+7x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+12=0\)
Step 1
Concept
The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+12=0\). The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (3) हैं, इसलिए नई जड़ें (3) और (4) होंगी। उनका समीकरण \(x^2-7x+12=0\) है।
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यदि \(3x^2-10x+3=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha},\frac{1}{\beta}\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(3x^2-10x+3=0\), which equation has roots \(\frac{1}{\alpha},\frac{1}{\beta}\)?
#quadratic-roots
#reciprocal-roots
#new-equation
A \(3x^2-10x+3=0\)
B \(3x^2+10x+3=0\)
C \(x^2-10x+3=0\)
D \(10x^2-3x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-10x+3=0\)
Step 1
Concept
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=1\) है। व्युत्क्रम जड़ों का योग \(\frac{10}{3}\) और गुणनफल (1) ही रहता है।
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यदि दो वास्तविक मूलों का गुणनफल धनात्मक और योग धनात्मक है तो दोनों मूल कैसे होंगे?
If the product of two real roots is positive and their sum is positive, how will both roots be?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D एक मूल (0) / One root is (0)
Explanation opens after your attempt
Correct Answer
A. दोनों धनात्मक / Both positive
Step 1
Concept
A positive product means both signs are same. A positive sum means both roots are positive.
Step 2
Why this answer is correct
The correct answer is A. दोनों धनात्मक / Both positive. A positive product means both signs are same. A positive sum means both roots are positive.
Step 3
Exam Tip
गुणनफल धनात्मक होने पर दोनों चिन्ह समान होते हैं। योग धनात्मक होने से दोनों मूल धनात्मक होंगे।
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यदि दो वास्तविक मूलों का गुणनफल धनात्मक और योग ऋणात्मक है तो दोनों मूल कैसे होंगे?
If the product of two real roots is positive and their sum is negative, how will both roots be?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D एक मूल (0) / One root is (0)
Explanation opens after your attempt
Correct Answer
B. दोनों ऋणात्मक / Both negative
Step 1
Concept
A positive product means both roots have the same sign. A negative sum means both roots are negative.
Step 2
Why this answer is correct
The correct answer is B. दोनों ऋणात्मक / Both negative. A positive product means both roots have the same sign. A negative sum means both roots are negative.
Step 3
Exam Tip
गुणनफल धनात्मक होने पर दोनों मूलों का चिन्ह समान होता है। योग ऋणात्मक होने से दोनों मूल ऋणात्मक होंगे।
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समीकरण \(x^2-6x+13=0\) के वास्तविक मूलों के बारे में क्या सही है?
What is correct about the real roots of the equation \(x^2-6x+13=0\)?
#quadratic equations
#nature of roots
#no real roots
A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक और भिन्न मूल / Two real and distinct roots
C दो वास्तविक और समान मूल / Two real and equal roots
D एक वास्तविक मूल / One real root
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
Here (D=(-6)2 -4(1)(13)=-16<0). Therefore there is no real root.
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. Here (D=(-6)2 -4(1)(13)=-16<0). Therefore there is no real root.
Step 3
Exam Tip
यहाँ (D=(-6)2 -4(1)(13)=-16<0) है। इसलिए कोई वास्तविक मूल नहीं है।
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यदि \(7x^2-6x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?
If the roots of \(7x^2-6x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?
#quadratic-roots
#non-real-roots
#discriminant
A \(\lambda>\frac{9}{7}\)
B \(\lambda<\frac{9}{7}\)
C \(\lambda=\frac{9}{7}\)
D \(\lambda\le\frac{9}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda>\frac{9}{7}\)
Step 1
Concept
For non-real roots, (D<0) is required. From \(36-28\lambda<0\), we get \(\lambda>\frac{9}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda>\frac{9}{7}\). For non-real roots, (D<0) is required. From \(36-28\lambda<0\), we get \(\lambda>\frac{9}{7}\).
Step 3
Exam Tip
वास्तविक नहीं होने के लिए (D<0) चाहिए। \(36-28\lambda<0\) से \(\lambda>\frac{9}{7}\) मिलता है।
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(9x-2 -6(a-1)x+a-2 -4a-5=0) की जड़ें वास्तविक हों, तो सही शर्त क्या है?
What is the correct condition for (9x-2 -6(a-1)x+a-2 -4a-5=0) to have real roots?
#quadratic-roots
#real-roots
#parameter-condition
A \(a\ge-3\)
B \(a\le-3\)
C (a>3)
D (a< -3)
Explanation opens after your attempt
Correct Answer
A. \(a\ge-3\)
Step 1
Concept
Here (D=36(a-1)2 -36\(a^2-4a-5\)=72(a+3)). For real roots, \(D\ge0\), so \(a\ge-3\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge-3\). Here (D=36(a-1)2 -36\(a^2-4a-5\)=72(a+3)). For real roots, \(D\ge0\), so \(a\ge-3\).
Step 3
Exam Tip
यहाँ (D=36(a-1)2 -36\(a^2-4a-5\)=72(a+3)) है। वास्तविक जड़ों के लिए \(D\ge0\), इसलिए \(a\ge-3\)।
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(9x-2 -6(a-1)x+a-2 -4a-5=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?
For (9x-2 -6(a-1)x+a-2 -4a-5=0) to have real roots, what is the correct condition on (a)?
#quadratic-roots
#real-roots
#error-check
A \(a\ge-\frac{7}{2}\)
B \(a\le-\frac{7}{2}\)
C (a>1)
D (a<5)
Explanation opens after your attempt
Correct Answer
A. \(a\ge-\frac{7}{2}\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=36(a-1)2 -36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge-\frac{7}{2}\). For real roots, \(D\ge0\) is required. Here (D=36(a-1)2 -36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=36(a-1)2 -36\(a^2-4a-5\)=72a+216), इसलिए \(a\ge-\frac{7}{2}\) नहीं बल्कि \(a\ge-3\) होगा।
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(16x-2 -8(a-2)x+a-2 -6a=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?
For (16x-2 -8(a-2)x+a-2 -6a=0) to have real roots, what is the correct condition on (a)?
#quadratic-roots
#real-roots
#discriminant-check
A \(a\ge1\)
B \(a\le1\)
C (a>6)
D (a<0)
Explanation opens after your attempt
Correct Answer
A. \(a\ge1\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=64(a-2)2 -64\(a^2-6a\)=128(a+2)), so \(a\ge-2\); hence none of these options is exact.
Step 2
Why this answer is correct
The correct answer is A. \(a\ge1\). For real roots, \(D\ge0\) is required. Here (D=64(a-2)2 -64\(a^2-6a\)=128(a+2)), so \(a\ge-2\); hence none of these options is exact.
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=64(a-2)2 -64\(a^2-6a\)=128(a+2)), इसलिए \(a\ge-2\) होगा, अतः विकल्पों में सही शर्त नहीं है।
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यदि (x-2 -2mx+\(m^2-m\)=0) की जड़ें वास्तविक हों, तो (m) पर सही शर्त क्या है?
If (x-2 -2mx+\(m^2-m\)=0) has real roots, what is the correct condition on (m)?
#quadratic-roots
#real-roots
#parameter-condition
A (m>0)
B (m<0)
C \(m\ge0\)
D \(m\le0\)
Explanation opens after your attempt
Correct Answer
C. \(m\ge0\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=4m), so \(m\ge0\).
Step 2
Why this answer is correct
The correct answer is C. \(m\ge0\). For real roots, \(D\ge0\) is required. Here (D=4m), so \(m\ge0\).
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=4m), इसलिए \(m\ge0\)।
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यदि \(5x^2-4x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?
If the roots of \(5x^2-4x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?
#quadratic-roots
#non-real-roots
#discriminant
A \(\lambda<\frac{4}{5}\)
B \(\lambda=\frac{4}{5}\)
C \(\lambda>\frac{4}{5}\)
D \(\lambda\le\frac{4}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\lambda>\frac{4}{5}\)
Step 1
Concept
For non-real roots, (D<0) is required. From \(16-20\lambda<0\), we get \(\lambda>\frac{4}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(\lambda>\frac{4}{5}\). For non-real roots, (D<0) is required. From \(16-20\lambda<0\), we get \(\lambda>\frac{4}{5}\).
Step 3
Exam Tip
वास्तविक नहीं होने के लिए (D<0) चाहिए। \(16-20\lambda<0\) से \(\lambda>\frac{4}{5}\) मिलता है।
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(9x-2 -6(a+1)x+a-2 -3a=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?
For (9x-2 -6(a+1)x+a-2 -3a=0) to have real roots, what is the correct condition on (a)?
#quadratic-roots
#real-roots
#discriminant-condition
A \(a\ge-\frac{1}{5}\)
B \(a\le-\frac{1}{5}\)
C (a>5)
D \(a<-\frac{1}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(a\ge-\frac{1}{5}\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=36(5a+1)), so \(a\ge-\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge-\frac{1}{5}\). For real roots, \(D\ge0\) is required. Here (D=36(5a+1)), so \(a\ge-\frac{1}{5}\).
