100 results found for "roots-parameter" in Class 10.
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 \(4x^2-3x+k=0\) 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 \(3x^2-2x+k=0\) 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 \(x^2-6x+c=0\) 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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\(x^2+12x+\lambda=0\) 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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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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\(x^2+10x+\lambda=0\) 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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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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\(x^2+2x+\lambda=0\) 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 \(x^2-6x+5=0\) 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 \(x^2-3x-2=0\) 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 \(x^2+px+64=0\) की जड़ें समान और धनात्मक हैं, तो (p) का मान क्या है?
If \(x^2+px+64=0\) has equal and positive roots, what is the value of (p)?
#quadratic-roots
#equal-positive-roots
#parameter
A (-16)
B (16)
C (-8)
D (8)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-256=0\), so \(p=\pm16\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-16).
Step 2
Why this answer is correct
The correct answer is A. (-16). For equal roots, \(p^2-256=0\), so \(p=\pm16\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-16).
Step 3
Exam Tip
समान जड़ों के लिए \(p^2-256=0\), इसलिए \(p=\pm16\)। समान जड़ \(-\frac{p}{2}\) धनात्मक होनी चाहिए, अतः (p=-16)।
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यदि (x-2 +(m-2 )x+25=0) की जड़ें समान हैं, तो (m) के मान क्या हैं?
If (x-2 +(m-2 )x+25=0) has equal roots, what are the values of (m)?
#quadratic-roots
#equal-roots
#parameter-values
A (12) और (-8) / (12) and (-8)
B (10) और (-10) / (10) and (-10)
C (7) और (-3) / (7) and (-3)
D (5) और (-5) / (5) and (-5)
Explanation opens after your attempt
Correct Answer
A. (12) और (-8) / (12) and (-8)
Step 1
Concept
For equal roots, ((m-2 )2 -100=0). Thus \(m-2=\pm10\), so (m=12) or (m=-8).
Step 2
Why this answer is correct
The correct answer is A. (12) और (-8) / (12) and (-8). For equal roots, ((m-2 )2 -100=0). Thus \(m-2=\pm10\), so (m=12) or (m=-8).
Step 3
Exam Tip
समान जड़ों के लिए ((m-2 )2 -100=0) होगा। इसलिए \(m-2=\pm10\), अतः (m=12) या (m=-8)।
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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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यदि \(3x^2+mx+10=0\) की जड़ें (2:5) के अनुपात में हैं, तो (m) के संभव मान क्या हैं?
If the roots of \(3x^2+mx+10=0\) are in the ratio (2:5), what are the possible values of (m)?
#quadratic-roots
#roots-ratio
#parameter
A \(7\sqrt{3}\) और \(-7\sqrt{3}\) / \(7\sqrt{3}\) and \(-7\sqrt{3}\)
B \(\frac{7\sqrt{3}}{2}\) और \(-\frac{7\sqrt{3}}{2}\) / \(\frac{7\sqrt{3}}{2}\) and \(-\frac{7\sqrt{3}}{2}\)
C \(14\sqrt{3}\) और \(-14\sqrt{3}\) / \(14\sqrt{3}\) and \(-14\sqrt{3}\)
D \(\frac{14}{\sqrt{3}}\) और \(-\frac{14}{\sqrt{3}}\) / \(\frac{14}{\sqrt{3}}\) and \(-\frac{14}{\sqrt{3}}\)
Explanation opens after your attempt
Correct Answer
A. \(7\sqrt{3}\) और \(-7\sqrt{3}\) / \(7\sqrt{3}\) and \(-7\sqrt{3}\)
Step 1
Concept
Let the roots be (2r) and (5r). From \(10r^2=\frac{10}{3}\), \(r=\pm\frac{1}{\sqrt{3}}\), so \(7r=-\frac{m}{3}\) gives \(m=\pm7\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(7\sqrt{3}\) और \(-7\sqrt{3}\) / \(7\sqrt{3}\) and \(-7\sqrt{3}\). Let the roots be (2r) and (5r). From \(10r^2=\frac{10}{3}\), \(r=\pm\frac{1}{\sqrt{3}}\), so \(7r=-\frac{m}{3}\) gives \(m=\pm7\sqrt{3}\).