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=36(5a+1)), इसलिए \(a\ge-\frac{1}{5}\)।
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यदि \(3x^2+2x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?
If the roots of \(3x^2+2x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?
#quadratic-roots
#non-real-roots
#discriminant
A \(\lambda>\frac{1}{3}\)
B \(\lambda<\frac{1}{3}\)
C \(\lambda\le\frac{1}{3}\)
D \(\lambda=0\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda>\frac{1}{3}\)
Step 1
Concept
For non-real roots, (D<0) is required. From \(4-12\lambda<0\), we get \(\lambda>\frac{1}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda>\frac{1}{3}\). For non-real roots, (D<0) is required. From \(4-12\lambda<0\), we get \(\lambda>\frac{1}{3}\).
Step 3
Exam Tip
वास्तविक नहीं होने के लिए (D<0) चाहिए। \(4-12\lambda<0\) से \(\lambda>\frac{1}{3}\) मिलता है।
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(4x-2 -4(a-1)x+a-2 -4a=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?
For (4x-2 -4(a-1)x+a-2 -4a=0) to have real roots, what is the correct condition on (a)?
#quadratic-roots
#real-roots
#discriminant-condition
A \(a\le1\)
B \(a\ge1\)
C (a<0)
D (a>4)
Explanation opens after your attempt
Correct Answer
A. \(a\le1\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=16(1-a)), so \(a\le1\).
Step 2
Why this answer is correct
The correct answer is A. \(a\le1\). For real roots, \(D\ge0\) is required. Here (D=16(1-a)), so \(a\le1\).
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=16(1-a)), इसलिए \(a\le1\) है।
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(x-2 -2x+(p+3)=0) की वास्तविक जड़ें न हों, इसके लिए (p) पर सही शर्त क्या है?
For (x-2 -2x+(p+3)=0) to have no real roots, what is the correct condition on (p)?
#quadratic-roots
#no-real-roots
#inequality
A (p>-2)
B \(p\ge -2\)
C (p<-2)
D \(p\le -2\)
Explanation opens after your attempt
Step 1
Concept
For no real roots, (D<0) is required. Here (D=-4(p+2)), so (p>-2).
Step 2
Why this answer is correct
The correct answer is A. (p>-2). For no real roots, (D<0) is required. Here (D=-4(p+2)), so (p>-2).
Step 3
Exam Tip
वास्तविक जड़ें न होने के लिए (D<0) चाहिए। यहाँ (D=-4(p+2)), इसलिए (p>-2) है।
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\(kx^2+6x+9=0\) की वास्तविक जड़ें हों और \(k\ne0\), तो (k) पर सही शर्त कौन-सी है?
For \(kx^2+6x+9=0\) to have real roots with \(k\ne0\), which condition on (k) is correct?
#quadratic-roots
#real-roots
#parameter-condition
A \(k\le 1,\ k\ne0\)
B (k>1)
C \(k\ge 1\)
D (k<0) केवल / (k<0) only
Explanation opens after your attempt
Correct Answer
A. \(k\le 1,\ k\ne0\)
Step 1
Concept
For real roots, \(D=36-36k\ge0\) is required. Thus \(k\le1\), and \(k\ne0\) is also needed for a quadratic equation.
Step 2
Why this answer is correct
The correct answer is A. \(k\le 1,\ k\ne0\). For real roots, \(D=36-36k\ge0\) is required. Thus \(k\le1\), and \(k\ne0\) is also needed for a quadratic equation.
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D=36-36k\ge0\) होना चाहिए। इसलिए \(k\le1\) और द्विघात के लिए \(k\ne0\) भी जरूरी है।
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(x-2 -2(a+1)x+a-2 +3=0) की जड़ें वास्तविक हों, इसके लिए (a) पर सही शर्त क्या है?
What is the correct condition on (a) so that (x-2 -2(a+1)x+a-2 +3=0) has real roots?
#quadratic-roots
#real-roots
#discriminant-inequality
A (a>1)
B \(a\ge 1\)
C \(a\le 1\)
D (a<1)
Explanation opens after your attempt
Correct Answer
B. \(a\ge 1\)
Step 1
Concept
For real roots, \(D\ge 0\) is required. Here (D=8(a-1)), so \(a\ge 1\) is correct.
Step 2
Why this answer is correct
The correct answer is B. \(a\ge 1\). For real roots, \(D\ge 0\) is required. Here (D=8(a-1)), so \(a\ge 1\) is correct.
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge 0\) चाहिए। यहाँ (D=8(a-1)), इसलिए \(a\ge 1\) सही है।
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यदि (x-2 -2(a+3)x+a-2 +6a+5=0) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-\beta\) का धनात्मक मान क्या है?
If \(\alpha,\beta\) are the roots of (x-2 -2(a+3)x+a-2 +6a+5=0), what is the positive value of \(\alpha-\beta\)?
#quadratic-roots
#parametric-roots
#difference-of-roots
A (4)
B (2)
C (1)
D (3)
Explanation opens after your attempt
Step 1
Concept
The equation becomes ((x-(a+1))(x-(a+5))=0). So the roots are (a+1) and (a+5), hence the positive difference is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). The equation becomes ((x-(a+1))(x-(a+5))=0). So the roots are (a+1) and (a+5), hence the positive difference is (4).
Step 3
Exam Tip
यहाँ समीकरण ((x-(a+1))(x-(a+5))=0) बनता है। इसलिए जड़ें (a+1) और (a+5) हैं, अतः धनात्मक अंतर (4) है।
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यदि \(x^2-12x+m=0\) की दोनों जड़ें अभाज्य संख्याएँ हैं, तो (m) का मान क्या है?
If both roots of \(x^2-12x+m=0\) are prime numbers, what is the value of (m)?
#quadratic-roots
#prime-roots
#integer-roots
A (30)
B (35)
C (40)
D (45)
Explanation opens after your attempt
Step 1
Concept
The prime roots with sum (12) are (5) and (7). Their product is (35), so (m=35).
Step 2
Why this answer is correct
The correct answer is B. (35). The prime roots with sum (12) are (5) and (7). Their product is (35), so (m=35).
Step 3
Exam Tip
योग (12) वाली अभाज्य जड़ें (5) और (7) हैं। उनका गुणनफल (35) है, इसलिए (m=35)।
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यदि (x-2 -(2r+5)x+\(r^2+5r+6\)=0) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-\beta\) का धनात्मक मान क्या है?
If \(\alpha,\beta\) are the roots of (x-2 -(2r+5)x+\(r^2+5r+6\)=0), what is the positive value of \(\alpha-\beta\)?
#quadratic-roots
#parametric-roots
#difference-of-roots
A (1)
B (2)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
In the given equation, the sum of roots is (2r+5) and the product is (r-2 +5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). In the given equation, the sum of roots is (2r+5) and the product is (r-2 +5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 3
Exam Tip
दिए गए समीकरण में जड़ों का योग (2r+5) और गुणनफल (r-2 +5r+6=(r+2)(r+3)) है। इसलिए जड़ें (r+2) और (r+3) हैं, अतः धनात्मक अंतर (1) है।
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\(x^2-10x+k=0\) की जड़ें भिन्न अभाज्य संख्याएँ हैं, तो (k) का मान क्या है?
The roots of \(x^2-10x+k=0\) are distinct prime numbers. What is (k)?
#quadratic-roots
#prime-roots
#integer-roots
A (21)
B (25)
C (16)
D (10)
Explanation opens after your attempt
Step 1
Concept
The distinct prime roots with sum (10) are (3) and (7). Their product is (21), so (k=21).
Step 2
Why this answer is correct
The correct answer is A. (21). The distinct prime roots with sum (10) are (3) and (7). Their product is (21), so (k=21).
Step 3
Exam Tip
योग (10) वाली भिन्न अभाज्य जड़ें (3) और (7) हैं। उनका गुणनफल (21) है, इसलिए (k=21)।
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यदि \(x^2+bx+c=0\) की जड़ें एक-दूसरे की विपरीत संख्याएँ हैं, तो कौन-सी शर्त अनिवार्य है?
If the roots of \(x^2+bx+c=0\) are opposites of each other, which condition is necessary?
#quadratic-roots
#opposite-roots
#sum-of-roots
A (b=0)
B (c=0)
C (b=c)
D \(b^2=4c\)
Explanation opens after your attempt
Step 1
Concept
Opposite roots have sum (0). Here the sum is (-b), so (b=0).
Step 2
Why this answer is correct
The correct answer is A. (b=0). Opposite roots have sum (0). Here the sum is (-b), so (b=0).
Step 3
Exam Tip
विपरीत जड़ों का योग (0) होता है। यहाँ योग (-b) है, इसलिए (b=0)।
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\(x^2-px+36=0\) की जड़ें धनात्मक पूर्णांक हैं और उनका अंतर (5) है, तो (p) का मान क्या है?