Step 3
Exam Tip
जड़ें (2r) और (5r) मानें। \(10r^2=\frac{10}{3}\) से \(r=\pm\frac{1}{\sqrt{3}}\), इसलिए \(7r=-\frac{m}{3}\) से \(m=\pm7\sqrt{3}\)।
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यदि \(5x^2-16x+p=0\) की जड़ों का धनात्मक अंतर \(\frac{6}{5}\) है, तो (p) का मान क्या है?
If the positive difference between the roots of \(5x^2-16x+p=0\) is \(\frac{6}{5}\), what is the value of (p)?
#quadratic-roots
#difference-of-roots
#parameter
A (9)
B (10)
C (11)
D (12)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=\frac{16}{5}\) and \(\alpha\beta=\frac{p}{5}\). Using (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta), we get (p=11).
Step 2
Why this answer is correct
The correct answer is C. (11). Here \(\alpha+\beta=\frac{16}{5}\) and \(\alpha\beta=\frac{p}{5}\). Using (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta), we get (p=11).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{16}{5}\) और \(\alpha\beta=\frac{p}{5}\) है। (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta) से (p=11) मिलता है।
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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 -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 \(x^2+px+36=0\) की जड़ें समान और धनात्मक हैं, तो (p) का मान क्या है?
If \(x^2+px+36=0\) has equal and positive roots, what is the value of (p)?
#quadratic-roots
#equal-positive-roots
#parameter
A (-12)
B (12)
C (6)
D (-6)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-144=0\), so \(p=\pm12\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-12).
Step 2
Why this answer is correct
The correct answer is A. (-12). For equal roots, \(p^2-144=0\), so \(p=\pm12\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-12).
Step 3
Exam Tip
समान जड़ों के लिए \(p^2-144=0\), इसलिए \(p=\pm12\)। समान जड़ \(-\frac{p}{2}\) धनात्मक होनी चाहिए, अतः (p=-12)।
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यदि (x-2 +(m+1)x+16=0) की जड़ें समान हैं, तो (m) के मान क्या हैं?
If (x-2 +(m+1)x+16=0) has equal roots, what are the values of (m)?
#quadratic-roots
#equal-roots
#parameter-values
A (7) और (-9) / (7) and (-9)
B (8) और (-8) / (8) and (-8)
C (9) और (-7) / (9) and (-7)
D (15) और (-17) / (15) and (-17)
Explanation opens after your attempt
Correct Answer
A. (7) और (-9) / (7) and (-9)
Step 1
Concept
For equal roots, ((m+1)2 -64=0). Thus \(m+1=\pm8\), so (m=7) or (m=-9).
Step 2
Why this answer is correct
The correct answer is A. (7) और (-9) / (7) and (-9). For equal roots, ((m+1)2 -64=0). Thus \(m+1=\pm8\), so (m=7) or (m=-9).
Step 3
Exam Tip
समान जड़ों के लिए ((m+1)2 -64=0) होगा। इसलिए \(m+1=\pm8\), अतः (m=7) या (m=-9)।
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यदि \(4x^2-12x+m=0\) की जड़ों का धनात्मक अंतर (1) है, तो (m) का मान क्या है?
If the positive difference between the roots of \(4x^2-12x+m=0\) is (1), what is the value of (m)?
#quadratic-roots
#difference-of-roots
#parameter
A (6)
B (8)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=3\) and \(\alpha\beta=\frac{m}{4}\). Putting (\(\alpha-\beta\)2 =1) gives (9-m=1), so (m=8).
Step 2
Why this answer is correct
The correct answer is B. (8). Here \(\alpha+\beta=3\) and \(\alpha\beta=\frac{m}{4}\). Putting (\(\alpha-\beta\)2 =1) gives (9-m=1), so (m=8).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=3\) और \(\alpha\beta=\frac{m}{4}\) है। (\(\alpha-\beta\)2 =1) रखने पर (9-m=1), इसलिए (m=8)।
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(x-2 -2x+\(a^2+1\)=0) की जड़ों की प्रकृति क्या है?
What is the nature of the roots of (x-2 -2x+\(a^2+1\)=0)?
#quadratic-roots
#nature-of-roots
#parameter
A (a=0) पर समान वास्तविक, अन्यथा वास्तविक नहीं / Equal real at (a=0), otherwise not real
B हर (a) पर दो वास्तविक भिन्न / Two real distinct for every (a)
C हर (a) पर समान वास्तविक / Equal real for every (a)
D हर (a) पर वास्तविक नहीं / Not real for every (a)
Explanation opens after your attempt
Correct Answer
A. (a=0) पर समान वास्तविक, अन्यथा वास्तविक नहीं / Equal real at (a=0), otherwise not real
Step 1
Concept
The discriminant is (D=4-4\(a^2+1\)=-4a-2 ). Thus (D=0) at (a=0), and (D<0) when \(a\neq0\).
Step 2
Why this answer is correct
The correct answer is A. (a=0) पर समान वास्तविक, अन्यथा वास्तविक नहीं / Equal real at (a=0), otherwise not real. The discriminant is (D=4-4\(a^2+1\)=-4a-2 ). Thus (D=0) at (a=0), and (D<0) when \(a\neq0\).
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+1\)=-4a-2 ) है। इसलिए (a=0) पर (D=0), और \(a\neq0\) पर (D<0)।
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यदि \(x^2+px+16=0\) की जड़ें समान और धनात्मक हैं, तो (p) का मान क्या है?
If \(x^2+px+16=0\) has equal and positive roots, what is the value of (p)?
#quadratic-roots
#equal-positive-roots
#parameter
A (-8)
B (8)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-64=0\), so \(p=\pm8\). The root \(-\frac{p}{2}\) must be positive, hence (p=-8).
Step 2
Why this answer is correct
The correct answer is A. (-8). For equal roots, \(p^2-64=0\), so \(p=\pm8\). The root \(-\frac{p}{2}\) must be positive, hence (p=-8).
Step 3
Exam Tip
समान जड़ों के लिए \(p^2-64=0\), इसलिए \(p=\pm8\)। जड़ \(-\frac{p}{2}\) धनात्मक होनी चाहिए, अतः (p=-8)।
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यदि (x-2 +(m-5 )x+9=0) की जड़ें बराबर पर विपरीत चिह्न वाली नहीं हैं और समान हैं, तो (m) के मान क्या हैं?
If (x-2 +(m-5 )x+9=0) has equal roots that are not of opposite signs, what are the values of (m)?
#quadratic-roots
#equal-roots
#parameter-values
A (11) और (-1) / (11) and (-1)
B (5) और (-5) / (5) and (-5)
C (8) और (2) / (8) and (2)
D (14) और (-4) / (14) and (-4)
Explanation opens after your attempt
Correct Answer
A. (11) और (-1) / (11) and (-1)
Step 1
Concept
For equal roots, ((m-5 )2 -36=0). Thus \(m-5=\pm6\), so (m=11) or (m=-1).
Step 2
Why this answer is correct
The correct answer is A. (11) और (-1) / (11) and (-1). For equal roots, ((m-5 )2 -36=0). Thus \(m-5=\pm6\), so (m=11) or (m=-1).
Step 3
Exam Tip
समान जड़ों के लिए ((m-5 )2 -36=0) होगा। इसलिए \(m-5=\pm6\), अतः (m=11) या (m=-1)।
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यदि \(x^2+px+12=0\) की जड़ें (r) और (r+1) हैं, तो (p) के संभव मान क्या हैं?
If the roots of \(x^2+px+12=0\) are (r) and (r+1), what are the possible values of (p)?
#quadratic-roots
#consecutive-roots
#parameter
A (7) और (-7) / (7) and (-7)
B (5) और (-5) / (5) and (-5)
C (6) और (-6) / (6) and (-6)
D (13) और (-13) / (13) and (-13)
Explanation opens after your attempt
Correct Answer
A. (7) और (-7) / (7) and (-7)
Step 1
Concept
From (r(r+1)=12), we get (r=3) or (r=-4). The sum is (7) or (-7), so (p=-7) or (p=7).
Step 2
Why this answer is correct
The correct answer is A. (7) और (-7) / (7) and (-7). From (r(r+1)=12), we get (r=3) or (r=-4). The sum is (7) or (-7), so (p=-7) or (p=7).
Step 3
Exam Tip
(r(r+1)=12) से (r=3) या (r=-4) मिलता है। जड़ों का योग (7) या (-7) है, इसलिए (p=-7) या (p=7)।
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यदि \(2x^2+mx+8=0\) की जड़ें (1:2) के अनुपात में हैं, तो (m) के संभव मान क्या हैं?
If the roots of \(2x^2+mx+8=0\) are in the ratio (1:2), what are the possible values of (m)?