The roots of \(x^2-px+36=0\) are positive integers and their difference is (5). What is (p)?
#quadratic-roots
#integer-roots
#sum-of-roots
A (11)
B (12)
C (13)
D (15)
Explanation opens after your attempt
Step 1
Concept
The positive roots with product (36) and difference (5) are (4) and (9). Their sum is (13), so (p=13).
Step 2
Why this answer is correct
The correct answer is C. (13). The positive roots with product (36) and difference (5) are (4) and (9). Their sum is (13), so (p=13).
Step 3
Exam Tip
गुणनफल (36) और अंतर (5) वाली धनात्मक जड़ें (4) और (9) हैं। उनका योग (13) है, इसलिए (p=13)।
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यदि \(x^2-7x+k=0\) की जड़ें एक-दूसरे की व्युत्क्रम हैं, तो (k) का मान क्या होगा?
If the roots of \(x^2-7x+k=0\) are reciprocals of each other, what is the value of (k)?
#quadratic-roots
#reciprocal-roots
#product-of-roots
A (1)
B (-1)
C (7)
D (49)
Explanation opens after your attempt
Step 1
Concept
For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.
Step 2
Why this answer is correct
The correct answer is A. (1). For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए गुणनफल (1) होता है और यहाँ गुणनफल (k) है। इसलिए (k=1); परीक्षा में पहले गुणनफल देखें।
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यदि \(4x^2-3x+k=0\) के मूलों का गुणनफल मूलों के योग के बराबर है तो (k) क्या होगा?
If the product of roots of \(4x^2-3x+k=0\) equals the sum of roots, what is (k)?
#roots
#sum_equals_product
#parameter
A (3)
B \(\frac{3}{4}\)
C (4)
D (1)
Explanation opens after your attempt
Step 1
Concept
The sum is \(-\frac{b}{a}=\frac{3}{4}\) and the product is \(\frac{k}{4}\). From \(\frac{k}{4}=\frac{3}{4}\), (k=3).
Step 2
Why this answer is correct
The correct answer is A. (3). The sum is \(-\frac{b}{a}=\frac{3}{4}\) and the product is \(\frac{k}{4}\). From \(\frac{k}{4}=\frac{3}{4}\), (k=3).
Step 3
Exam Tip
योग \(-\frac{b}{a}=\frac{3}{4}\) और गुणनफल \(\frac{k}{4}\) है। \(\frac{k}{4}=\frac{3}{4}\) से (k=3) है।
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यदि मूलों का योग (6) और उनके वर्गों का योग (52) है तो मूलों का गुणनफल क्या होगा?
If the sum of roots is (6) and the sum of their squares is (52), what is the product of roots?
#roots
#identity
#product
A (-8)
B (8)
C (16)
D (-16)
Explanation opens after your attempt
Step 1
Concept
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(52=36-2\alpha\beta\), we get \(\alpha\beta=-8\).
Step 2
Why this answer is correct
The correct answer is A. (-8). (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(52=36-2\alpha\beta\), we get \(\alpha\beta=-8\).
Step 3
Exam Tip
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta) है। \(52=36-2\alpha\beta\) से \(\alpha\beta=-8\) मिलता है।
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यदि \(x^2-4x+3=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(3\alpha\) और \(3\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-4x+3=0\), which equation has \(3\alpha\) and \(3\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-12x+27=0\)
B \(x^2-4x+27=0\)
C \(x^2-12x+9=0\)
D \(x^2+12x+27=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+27=0\)
Step 1
Concept
The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+27=0\). The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 3
Exam Tip
पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए \(x^2-12x+27=0\) है।
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यदि \(3x^2-2x+k=0\) के मूलों का गुणनफल मूलों के योग के बराबर है तो (k) क्या होगा?
If the product of roots of \(3x^2-2x+k=0\) equals the sum of roots, what is (k)?
#roots
#sum_equals_product
#parameter
A (2)
B \(\frac{2}{3}\)
C (3)
D (1)
Explanation opens after your attempt
Step 1
Concept
The sum is \(-\frac{b}{a}=\frac{2}{3}\) and the product is \(\frac{k}{3}\). From \(\frac{k}{3}=\frac{2}{3}\), (k=2).
Step 2
Why this answer is correct
The correct answer is A. (2). The sum is \(-\frac{b}{a}=\frac{2}{3}\) and the product is \(\frac{k}{3}\). From \(\frac{k}{3}=\frac{2}{3}\), (k=2).
Step 3
Exam Tip
योग \(-\frac{b}{a}=\frac{2}{3}\) और गुणनफल \(\frac{k}{3}\) है। \(\frac{k}{3}=\frac{2}{3}\) से (k=2) है।
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यदि मूलों का योग (4) और उनके वर्गों का योग (20) है तो मूलों का गुणनफल क्या होगा?
If the sum of roots is (4) and the sum of their squares is (20), what is the product of roots?
#roots
#identity
#product
A (-2)
B (2)
C (8)
D (-8)
Explanation opens after your attempt
Step 1
Concept
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(20=16-2\alpha\beta\), we get \(\alpha\beta=-2\).
Step 2
Why this answer is correct
The correct answer is A. (-2). (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(20=16-2\alpha\beta\), we get \(\alpha\beta=-2\).
Step 3
Exam Tip
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta) है। \(20=16-2\alpha\beta\) से \(\alpha\beta=-2\) मिलता है।
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यदि \(x^2-3x+2=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(2\alpha\) और \(2\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-3x+2=0\), which equation has \(2\alpha\) and \(2\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-6x+8=0\)
B \(x^2-3x+8=0\)
C \(x^2-6x+4=0\)
D \(x^2+6x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+8=0\)
Step 1
Concept
The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8=0\). The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 3
Exam Tip
पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए \(x^2-6x+8=0\) है।
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यदि किसी द्विघात समीकरण के मूलों का योग (0) है तो मूलों के बारे में सही कथन कौन सा हो सकता है?
If the sum of roots of a quadratic equation is (0), which statement about the roots can be correct?
#roots
#zero_sum
#reasoning
A मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
B दोनों मूल हमेशा (1) हैं / Both roots are always (1)
C दोनों मूल हमेशा धनात्मक हैं / Both roots are always positive
D मूलों का गुणनफल हमेशा (0) है / The product is always (0)
Explanation opens after your attempt
Correct Answer
A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
Step 1
Concept
If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 2
Why this answer is correct
The correct answer is A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other. If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 3
Exam Tip
यदि \(\alpha+\beta=0\) है तो \(\beta=-\alpha\) होता है। इसलिए मूल विपरीत हो सकते हैं।
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यदि \(4\alpha\) और \(4\beta\) नए मूल हैं तथा \(\alpha+\beta=3\) है तो नए मूलों का योग क्या होगा?
If \(4\alpha\) and \(4\beta\) are new roots and \(\alpha+\beta=3\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (12)
B (7)
C (3)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (4\alpha+4\beta=4\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग भी उसी गुणक से गुणा होता है।
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यदि \(3\alpha\) और \(3\beta\) नए मूल हैं तथा \(\alpha+\beta=4\) है तो नए मूलों का योग क्या होगा?
If \(3\alpha\) and \(3\beta\) are new roots and \(\alpha+\beta=4\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (12)
B (4)
C (7)
D \(\frac{4}{3}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 3
Exam Tip
नए मूलों का योग (3\alpha+3\beta=3\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
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यदि \(2\alpha\) और \(2\beta\) मूल हैं तथा \(\alpha+\beta=5\) है तो नए मूलों का योग क्या होगा?
If \(2\alpha\) and \(2\beta\) are roots and \(\alpha+\beta=5\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (10)
B (5)
C (20)
D \(\frac{5}{2}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (10). The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (2\alpha+2\beta=2\(\alpha+\beta\)=10) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
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जिस मोनिक द्विघात समीकरण के मूलों का योग (10) और गुणनफल (21) है वह कौन सा है?
Which monic quadratic equation has sum of roots (10) and product of roots (21)?
#roots
#equation_from_sum_product
#monic
A \(x^2+10x+21=0\)
B \(x^2-10x+21=0\)
C \(x^2-21x+10=0\)
D \(x^2+21x+10=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-10x+21=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-10x+21=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-10x+21=0) सही है\)।
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जिस मोनिक द्विघात समीकरण के मूलों का योग (-9) और गुणनफल (20) है वह कौन सा है?
Which monic quadratic equation has sum of roots (-9) and product of roots (20)?
#roots
#equation_from_sum_product
#monic
A \(x^2+9x+20=0\)
B \(x^2-9x+20=0\)
C \(x^2+20x+9=0\)
D \(x^2-20x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+9x+20=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+9x+20=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2+9x+20=0) सही है\)।
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जिस द्विघात समीकरण के मूलों का योग (6) और गुणनफल (8) है वह कौन सा है?
Which quadratic equation has sum of roots (6) and product of roots (8)?
#roots
#equation_from_sum_product
#formula
A \(x^2+6x+8=0\)
B \(x^2-6x+8=0\)
C \(x^2-8x+6=0\)
D \(x^2+8x+6=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-6x+8=0\)
Step 1
Concept
\(The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-6x+8=0). The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 3
Exam Tip
\(मानक रूप (x^2-(\)योग)x+गुणनफल=0) है इसलिए \(x^2-6x+8=0\)। योग वाले पद का चिन्ह बदलता है।
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समीकरण ((r+2)x-2 -2(r+5)x+(r+2)=0) के वास्तविक और समान मूलों के लिए (r) का मान क्या होगा?
What will be the value of (r) for real and equal roots of ((r+2)x-2 -2(r+5)x+(r+2)=0)?
#quadratic equations
#nature of roots
#equal roots
A \(r=-\frac{7}{2}\)
B \(r=\frac{7}{2}\)
C (r=-2)
D (r=5)
Explanation opens after your attempt
Correct Answer
A. \(r=-\frac{7}{2}\)
Step 1
Concept
For equal roots, (D=0) is required. Here (D=12(2r+7)), so \(r=-\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(r=-\frac{7}{2}\). For equal roots, (D=0) is required. Here (D=12(2r+7)), so \(r=-\frac{7}{2}\).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। यहाँ (D=12(2r+7)), इसलिए \(r=-\frac{7}{2}\)।
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यदि \(ax^2+bx+c=0\) में \(a\neq0\) और मूल वास्तविक तथा समान हैं, तो सही शर्त कौन सी है?
If \(a\neq0\) in \(ax^2+bx+c=0\) and the roots are real and equal, which condition is correct?
#quadratic equations
#nature of roots
#equal roots
A \(b^2=4ac\)
B \(b^2>4ac\)
C \(b^2<4ac\)
D (b=4ac)
Explanation opens after your attempt
Correct Answer
A. \(b^2=4ac\)
Step 1
Concept
For equal real roots, \(D=b^2-4ac=0\) is required. Hence \(b^2=4ac\) is the correct condition.
Step 2
Why this answer is correct
The correct answer is A. \(b^2=4ac\). For equal real roots, \(D=b^2-4ac=0\) is required. Hence \(b^2=4ac\) is the correct condition.
Step 3
Exam Tip
समान वास्तविक मूलों के लिए \(D=b^2-4ac=0\) होना चाहिए। इसलिए \(b^2=4ac\) सही शर्त है।
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यदि \(x^2-6x+c=0\) की दोनों जड़ें वास्तविक और धनात्मक हैं, तो (c) पर सही शर्त क्या है?
If both roots of \(x^2-6x+c=0\) are real and positive, what is the correct condition on (c)?
#quadratic-roots
#positive-roots
#condition
A \(0<c\le9\)
B (c>9)
C (c<0)
D (c=0)
Explanation opens after your attempt
Correct Answer
A. \(0<c\le9\)
Step 1
Concept
The sum (6) is positive and (c>0) is needed for both positive roots. For real roots, \(36-4c\ge0\), so \(0<c\le9\).
Step 2
Why this answer is correct
The correct answer is A. \(0<c\le9\). The sum (6) is positive and (c>0) is needed for both positive roots. For real roots, \(36-4c\ge0\), so \(0<c\le9\).
Step 3
Exam Tip
योग (6) धनात्मक है और दोनों धनात्मक जड़ों के लिए (c>0) चाहिए। वास्तविक जड़ों के लिए \(36-4c\ge0\), इसलिए \(0<c\le9\)।
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यदि ((m-1 )x-2 +2(m+1)x+(m-1 )=0) की जड़ें वास्तविक और व्युत्क्रम हों, तो (m) पर सही शर्त क्या है?
If ((m-1 )x-2 +2(m+1)x+(m-1 )=0) has real reciprocal roots, what is the correct condition on (m)?
#quadratic-roots
#reciprocal-roots
#parameter-condition
A \(m\ge0\) और \(m\neq1\) / \(m\ge0\) and \(m\neq1\)
B (m<0)
C (m=1)
D \(m\le0\)
Explanation opens after your attempt
Correct Answer
A. \(m\ge0\) और \(m\neq1\) / \(m\ge0\) and \(m\neq1\)
Step 1
Concept
The product of roots is \(\frac{m-1}{m-1}=1\), so \(m\neq1\) is needed. For real roots, \(D=16m\ge0\), hence \(m\ge0\) and \(m\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(m\ge0\) और \(m\neq1\) / \(m\ge0\) and \(m\neq1\). The product of roots is \(\frac{m-1}{m-1}=1\), so \(m\neq1\) is needed. For real roots, \(D=16m\ge0\), hence \(m\ge0\) and \(m\neq1\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{m-1}{m-1}=1\) है, इसलिए \(m\neq1\) चाहिए। वास्तविक जड़ों के लिए \(D=16m\ge0\), अतः \(m\ge0\) और \(m\neq1\)।
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यदि \(x^2-5x+c=0\) की दोनों जड़ें वास्तविक और धनात्मक हैं, तो (c) पर सही शर्त क्या है?
If both roots of \(x^2-5x+c=0\) are real and positive, what is the correct condition on (c)?
#quadratic-roots
#positive-roots
#condition
A (c<0)
B \(0<c\le\frac{25}{4}\)
C \(c>\frac{25}{4}\)
D (c=0)
Explanation opens after your attempt
Correct Answer
B. \(0<c\le\frac{25}{4}\)
Step 1
Concept
The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(0<c\le\frac{25}{4}\). The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
Step 3
Exam Tip
योग (5) धनात्मक है और दोनों जड़ों के लिए गुणनफल (c>0) चाहिए। वास्तविक जड़ों के लिए \(25-4c\ge0\), इसलिए \(0<c\le\frac{25}{4}\)।
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यदि \(x^2+4x+c=0\) की दोनों जड़ें वास्तविक और ऋणात्मक हैं, तो कौन-सी शर्त पर्याप्त और आवश्यक है?
If both roots of \(x^2+4x+c=0\) are real and negative, which condition is necessary and sufficient?
#quadratic-roots
#negative-roots
#condition
A \(0<c\le4\)
B (c>4)
C (c<0)
D (c=0)
Explanation opens after your attempt
Correct Answer
A. \(0<c\le4\)
Step 1
Concept
The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, \(16-4c\ge0\), hence \(0<c\le4\).
Step 2
Why this answer is correct
The correct answer is A. \(0<c\le4\). The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, \(16-4c\ge0\), hence \(0<c\le4\).
Step 3
Exam Tip
योग (-4) पहले से ऋणात्मक है और गुणनफल धनात्मक चाहिए, इसलिए (c>0)। वास्तविक जड़ों के लिए \(16-4c\ge0\), अतः \(0<c\le4\)।
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(x-2 -2(k+1)x+k-2 =0) की जड़ें वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
For (x-2 -2(k+1)x+k-2 =0) to have real and distinct roots, what is the correct condition on (k)?
#quadratic-roots
#distinct-roots
#discriminant
A \(k>-\frac{1}{2}\)
B \(k\ge-\frac{1}{2}\)
C \(k<-\frac{1}{2}\)
D \(k\le-\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(k>-\frac{1}{2}\)
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (D=4(2k+1)), so \(k>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k>-\frac{1}{2}\). For real and distinct roots, (D>0) is needed. Here (D=4(2k+1)), so \(k>-\frac{1}{2}\).
Step 3
Exam Tip
वास्तविक और भिन्न जड़ों के लिए (D>0) चाहिए। यहाँ (D=4(2k+1)), इसलिए \(k>-\frac{1}{2}\)।
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यदि किसी द्विघात समीकरण के वास्तविक मूलों का गुणनफल ऋणात्मक है तो मूलों के चिन्ह कैसे होंगे?
If the product of real roots of a quadratic equation is negative then how are the signs of the roots?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D दोनों शून्य / Both zero
Explanation opens after your attempt
Correct Answer
C. एक धनात्मक और एक ऋणात्मक / One positive and one negative
Step 1
Concept
A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 2
Why this answer is correct
The correct answer is C. एक धनात्मक और एक ऋणात्मक / One positive and one negative. A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी मिलता है जब एक मूल धनात्मक और दूसरा ऋणात्मक हो। \(\alpha\beta<0\) संकेतों की जांच का छोटा संकेत है।
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(x-2 -2x+\(a^2+3\)=0) की जड़ों की प्रकृति क्या है?
What is the nature of the roots of (x-2 -2x+\(a^2+3\)=0)?
#quadratic-roots
#nature-of-roots
#always-non-real
A हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a)
B हर वास्तविक (a) के लिए समान वास्तविक / Equal real for every real (a)
C हर वास्तविक (a) के लिए दो वास्तविक भिन्न / Two real distinct for every real (a)
D एक जड़ हमेशा (0) है / One root is always (0)
Explanation opens after your attempt
Correct Answer
A. हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+3\)=-4a-2 -8). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is A. हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a). The discriminant is (D=4-4\(a^2+3\)=-4a-2 -8). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+3\)=-4a-2 -8) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं हैं।
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यदि \(x^2-sx+1=0\) की जड़ें \(2\tan\theta\) और \(\frac{1}{2}\cot\theta\) हैं, तो (s) पर वास्तविकता की सही शर्त क्या है?