#quadratic-roots
#roots-ratio
#parameter
A \(6\sqrt{2}\) और \(-6\sqrt{2}\) / \(6\sqrt{2}\) and \(-6\sqrt{2}\)
B \(3\sqrt{2}\) और \(-3\sqrt{2}\) / \(3\sqrt{2}\) and \(-3\sqrt{2}\)
C (12) और (-12) / (12) and (-12)
D (8) और (-8) / (8) and (-8)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\) और \(-6\sqrt{2}\) / \(6\sqrt{2}\) and \(-6\sqrt{2}\)
Step 1
Concept
Let the roots be (r) and (2r), then \(2r^2=4\) gives \(r=\pm\sqrt{2}\). Since \(3r=-\frac{m}{2}\), we get \(m=\pm6\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\) और \(-6\sqrt{2}\) / \(6\sqrt{2}\) and \(-6\sqrt{2}\). Let the roots be (r) and (2r), then \(2r^2=4\) gives \(r=\pm\sqrt{2}\). Since \(3r=-\frac{m}{2}\), we get \(m=\pm6\sqrt{2}\).
Step 3
Exam Tip
जड़ें (r) और (2r) मानें, तब \(2r^2=4\) से \(r=\pm\sqrt{2}\) मिलता है। योग \(3r=-\frac{m}{2}\), इसलिए \(m=\pm6\sqrt{2}\)।
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यदि \(3x^2-11x+p=0\) की जड़ों का अंतर \(\frac{5}{3}\) है, तो (p) का मान क्या है?
If the difference between the roots of \(3x^2-11x+p=0\) is \(\frac{5}{3}\), what is the value of (p)?
#quadratic-roots
#difference-of-roots
#parameter
A (8)
B (7)
C (6)
D (5)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=\frac{11}{3}\) and (\(\alpha-\beta\)2 =\frac{25}{9}). Using (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta), we get \(\alpha\beta=\frac{8}{3}\), so (p=8).
Step 2
Why this answer is correct
The correct answer is A. (8). Here \(\alpha+\beta=\frac{11}{3}\) and (\(\alpha-\beta\)2 =\frac{25}{9}). Using (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta), we get \(\alpha\beta=\frac{8}{3}\), so (p=8).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{11}{3}\) और (\(\alpha-\beta\)2 =\frac{25}{9}) है। सूत्र (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta) से \(\alpha\beta=\frac{8}{3}\), इसलिए (p=8)।
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((k-2)x-2 +4x+1=0) की जड़ें समान हों, तो (k) का मान क्या है?
If ((k-2)x-2 +4x+1=0) has equal roots, what is (k)?
#quadratic-roots
#equal-roots
#parameter
A (6)
B (4)
C (2)
D (-6)
Explanation opens after your attempt
Step 1
Concept
For equal roots, put (D=0). From (16-4(k-2)=0), we get (k=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal roots, put (D=0). From (16-4(k-2)=0), we get (k=6).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। (16-4(k-2)=0) से (k=6) मिलता है।
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\(x^2-sx+9=0\) की जड़ें समान और धनात्मक हों, तो (s) का मान क्या होगा?
If \(x^2-sx+9=0\) has equal and positive roots, what is (s)?
#quadratic-roots
#equal-positive-roots
#parameter
A (6)
B (-6)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(s^2-36=0\), so \(s=\pm6\). The equal root is \(\frac{s}{2}\), which is positive when (s=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal roots, \(s^2-36=0\), so \(s=\pm6\). The equal root is \(\frac{s}{2}\), which is positive when (s=6).
Step 3
Exam Tip
समान जड़ों के लिए \(s^2-36=0\), इसलिए \(s=\pm6\)। समान जड़ \(\frac{s}{2}\) है, जो धनात्मक होने पर (s=6) देता है।
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\(x^2+6x+r=0\) की जड़ों का अंतर \(2\sqrt{5}\) है, तो (r) का मान क्या है?
If the difference between the roots of \(x^2+6x+r=0\) is \(2\sqrt{5}\), what is (r)?
#quadratic-roots
#difference-of-roots
#parameter
A (4)
B (5)
C (6)
D (9)
Explanation opens after your attempt
Step 1
Concept
Using (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta), we get (20=36-4r). Hence (r=4).
Step 2
Why this answer is correct
The correct answer is A. (4). Using (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta), we get (20=36-4r). Hence (r=4).