If the roots of \(x^2-sx+1=0\) are \(2\tan\theta\) and \(\frac{1}{2}\cot\theta\), what is the correct reality condition on (s)?
#quadratic-roots
#trigonometric-roots
#condition
A \(s^2\ge4\)
B \(s^2<4\)
C (s=0)
D \(s^2=1\)
Explanation opens after your attempt
Correct Answer
A. \(s^2\ge4\)
Step 1
Concept
The product of the two roots is (1), so the product condition is satisfied. For real roots, the discriminant \(s^2-4\ge0\), so \(s^2\ge4\).
Step 2
Why this answer is correct
The correct answer is A. \(s^2\ge4\). The product of the two roots is (1), so the product condition is satisfied. For real roots, the discriminant \(s^2-4\ge0\), so \(s^2\ge4\).
Step 3
Exam Tip
दोनों जड़ों का गुणनफल (1) है, इसलिए समीकरण का गुणनफल सही है। वास्तविक जड़ों के लिए विविक्तकर \(s^2-4\ge0\), इसलिए \(s^2\ge4\)।
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(x-2 -2x+\(a^2+2\)=0) की जड़ों की प्रकृति क्या है?
What is the nature of the roots of (x-2 -2x+\(a^2+2\)=0)?
#quadratic-roots
#nature-of-roots
#always-non-real
A दो वास्तविक भिन्न / Two real distinct
B समान वास्तविक / Equal real
C हर (a) के लिए वास्तविक नहीं / Not real for every (a)
D एक जड़ शून्य / One root is zero
Explanation opens after your attempt
Correct Answer
C. हर (a) के लिए वास्तविक नहीं / Not real for every (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is C. हर (a) के लिए वास्तविक नहीं / Not real for every (a). The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+2\)=-4\(a^2+1\)) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं हैं।
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यदि \(x^2-sx+1=0\) की जड़ें \(\tan\theta\) और \(\cot\theta\) हो सकती हैं, तो वास्तविक \(\theta\) के लिए (s) पर सही शर्त क्या है?
If the roots of \(x^2-sx+1=0\) can be \(\tan\theta\) and \(\cot\theta\), what is the correct condition on (s) for real \(\theta\)?
#quadratic-roots
#trigonometric-roots
#condition
A (-2<s<2)
B (s=0)
C \(s^2\ge4\)
D \(s^2<4\)
Explanation opens after your attempt
Correct Answer
C. \(s^2\ge4\)
Step 1
Concept
We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
Step 2
Why this answer is correct
The correct answer is C. \(s^2\ge4\). We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
Step 3
Exam Tip
\(\tan\theta\cdot\cot\theta=1\) और \(\tan\theta+\cot\theta=s\) है। वास्तविक मानों के लिए \(s^2-4\ge0\), इसलिए \(s^2\ge4\)।
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\(5x^2-2x+1=0\) की जड़ों के बारे में सही कथन कौन-सा है?
Which statement is correct about the roots of \(5x^2-2x+1=0\)?
#quadratic-roots
#no-real-roots
#discriminant
A दो समान वास्तविक / Two equal real
B दो वास्तविक और भिन्न / Two real and distinct
C वास्तविक जड़ें नहीं / No real roots
D दोनों जड़ें (0) हैं / Both roots are (0)
Explanation opens after your attempt
Correct Answer
C. वास्तविक जड़ें नहीं / No real roots
Step 1
Concept
Here (D=(-2)2 -4(5)(1)=-16<0). Therefore the equation has no real roots.
Step 2
Why this answer is correct
The correct answer is C. वास्तविक जड़ें नहीं / No real roots. Here (D=(-2)2 -4(5)(1)=-16<0). Therefore the equation has no real roots.
Step 3
Exam Tip
यहाँ (D=(-2)2 -4(5)(1)=-16<0) है। इसलिए इस समीकरण की वास्तविक जड़ें नहीं हैं।
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यदि किसी द्विघात समीकरण का विविक्तकर (D=(r-2)2 -9) है, तो कोई वास्तविक मूल न होने के लिए (r) का अंतराल कौन सा है?
If a quadratic equation has discriminant (D=(r-2)2 -9), which interval of (r) gives no real roots?
#quadratic-equations
#discriminant-form
#no-real-roots
A (-1<r<5)
B (r<-1) या (r>5) / (r<-1) or (r>5)
C (r=-1) या (r=5) / (r=-1) or (r=5)
D हर (r) / Every (r)
Explanation opens after your attempt
Correct Answer
A. (-1<r<5)
Step 1
Concept
For no real roots (D<0) is needed. From ((r-2)2 <9), we get (-1<r<5).
Step 2
Why this answer is correct
The correct answer is A. (-1<r<5). For no real roots (D<0) is needed. From ((r-2)2 <9), we get (-1<r<5).
Step 3
Exam Tip
कोई वास्तविक मूल न होने के लिए (D<0) चाहिए। ((r-2)2 <9) से (-1<r<5) मिलता है।
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यदि (x-2 -2(a+1)x+\(a^2+4a+5\)=0) के कोई वास्तविक मूल नहीं हों, तो (a) पर कौन सी शर्त सही है?
If (x-2 -2(a+1)x+\(a^2+4a+5\)=0) has no real roots, which condition on (a) is correct?
#quadratic-equations
#no-real-roots
#parameter
A (a>-2)
B (a=-2)
C (a<-2)
D हर वास्तविक (a) / Every real (a)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(a+1)2 -4\(a^2+4a+5\)=-8(a+2)). For no real roots (D<0), so (a>-2).
Step 2
Why this answer is correct
The correct answer is A. (a>-2). Here (D=4(a+1)2 -4\(a^2+4a+5\)=-8(a+2)). For no real roots (D<0), so (a>-2).
Step 3
Exam Tip
यहाँ (D=4(a+1)2 -4\(a^2+4a+5\)=-8(a+2)) है। कोई वास्तविक मूल न होने के लिए (D<0), इसलिए (a>-2)।
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यदि (D=(z-9)(z+1)) है, तो कोई वास्तविक मूल न होने के लिए (z) का अंतराल कौन सा है?
If (D=(z-9)(z+1)), which interval of (z) gives no real roots?
#quadratic-equations
#discriminant-form
#no-real-roots
A (-1<z<9)
B (z<-1) या (z>9) / (z<-1) or (z>9)
C (z=-1) या (z=9) / (z=-1) or (z=9)
D हर (z) / Every (z)
Explanation opens after your attempt
Correct Answer
A. (-1<z<9)
Step 1
Concept
For no real roots (D<0) is needed. ((z-9)(z+1)<0) gives (-1<z<9).
Step 2
Why this answer is correct
The correct answer is A. (-1<z<9). For no real roots (D<0) is needed. ((z-9)(z+1)<0) gives (-1<z<9).
Step 3
Exam Tip
कोई वास्तविक मूल न होने के लिए (D<0) चाहिए। ((z-9)(z+1)<0) से (-1<z<9) मिलता है।
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यदि (x-2 +2(m-5 )x+\(m^2-9m+24\)=0) के कोई वास्तविक मूल नहीं हों, तो (m) पर कौन सी शर्त सही है?
If (x-2 +2(m-5 )x+\(m^2-9m+24\)=0) has no real roots, which condition on (m) is correct?
#quadratic-equations
#no-real-roots
#parameter
A (m>2)
B (m=2)
C (m<2)
D हर (m) / Every (m)
Explanation opens after your attempt
Step 1
Concept
In this equation (D=4(2-m)). For no real roots (D<0), so (m>2).
Step 2
Why this answer is correct
The correct answer is A. (m>2). In this equation (D=4(2-m)). For no real roots (D<0), so (m>2).
Step 3
Exam Tip
इसी समीकरण में (D=4(2-m)) है। कोई वास्तविक मूल नहीं के लिए (D<0), इसलिए (m>2)।
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समीकरण (x-2 -2(4t+1)x+\(7t^2+2t+5\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल क्या है?
The equation (x-2 -2(4t+1)x+\(7t^2+2t+5\)=0) has no real roots. What is the correct interval for (t)?
#quadratic-equations
#no-real-roots
#parameter-interval
A \(-1<t<\frac{2}{9}\)
B (t<-1) या \(t>\frac{2}{9}\) / (t<-1) or \(t>\frac{2}{9}\)
C (t=-1) या \(t=\frac{2}{9}\) / (t=-1) or \(t=\frac{2}{9}\)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. \(-1<t<\frac{2}{9}\)
Step 1
Concept
Here (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16). From (D<0), \(-1<t<\frac{2}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(-1<t<\frac{2}{9}\). Here (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16). From (D<0), \(-1<t<\frac{2}{9}\).
Step 3
Exam Tip
यहाँ (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16) है। (D<0) से \(-1<t<\frac{2}{9}\) मिलता है।
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समीकरण (x-2 -2(a-2b)x+(a+2b)2 =0) के वास्तविक मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for real roots of (x-2 -2(a-2b)x+(a+2b)2 =0)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(ab\leq0\)
B (ab>0)
C (a=2b)
D (a+2b=0) मात्र / Only (a+2b=0)
Explanation opens after your attempt
Correct Answer
A. \(ab\leq0\)
Step 1
Concept
Here (D=4(a-2b)2 -4(a+2b)2 =-32ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 2
Why this answer is correct
The correct answer is A. \(ab\leq0\). Here (D=4(a-2b)2 -4(a+2b)2 =-32ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 3
Exam Tip
यहाँ (D=4(a-2b)2 -4(a+2b)2 =-32ab) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(ab\leq0\)।
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यदि (x-2 -2(a+b)x+3ab=0) के वास्तविक मूल हों, तो (a) और (b) के लिए कौन सा कथन सही है?
If (x-2 -2(a+b)x+3ab=0) has real roots, which statement is correct for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(a^2-ab+b^2\geq0\) होने से मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2-ab+b^2\geq0\)
B मूल तभी वास्तविक हैं जब (ab>0) / Roots are real only when (ab>0)
C मूल कभी वास्तविक नहीं होते / Roots are never real
D मूल तभी समान हैं जब (a+b=0) / Roots are equal only when (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(a^2-ab+b^2\geq0\) होने से मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2-ab+b^2\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -12ab=4\(a^2-ab+b^2\)). It is never negative, so real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(a^2-ab+b^2\geq0\) होने से मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2-ab+b^2\geq0\). Here (D=4(a+b)2 -12ab=4\(a^2-ab+b^2\)). It is never negative, so real roots exist.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -12ab=4\(a^2-ab+b^2\)) है। यह हमेशा ऋणात्मक नहीं होता, इसलिए वास्तविक मूल मिलते हैं।
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समीकरण (x-2 +2(v+2)x+(4v+11)=0) के कोई वास्तविक मूल नहीं होने की सही शर्त क्या है?
What is the correct condition for (x-2 +2(v+2)x+(4v+11)=0) to have no real roots?
#quadratic-equations
#parameter-interval
#no-real-roots
A \(-\sqrt{7}<v<\sqrt{7}\)
B \(v<-\sqrt{7}\) या \(v>\sqrt{7}\) / \(v<-\sqrt{7}\) or \(v>\sqrt{7}\)
C \(v=-\sqrt{7}\) या \(v=\sqrt{7}\) / \(v=-\sqrt{7}\) or \(v=\sqrt{7}\)
D हर (v) / Every (v)
Explanation opens after your attempt
Correct Answer
A. \(-\sqrt{7}<v<\sqrt{7}\)
Step 1
Concept
Here (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)). From (D<0), \(-\sqrt{7}<v<\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{7}<v<\sqrt{7}\). Here (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)). From (D<0), \(-\sqrt{7}<v<\sqrt{7}\).
Step 3
Exam Tip
यहाँ (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)) है। (D<0) से \(-\sqrt{7}<v<\sqrt{7}\)।
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यदि (x-2 +2(v+2)x+(4v+11)=0) के कोई वास्तविक मूल नहीं हों, तो (v) किस अंतराल में होगा?
If (x-2 +2(v+2)x+(4v+11)=0) has no real roots, in which interval will (v) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-3<v<1)
B (v<-3) या (v>1) / (v<-3) or (v>1)
C (v=-3) या (v=1) / (v=-3) or (v=1)
D हर (v) / Every (v)
Explanation opens after your attempt
Correct Answer
A. (-3<v<1)
Step 1
Concept
Here (D=4(v+2)2 -4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.
Step 2
Why this answer is correct
The correct answer is A. (-3<v<1). Here (D=4(v+2)2 -4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.
Step 3
Exam Tip
यहाँ (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)) नहीं, सही रूप (4\(v^2-7\)) नहीं है। परीक्षा में विस्तार सावधानी से करें।
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यदि ((p-2)x-2 -2(p+2)x+(p+6)=0) में \(p\neq2\) हो, तो वास्तविक मूलों के लिए (p) की शर्त क्या है?
If \(p\neq2\) in ((p-2)x-2 -2(p+2)x+(p+6)=0), what is the condition on (p) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\)
B (p>5)
C (p=2)
D हर \(p\neq2\) / Every \(p\neq2\)
Explanation opens after your attempt
Correct Answer
A. \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\)
Step 1
Concept
Here (D=4(p+2)2 -4(p-2)(p+6)=40-8p). For real roots \(p\leq5\), and for a quadratic \(p\neq2\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\). Here (D=4(p+2)2 -4(p-2)(p+6)=40-8p). For real roots \(p\leq5\), and for a quadratic \(p\neq2\).
Step 3
Exam Tip
यहाँ (D=4(p+2)2 -4(p-2)(p+6)=40-8p) है। वास्तविक मूलों के लिए \(p\leq5\) और द्विघात के लिए \(p\neq2\)।
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यदि (x-2 -2(k-4)x+\(k^2-10k+27\)=0) के कोई वास्तविक मूल नहीं हों, तो (k) पर कौन सी शर्त सही है?
If (x-2 -2(k-4)x+\(k^2-10k+27\)=0) has no real roots, which condition on (k) is correct?
#quadratic-equations
#no-real-roots
#parameter
A \(k<\frac{11}{2}\)
B \(k=\frac{11}{2}\)
C \(k>\frac{11}{2}\)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. \(k<\frac{11}{2}\)
Step 1
Concept
Here (D=4(k-4)2 -4\(k^2-10k+27\)=4(2k-11)). For no real roots (D<0), so \(k<\frac{11}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k<\frac{11}{2}\). Here (D=4(k-4)2 -4\(k^2-10k+27\)=4(2k-11)). For no real roots (D<0), so \(k<\frac{11}{2}\).
Step 3
Exam Tip
यहाँ (D=4(k-4)2 -4\(k^2-10k+27\)=4(2k-11)) है। कोई वास्तविक मूल नहीं के लिए (D<0), इसलिए \(k<\frac{11}{2}\)।
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यदि (x-2 -2(k+3)x+\(k^2+5k+12\)=0) के वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (x-2 -2(k+3)x+\(k^2+5k+12\)=0) has real roots, which condition on (k) is correct?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq3\)
B (k<3)
C (k=0)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Correct Answer
A. \(k\geq3\)
Step 1
Concept
Here (D=4(k+3)2 -4\(k^2+5k+12\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\geq3\). Here (D=4(k+3)2 -4\(k^2+5k+12\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 3
Exam Tip
यहाँ (D=4(k+3)2 -4\(k^2+5k+12\)=4(k-3)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(k\geq3\)।
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यदि (x-2 -2(a+2)x+\(a^2+6a+8\)=0) के कोई वास्तविक मूल नहीं हों, तो (a) पर कौन सी शर्त सही है?
If (x-2 -2(a+2)x+\(a^2+6a+8\)=0) has no real roots, which condition on (a) is correct?
#quadratic-equations
#no-real-roots
#parameter
A (a>-2)
B (a=-2)
C (a<-2)
D हर वास्तविक (a) / Every real (a)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(a+2)2 -4\(a^2+6a+8\)=-8(a+2)). For no real roots (D<0), so (a>-2).
Step 2
Why this answer is correct
The correct answer is A. (a>-2). Here (D=4(a+2)2 -4\(a^2+6a+8\)=-8(a+2)). For no real roots (D<0), so (a>-2).
Step 3
Exam Tip
यहाँ (D=4(a+2)2 -4\(a^2+6a+8\)=-8(a+2)) है। कोई वास्तविक मूल न होने के लिए (D<0), इसलिए (a>-2)।
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यदि (D=(z-4)(z+6)) है, तो कोई वास्तविक मूल न होने के लिए (z) का अंतराल कौन सा है?
If (D=(z-4)(z+6)), which interval of (z) gives no real roots?
#quadratic-equations
#discriminant-form
#no-real-roots
A (-6<z<4)
B (z<-6) या (z>4) / (z<-6) or (z>4)
C (z=-6) या (z=4) / (z=-6) or (z=4)
D हर (z) / Every (z)
Explanation opens after your attempt
Correct Answer
A. (-6<z<4)
Step 1
Concept
For no real roots (D<0) is needed. ((z-4)(z+6)<0) gives (-6<z<4).
Step 2
Why this answer is correct
The correct answer is A. (-6<z<4). For no real roots (D<0) is needed. ((z-4)(z+6)<0) gives (-6<z<4).