Step 3
Exam Tip
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta) से (20=36-4r) मिलता है। इसलिए (r=4)।
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\(2x^2+kx+2=0\) की जड़ें समान और ऋणात्मक हों, तो (k) का मान क्या होगा?
If \(2x^2+kx+2=0\) has equal and negative roots, what is (k)?
#quadratic-roots
#equal-negative-roots
#parameter
A (4)
B (-4)
C (2)
D (-2)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(k^2-16=0\), so \(k=\pm4\). The equal root is \(-\frac{k}{4}\), which is negative only when (k=4).
Step 2
Why this answer is correct
The correct answer is A. (4). For equal roots, \(k^2-16=0\), so \(k=\pm4\). The equal root is \(-\frac{k}{4}\), which is negative only when (k=4).
Step 3
Exam Tip
समान जड़ों के लिए \(k^2-16=0\), इसलिए \(k=\pm4\)। समान जड़ \(-\frac{k}{4}\) है, जो ऋणात्मक तभी होगी जब (k=4)।
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(x-2 -(m+1)x+m=0) की जड़ें एक-दूसरे की व्युत्क्रम हों, तो (m) का मान क्या है?
If the roots of (x-2 -(m+1)x+m=0) are reciprocals of each other, what is (m)?
#quadratic-roots
#reciprocal-roots
#parameter
A (1)
B (-1)
C (0)
D (2)
Explanation opens after your attempt
Step 1
Concept
For reciprocal roots, \(\alpha\beta=1\). Here \(\alpha\beta=m\), so (m=1).
Step 2
Why this answer is correct
The correct answer is A. (1). For reciprocal roots, \(\alpha\beta=1\). Here \(\alpha\beta=m\), so (m=1).
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए \(\alpha\beta=1\) होता है। यहाँ \(\alpha\beta=m\), इसलिए (m=1)।
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\(x^2+kx+16=0\) की जड़ें समान हों, तो (k) के संभव मान क्या हैं?
If \(x^2+kx+16=0\) has equal roots, what are the possible values of (k)?
#quadratic-roots
#equal-roots
#parameter
A (8) और (-8) / (8) and (-8)
B (4) और (-4) / (4) and (-4)
C (16) और (-16) / (16) and (-16)
D (2) और (-2) / (2) and (-2)
Explanation opens after your attempt
Correct Answer
A. (8) और (-8) / (8) and (-8)
Step 1
Concept
For equal roots, \(k^2-64=0\) must hold. Hence \(k=\pm8\).
Step 2
Why this answer is correct
The correct answer is A. (8) और (-8) / (8) and (-8). For equal roots, \(k^2-64=0\) must hold. Hence \(k=\pm8\).
Step 3
Exam Tip
समान जड़ों के लिए \(k^2-64=0\) होना चाहिए। अतः \(k=\pm8\) है।
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(x-2 -(k+2)x+2k=0) की जड़ों का अंतर (2) है, तो (k) के मान क्या हैं?
If the difference between the roots of (x-2 -(k+2)x+2k=0) is (2), what are the values of (k)?
#quadratic-roots
#difference-of-roots
#parameter
A (0) और (4) / (0) and (4)
B (2) और (4) / (2) and (4)
C (-2) और (4) / (-2) and (4)
D (0) और (2) / (0) and (2)
Explanation opens after your attempt
Correct Answer
A. (0) और (4) / (0) and (4)
Step 1
Concept
(\(\alpha-\beta\)2 =(k+2)2 -8k=(k-2)2 ). Setting it equal to (4) gives (k=0) or (k=4).
Step 2
Why this answer is correct
The correct answer is A. (0) और (4) / (0) and (4). (\(\alpha-\beta\)2 =(k+2)2 -8k=(k-2)2 ). Setting it equal to (4) gives (k=0) or (k=4).
Step 3
Exam Tip
(\(\alpha-\beta\)2 =(k+2)2 -8k=(k-2)2 ) है। इसे (4) रखने पर (k=0) या (k=4) मिलता 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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यदि (x-2 +(a-2)x+a=0) की जड़ों का अंतर (3) है, तो (a) का मान क्या है?
If the difference between the roots of (x-2 +(a-2)x+a=0) is (3), what is the value of (a)?