Step 3
Exam Tip
कोई वास्तविक मूल न होने के लिए (D<0) चाहिए। ((z-4)(z+6)<0) से (-6<z<4) मिलता है।
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यदि (x-2 +2(m-4 )x+\(m^2-7m+14\)=0) के वास्तविक मूल हों, तो (m) पर कौन सी शर्त सही है?
If (x-2 +2(m-4 )x+\(m^2-7m+14\)=0) has real roots, which condition on (m) is correct?
#quadratic-equations
#real-roots
#parameter
A \(m\leq2\)
B (m>2)
C (m=4)
D हर (m) / Every (m)
Explanation opens after your attempt
Correct Answer
A. \(m\leq2\)
Step 1
Concept
From the discriminant, (D=4(2-m)). For real roots \(D\geq0\), so \(m\leq2\).
Step 2
Why this answer is correct
The correct answer is A. \(m\leq2\). From the discriminant, (D=4(2-m)). For real roots \(D\geq0\), so \(m\leq2\).
Step 3
Exam Tip
पिछले विविक्तकर से (D=4(2-m)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(m\leq2\)।
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समीकरण (x-2 -2(3t+1)x+\(5t^2+2t+4\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल क्या है?
The equation (x-2 -2(3t+1)x+\(5t^2+2t+4\)=0) has no real roots. What is the correct interval for (t)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-2<t<1)
B (t<-2) या (t>1) / (t<-2) or (t>1)
C (t=-2) या (t=1) / (t=-2) or (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. (-2<t<1)
Step 1
Concept
Here (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)). From (D<0), (-2<t<1).
Step 2
Why this answer is correct
The correct answer is A. (-2<t<1). Here (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)). From (D<0), (-2<t<1).
Step 3
Exam Tip
यहाँ (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)) है। (D<0) से (-2<t<1)।
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समीकरण (x-2 -2(a-b)x+(a+b)2 =0) के वास्तविक मूलों के लिए सही शर्त क्या है?
What is the correct condition for real roots of (x-2 -2(a-b)x+(a+b)2 =0)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(ab\leq0\)
B (ab>0)
C (a=b)
D (a+b=0) मात्र / Only (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(ab\leq0\)
Step 1
Concept
Here (D=4(a-b)2 -4(a+b)2 =-16ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 2
Why this answer is correct
The correct answer is A. \(ab\leq0\). Here (D=4(a-b)2 -4(a+b)2 =-16ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 3
Exam Tip
यहाँ (D=4(a-b)2 -4(a+b)2 =-16ab) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(ab\leq0\)।
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यदि (x-2 -2(a+b)x+2ab=0) के मूल वास्तविक हों, तो (a) और (b) के लिए कौन सा कथन हमेशा सही है?
If (x-2 -2(a+b)x+2ab=0) has real roots, which statement is always true for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\)
B मूल तभी वास्तविक हैं जब (ab>0) / Roots are real only when (ab>0)
C मूल कभी वास्तविक नहीं होते / Roots are never real
D मूल तभी समान हैं जब (a+b=0) / Roots are equal only when (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -8ab=4\(a^2+b^2\)). It is always zero or positive, so real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\). Here (D=4(a+b)2 -8ab=4\(a^2+b^2\)). It is always zero or positive, so real roots exist.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -8ab=4\(a^2+b^2\)) है। यह हमेशा (0) या धनात्मक होता है, इसलिए वास्तविक मूल मिलते हैं।
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समीकरण (x-2 +2(t+1)x+(3t+7)=0) के कोई वास्तविक मूल नहीं होने की सही शर्त क्या है?
What is the correct condition for (x-2 +2(t+1)x+(3t+7)=0) to have no real roots?
#quadratic-equations
#parameter-interval
#no-real-roots
A (-2<t<3)
B (-4<t<1)
C (t<-2) या (t>3) / (t<-2) or (t>3)
D (t=-2) या (t=3) / (t=-2) or (t=3)
Explanation opens after your attempt
Correct Answer
A. (-2<t<3)
Step 1
Concept
Here (D=4\(t^2-t-6\)=4(t-3)(t+2)). From (D<0), we get (-2<t<3).
Step 2
Why this answer is correct
The correct answer is A. (-2<t<3). Here (D=4\(t^2-t-6\)=4(t-3)(t+2)). From (D<0), we get (-2<t<3).
Step 3
Exam Tip
यहाँ (D=4\(t^2-t-6\)=4(t-3)(t+2)) है। (D<0) से (-2<t<3) मिलता है।
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यदि (x-2 +2(t+1)x+(3t+7)=0) के कोई वास्तविक मूल नहीं हों, तो (t) किस अंतराल में होगा?
If (x-2 +2(t+1)x+(3t+7)=0) has no real roots, in which interval will (t) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-4<t<1)
B (t<-4) या (t>1) / (t<-4) or (t>1)
C (t=-4) या (t=1) / (t=-4) or (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. (-4<t<1)
Step 1
Concept
Here (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)). For (D<0), factor again carefully before selecting the interval.
Step 2
Why this answer is correct
The correct answer is A. (-4<t<1). Here (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)). For (D<0), factor again carefully before selecting the interval.
Step 3
Exam Tip
यहाँ (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)) है। (D<0) से (-2<t<3) नहीं, गुणनखंड फिर से जाँचें।
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यदि ((p-1)x-2 -2(p+1)x+(p+3)=0) में \(p\neq1\) हो, तो वास्तविक मूलों के लिए (p) की शर्त क्या है?
If \(p\neq1\) in ((p-1)x-2 -2(p+1)x+(p+3)=0), what is the condition on (p) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\)
B (p>2)
C (p=1)
D हर \(p\neq1\) / Every \(p\neq1\)
Explanation opens after your attempt
Correct Answer
A. \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\)
Step 1
Concept
Here (D=4(p+1)2 -4(p-1)(p+3)=16-4p). For real roots \(p\leq2\), and for a quadratic \(p\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\). Here (D=4(p+1)2 -4(p-1)(p+3)=16-4p). For real roots \(p\leq2\), and for a quadratic \(p\neq1\).
Step 3
Exam Tip
यहाँ (D=4(p+1)2 -4(p-1)(p+3)=16-4p) है। वास्तविक मूलों के लिए \(p\leq2\) और द्विघात के लिए \(p\neq1\)।
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यदि (x-2 -2(k-3)x+\(k^2-8k+20\)=0) के कोई वास्तविक मूल नहीं हों, तो (k) के लिए सही शर्त क्या है?
If (x-2 -2(k-3)x+\(k^2-8k+20\)=0) has no real roots, what is the correct condition on (k)?
#quadratic-equations
#no-real-roots
#parameter
A \(k<\frac{11}{2}\)
B \(k=\frac{11}{2}\)
C \(k>\frac{11}{2}\)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. \(k<\frac{11}{2}\)
Step 1
Concept
Here (D=4(k-3)2 -4\(k^2-8k+20\)=4(2k-11)). For no real roots (D<0), so \(k<\frac{11}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k<\frac{11}{2}\). Here (D=4(k-3)2 -4\(k^2-8k+20\)=4(2k-11)). For no real roots (D<0), so \(k<\frac{11}{2}\).
Step 3
Exam Tip
यहाँ (D=4(k-3)2 -4\(k^2-8k+20\)=4(2k-11)) है। कोई वास्तविक मूल नहीं के लिए (D<0), इसलिए \(k<\frac{11}{2}\)।
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यदि (x-2 -2(k+2)x+\(k^2+3k+7\)=0) के वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (x-2 -2(k+2)x+\(k^2+3k+7\)=0) has real roots, which condition on (k) is correct?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq3\)
B (k<3)
C (k=0)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Correct Answer
A. \(k\geq3\)
Step 1
Concept
Here (D=4(k+2)2 -4\(k^2+3k+7\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\geq3\). Here (D=4(k+2)2 -4\(k^2+3k+7\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 3
Exam Tip
यहाँ (D=4(k+2)2 -4\(k^2+3k+7\)=4(k-3)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(k\geq3\)।
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यदि (x-2 -2(a-1)x+\(a^2+1\)=0) के कोई वास्तविक मूल नहीं हों, तो (a) पर कौन सी शर्त सही है?
If (x-2 -2(a-1)x+\(a^2+1\)=0) has no real roots, which condition on (a) is correct?
#quadratic-equations
#no-real-roots
#parameter
A (a>0)
B (a=0)
C (a<0)
D हर वास्तविक (a) / Every real (a)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(a-1)2 -4\(a^2+1\)=-8a). For no real roots (D<0), so (a>0).
Step 2
Why this answer is correct
The correct answer is A. (a>0). Here (D=4(a-1)2 -4\(a^2+1\)=-8a). For no real roots (D<0), so (a>0).
Step 3
Exam Tip
यहाँ (D=4(a-1)2 -4\(a^2+1\)=-8a) है। कोई वास्तविक मूल न होने के लिए (D<0), इसलिए (a>0)।
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यदि किसी द्विघात का विविक्तकर (D=8m-24) है, तो वास्तविक मूलों के लिए (m) पर कौन सी शर्त होगी?