#quadratic-roots
#difference-of-roots
#parameter
A \(4+\sqrt{21}\) या \(4-\sqrt{21}\) / \(4+\sqrt{21}\) or \(4-\sqrt{21}\)
B \(2+\sqrt{21}\) या \(2-\sqrt{21}\) / \(2+\sqrt{21}\) or \(2-\sqrt{21}\)
C \(4+\sqrt{5}\) या \(4-\sqrt{5}\) / \(4+\sqrt{5}\) or \(4-\sqrt{5}\)
D \(-4+\sqrt{21}\) या \(-4-\sqrt{21}\) / \(-4+\sqrt{21}\) or \(-4-\sqrt{21}\)
Explanation opens after your attempt
Correct Answer
A. \(4+\sqrt{21}\) या \(4-\sqrt{21}\) / \(4+\sqrt{21}\) or \(4-\sqrt{21}\)
Step 1
Concept
Put (\(\alpha-\beta\)2 =9) in (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta). This gives \(a^2-8a-5=0\), so \(a=4\pm\sqrt{21}\).
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{21}\) या \(4-\sqrt{21}\) / \(4+\sqrt{21}\) or \(4-\sqrt{21}\). Put (\(\alpha-\beta\)2 =9) in (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta). This gives \(a^2-8a-5=0\), so \(a=4\pm\sqrt{21}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta) में (\(\alpha-\beta\)2 =9) रखें। इससे \(a^2-8a-5=0\) और \(a=4\pm\sqrt{21}\) मिलता है।
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(x-2 +2(k-1)x+k+5=0) की जड़ें समान हों, तो (k) के मान कौन-से हैं?
If (x-2 +2(k-1)x+k+5=0) has equal roots, what are the values of (k)?
#quadratic-roots
#equal-roots
#parameter-values
A (4) और (-1) / (4) and (-1)
B (4) और (1) / (4) and (1)
C (-4) और (1) / (-4) and (1)
D (-5) और (4) / (-5) and (4)
Explanation opens after your attempt
Correct Answer
A. (4) और (-1) / (4) and (-1)
Step 1
Concept
For equal roots, put (D=0). This gives \(k^2-3k-4=0\), so (k=4) or (k=-1).
Step 2
Why this answer is correct
The correct answer is A. (4) और (-1) / (4) and (-1). For equal roots, put (D=0). This gives \(k^2-3k-4=0\), so (k=4) or (k=-1).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। इससे \(k^2-3k-4=0\) मिलता है, इसलिए (k=4) या (k=-1)।
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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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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 (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 \(kx^2-10x+k=0\) 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 (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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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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\(x^2-10x+k=0\) 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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\(x^2-4x+k=0\) 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 \(x^2+bx+c=0\) 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 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 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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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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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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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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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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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 (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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यदि (x-2 +2(m-5 )x+\(m^2-9m+24\)=0) के समान मूल हों, तो (m) का मान क्या होगा?
If (x-2 +2(m-5 )x+\(m^2-9m+24\)=0) has equal roots, what is the value of (m)?
#quadratic-equations
#equal-roots
#parameter
A (2)
B (5)
C (9)
D (24)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(m-5 )2 -4\(m^2-9m+24\)=4(2-m)). From (D=0), (m=2).
Step 2
Why this answer is correct
The correct answer is A. (2). Here (D=4(m-5 )2 -4\(m^2-9m+24\)=4(2-m)). From (D=0), (m=2).
Step 3
Exam Tip
यहाँ (D=4(m-5 )2 -4\(m^2-9m+24\)=4(2-m)) है। (D=0) से (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 \(x^2-2\theta x+3\theta=0\) के समान मूलों के लिए \(\theta\) के मान कौन से होंगे?
What are the values of \(\theta\) for equal roots of \(x^2-2\theta x+3\theta=0\)?
#quadratic-equations
#equal-roots
#parameter
A \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
B केवल \(\theta=3\) / Only \(\theta=3\)
C केवल \(\theta=0\) / Only \(\theta=0\)
D \(\theta=-3\) या \(\theta=3\) / \(\theta=-3\) or \(\theta=3\)
Explanation opens after your attempt
Correct Answer
A. \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
Step 1
Concept
For equal roots (D=4\theta\(\theta-3\)=0). Hence \(\theta=0\) or \(\theta=3\).
Step 2
Why this answer is correct
The correct answer is A. \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\). For equal roots (D=4\theta\(\theta-3\)=0). Hence \(\theta=0\) or \(\theta=3\).
Step 3
Exam Tip
समान मूलों के लिए (D=4\theta\(\theta-3\)=0) है। अतः \(\theta=0\) या \(\theta=3\)।
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यदि \(x^2-2\theta x+3\theta=0\) के दो वास्तविक और असमान मूल हों, तो \(\theta\) पर कौन सी शर्त सही है?
If \(x^2-2\theta x+3\theta=0\) has two real and distinct roots, which condition on \(\theta\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\)
B \(0<\theta<3\)
C \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
D हर \(\theta\) / Every \(\theta\)
Explanation opens after your attempt
Correct Answer
A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\)
Step 1
Concept
Here (D=4\theta-2 -12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 2
Why this answer is correct
The correct answer is A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\). Here (D=4\theta-2 -12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 3
Exam Tip
यहाँ (D=4\theta-2 -12\theta=4\theta\(\theta-3\)) है। (D>0) से \(\theta<0\) या \(\theta>3\)।
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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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समीकरण ((q+3)x-2 -2(q-2)x+q=0) में समान मूलों के लिए (q) का सही मान क्या है, जबकि \(q\neq-3\)?
What is the correct value of (q) for equal roots in ((q+3)x-2 -2(q-2)x+q=0), where \(q\neq-3\)?
#quadratic-equations
#parameter
#equal-roots
A \(q=\frac{4}{7}\)
B \(q=\frac{7}{4}\)
C (q=4)
D (q=-3)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{4}{7}\)
Step 1
Concept
Here (D=4(q-2)2 -4q(q+3)=4(4-7q)). From (D=0), \(q=\frac{4}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{4}{7}\). Here (D=4(q-2)2 -4q(q+3)=4(4-7q)). From (D=0), \(q=\frac{4}{7}\).
Step 3
Exam Tip
यहाँ (D=4(q-2)2 -4q(q+3)=4(4-7q)) है। (D=0) से \(q=\frac{4}{7}\)।
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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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समीकरण (x-2 -(u+6)x+6u=0) में समान मूलों के लिए (u) का मान क्या होगा?
What is the value of (u) for equal roots in (x-2 -(u+6)x+6u=0)?
#quadratic-equations
#equal-roots
#parameter
A (u=6)
B (u=-6)
C (u=0)
D (u=12)
Explanation opens after your attempt
Step 1
Concept
Here (D=(u+6)2 -24u=(u-6)2 ). From (D=0), (u=6).
Step 2
Why this answer is correct
The correct answer is A. (u=6). Here (D=(u+6)2 -24u=(u-6)2 ). From (D=0), (u=6).
Step 3
Exam Tip
यहाँ (D=(u+6)2 -24u=(u-6)2 ) है। (D=0) से (u=6) मिलता है।
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समीकरण (x-2 -(t+7)x+7t=0) के दो वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two real and distinct roots of (x-2 -(t+7)x+7t=0)?
#quadratic-equations
#parameter
#distinct-roots
A \(t\neq7\)
B (t=7)
C (t>7) मात्र / Only (t>7)
D (t<7) मात्र / Only (t<7)
Explanation opens after your attempt
Correct Answer
A. \(t\neq7\)
Step 1
Concept
Here (D=(t+7)2 -28t=(t-7)2 ). For two distinct roots (D>0), so \(t\neq7\).
Step 2
Why this answer is correct
The correct answer is A. \(t\neq7\). Here (D=(t+7)2 -28t=(t-7)2 ). For two distinct roots (D>0), so \(t\neq7\).
Step 3
Exam Tip
यहाँ (D=(t+7)2 -28t=(t-7)2 ) है। दो असमान मूलों के लिए (D>0), इसलिए \(t\neq7\)।
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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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समीकरण (2x-2 -2(3k-1)x+(k+1)2 =0) में समान मूलों के लिए (k) के मान क्या होंगे?
For equal roots in (2x-2 -2(3k-1)x+(k+1)2 =0), what are the values of (k)?
#quadratic-equations
#equal-roots
#parameter
A (k=1) या (k=-3) / (k=1) or (k=-3)
B (k=3) या (k=-1) / (k=3) or (k=-1)
C (k=0) या (k=-2) / (k=0) or (k=-2)
D (k=2) या (k=-3) / (k=2) or (k=-3)
Explanation opens after your attempt
Correct Answer
A. (k=1) या (k=-3) / (k=1) or (k=-3)
Step 1
Concept
Here (D=4(3k-1)2 -8(k+1)2 ). For equal roots, solve (D=0) directly and avoid mental shortcuts.