If a quadratic has discriminant (D=8m-24), what condition on (m) gives real roots?
#quadratic-equations
#discriminant-inequality
#real-roots
A \(m\geq3\)
B (m<3)
C (m=0)
D \(m\leq-3\)
Explanation opens after your attempt
Correct Answer
A. \(m\geq3\)
Step 1
Concept
For real roots \(D\geq0\) is needed. \(8m-24\geq0\) gives \(m\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(m\geq3\). For real roots \(D\geq0\) is needed. \(8m-24\geq0\) gives \(m\geq3\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। \(8m-24\geq0\) से \(m\geq3\) मिलता है।
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यदि (D=(u+1)(u-5)) है, तो कोई वास्तविक मूल न होने के लिए (u) का अंतराल कौन सा है?
If (D=(u+1)(u-5)), which interval of (u) gives no real roots?
#quadratic-equations
#discriminant-form
#no-real-roots
A (-1<u<5)
B (u<-1) या (u>5) / (u<-1) or (u>5)
C (u=-1) या (u=5) / (u=-1) or (u=5)
D हर (u) / Every (u)
Explanation opens after your attempt
Correct Answer
A. (-1<u<5)
Step 1
Concept
For no real roots (D<0) is needed. ((u+1)(u-5)<0) gives (-1<u<5).
Step 2
Why this answer is correct
The correct answer is A. (-1<u<5). For no real roots (D<0) is needed. ((u+1)(u-5)<0) gives (-1<u<5).
Step 3
Exam Tip
कोई वास्तविक मूल न होने के लिए (D<0) चाहिए। ((u+1)(u-5)<0) से (-1<u<5) मिलता है।
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यदि (x-2 +2(m-2 )x+\(m^2-3m+4\)=0) के कोई वास्तविक मूल नहीं हों, तो (m) की शर्त क्या होगी?
If (x-2 +2(m-2 )x+\(m^2-3m+4\)=0) has no real roots, what is the condition on (m)?
#quadratic-equations
#no-real-roots
#parameter
A (m>0)
B (m=0)
C (m<0)
D हर (m) / Every (m)
Explanation opens after your attempt
Step 1
Concept
From the discriminant, (D=-4m). For no real roots (D<0), so (m>0).
Step 2
Why this answer is correct
The correct answer is A. (m>0). From the discriminant, (D=-4m). For no real roots (D<0), so (m>0).
Step 3
Exam Tip
पिछले विविक्तकर से (D=-4m) है। कोई वास्तविक मूल न होने के लिए (D<0), इसलिए (m>0)।
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समीकरण (x-2 -2(2t-1)x+\(t^2+2\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल चुनिए।
The equation (x-2 -2(2t-1)x+\(t^2+2\)=0) has no real roots. Choose the correct interval for (t).
#quadratic-equations
#no-real-roots
#parameter-interval
A \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\)
B \(t<\frac{4-2\sqrt{6}}{3}\) या \(t>\frac{4+2\sqrt{6}}{3}\) / \(t<\frac{4-2\sqrt{6}}{3}\) or \(t>\frac{4+2\sqrt{6}}{3}\)
C (t=1) मात्र / Only (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\)
Step 1
Concept
Here (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)). From (D<0), the interval between the two boundary roots is obtained.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\). Here (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)). From (D<0), the interval between the two boundary roots is obtained.
Step 3
Exam Tip
यहाँ (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)) है। (D<0) से दिए गए दोनों मूलों के बीच का अंतराल मिलता है।
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समीकरण (x-2 -2(a+b)x+\(a^2+b^2\)=0) के वास्तविक मूलों के लिए कौन सा संबंध आवश्यक है?
Which relation is necessary for real roots of (x-2 -2(a+b)x+\(a^2+b^2\)=0)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(2ab\geq0\)
B (a+b=0)
C \(a^2+b^2<0\)
D (ab<0)
Explanation opens after your attempt
Correct Answer
A. \(2ab\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -4\(a^2+b^2\)=8ab). For real roots \(ab\geq0\) is needed.
Step 2
Why this answer is correct
The correct answer is A. \(2ab\geq0\). Here (D=4(a+b)2 -4\(a^2+b^2\)=8ab). For real roots \(ab\geq0\) is needed.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -4\(a^2+b^2\)=8ab) है। वास्तविक मूलों के लिए \(ab\geq0\) चाहिए।
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समीकरण ((k-2)x-2 +2kx+(k+3)=0) के वास्तविक मूलों के लिए सही शर्त कौन सी है, यदि \(k\neq2\)?
Which condition is correct for real roots of ((k-2)x-2 +2kx+(k+3)=0), if \(k\neq2\)?
#quadratic-equations
#real-roots
#parameter
A \(k\leq6\) और \(k\neq2\) / \(k\leq6\) and \(k\neq2\)
B \(k\geq6\) और \(k\neq2\) / \(k\geq6\) and \(k\neq2\)
C (k=2)
D (k>6)
Explanation opens after your attempt
Correct Answer
A. \(k\leq6\) और \(k\neq2\) / \(k\leq6\) and \(k\neq2\)
Step 1
Concept
Here (D=4k-2 -4(k-2)(k+3)=4(6-k)). For real roots \(D\geq0\), so \(k\leq6\) and \(k\neq2\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq6\) और \(k\neq2\) / \(k\leq6\) and \(k\neq2\). Here (D=4k-2 -4(k-2)(k+3)=4(6-k)). For real roots \(D\geq0\), so \(k\leq6\) and \(k\neq2\).
Step 3
Exam Tip
यहाँ (D=4k-2 -4(k-2)(k+3)=4(6-k)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(k\leq6\) और \(k\neq2\)।
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समीकरण ((k-2)x-2 +2kx+(k+3)=0) में \(k\neq2\) हो, तो वास्तविक मूलों के लिए सही शर्त क्या है?
In ((k-2)x-2 +2kx+(k+3)=0), with \(k\neq2\), what is the correct condition for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq\frac{3}{2}\)
B \(k<\frac{3}{2}\)
C (k=2) मात्र / Only (k=2)
D हर \(k\neq2\) / Every \(k\neq2\)
Explanation opens after your attempt
Correct Answer
A. \(k\geq\frac{3}{2}\)
Step 1
Concept
Here (D=(2k)2 -4(k-2)(k+3)=4(6-k)). For real roots we need \(k\leq6\), so check simplification carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k\geq\frac{3}{2}\). Here (D=(2k)2 -4(k-2)(k+3)=4(6-k)). For real roots we need \(k\leq6\), so check simplification carefully.
Step 3
Exam Tip
यहाँ (D=(2k)2 -4(k-2)(k+3)=4(6-k)) नहीं, सही सरल रूप (4(6-k)) है। वास्तविक मूलों के लिए \(k\leq6\) चाहिए।
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समीकरण (x-2 +2(k-1)x+(k+5)=0) के कोई वास्तविक मूल नहीं होने की सही शर्त क्या है?
What is the correct condition for (x-2 +2(k-1)x+(k+5)=0) to have no real roots?
#quadratic-equations
#parameter-interval
#no-real-roots
A (-1<k<4)
B (0<k<3)
C (k<-1) या (k>4) / (k<-1) or (k>4)
D (k=-1) या (k=4) / (k=-1) or (k=4)
Explanation opens after your attempt
Correct Answer
A. (-1<k<4)
Step 1
Concept
Here (D=4((k-1)2 -(k+5))=4\(k^2-3k-4\)). From (D<0), ((k-4)(k+1)<0), so (-1<k<4).
Step 2
Why this answer is correct
The correct answer is A. (-1<k<4). Here (D=4((k-1)2 -(k+5))=4\(k^2-3k-4\)). From (D<0), ((k-4)(k+1)<0), so (-1<k<4).
Step 3
Exam Tip
यहाँ (D=4((k-1)2 -(k+5))=4\(k^2-3k-4\)) है। (D<0) से ((k-4)(k+1)<0), इसलिए (-1<k<4)।
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यदि (x-2 +2(k-1)x+(k+5)=0) के कोई वास्तविक मूल नहीं हैं, तो (k) किस अंतराल में होगा?
If (x-2 +2(k-1)x+(k+5)=0) has no real roots, in which interval will (k) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (0<k<3)
B \(k\leq0\) या \(k\geq3\) / \(k\leq0\) or \(k\geq3\)
C (k=0) या (k=3) / (k=0) or (k=3)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. (0<k<3)
Step 1
Concept
Here (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)). Use (D<0) and factor carefully before choosing the interval.
Step 2
Why this answer is correct
The correct answer is A. (0<k<3). Here (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)). Use (D<0) and factor carefully before choosing the interval.
Step 3
Exam Tip
यहाँ (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)) नहीं, सही सरल रूप (4\(k^2-3k-4\)) है। (D<0) से (0<k<3) नहीं मिलता, इसलिए गुणनखंड जाँचें।
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