Step 2
Why this answer is correct
The correct answer is A. (k=1) या (k=-3) / (k=1) or (k=-3). Here (D=4(3k-1)2 -8(k+1)2 ). For equal roots, solve (D=0) directly and avoid mental shortcuts.
Step 3
Exam Tip
यहाँ (D=4(3k-1)2 -8(k+1)2 =4\(7k^2-10k-1\)) नहीं, यह विकल्प जाँचने योग्य है। सही समान मूल के लिए सीधे (D=0) हल करें।
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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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यदि (x-2 +2(m-4 )x+\(m^2-7m+14\)=0) के समान मूल हों, तो (m) का मान क्या होगा?
If (x-2 +2(m-4 )x+\(m^2-7m+14\)=0) has equal roots, what is the value of (m)?
#quadratic-equations
#equal-roots
#parameter
A (2)
B (4)
C (7)
D (14)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(m-4 )2 -4\(m^2-7m+14\)=4(2-m)). From (D=0), (m=2).
Step 2
Why this answer is correct
The correct answer is A. (2). Here (D=4(m-4 )2 -4\(m^2-7m+14\)=4(2-m)). From (D=0), (m=2).
Step 3
Exam Tip
यहाँ (D=4(m-4 )2 -4\(m^2-7m+14\)=4(2-m)) है। (D=0) से (m=2) मिलता है।
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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 \(x^2-2\mu x+2\mu=0\) के समान मूलों के लिए \(\mu\) के मान कौन से होंगे?
What are the values of \(\mu\) for equal roots of \(x^2-2\mu x+2\mu=0\)?
#quadratic-equations
#equal-roots
#parameter
A \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
B केवल \(\mu=2\) / Only \(\mu=2\)
C केवल \(\mu=0\) / Only \(\mu=0\)
D \(\mu=-2\) या \(\mu=2\) / \(\mu=-2\) or \(\mu=2\)
Explanation opens after your attempt
Correct Answer
A. \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
Step 1
Concept
For equal roots (D=4\mu\(\mu-2\)=0). Therefore \(\mu=0\) or \(\mu=2\).
Step 2
Why this answer is correct
The correct answer is A. \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\). For equal roots (D=4\mu\(\mu-2\)=0). Therefore \(\mu=0\) or \(\mu=2\).
Step 3
Exam Tip
समान मूलों के लिए (D=4\mu\(\mu-2\)=0) है। इसलिए \(\mu=0\) या \(\mu=2\)।
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यदि \(x^2-2\mu x+2\mu=0\) के दो वास्तविक और असमान मूल हों, तो \(\mu\) पर कौन सी शर्त सही है?
If \(x^2-2\mu x+2\mu=0\) has two real and distinct roots, which condition on \(\mu\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\)
B \(0<\mu<2\)
C \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
D हर \(\mu\) / Every \(\mu\)
Explanation opens after your attempt
Correct Answer
A. \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\)
Step 1
Concept
Here (D=4\mu-2 -8\mu=4\mu\(\mu-2\)). From (D>0), \(\mu<0\) or \(\mu>2\).
Step 2
Why this answer is correct
The correct answer is A. \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\). Here (D=4\mu-2 -8\mu=4\mu\(\mu-2\)). From (D>0), \(\mu<0\) or \(\mu>2\).
Step 3
Exam Tip
यहाँ (D=4\mu-2 -8\mu=4\mu\(\mu-2\)) है। (D>0) से \(\mu<0\) या \(\mu>2\)।
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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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समीकरण ((q+2)x-2 -2(q-1)x+q=0) में समान मूलों के लिए सही (q) कौन सा है, यदि \(q\neq-2\)?
Which value of (q) gives equal roots in ((q+2)x-2 -2(q-1)x+q=0), if \(q\neq-2\)?
#quadratic-equations
#parameter
#equal-roots
A \(q=\frac{1}{4}\)
B \(q=\frac{1}{2}\)
C (q=1)
D (q=-2)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{1}{4}\)
Step 1
Concept
Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). For equal roots (D=0), so \(q=\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{1}{4}\). Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). For equal roots (D=0), so \(q=\frac{1}{4}\).
Step 3
Exam Tip
यहाँ (D=4(q-1)2 -4q(q+2)=4(1-4q)) है। समान मूलों के लिए (D=0), इसलिए \(q=\frac{1}{4}\)।
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समीकरण ((q+2)x-2 -2(q-1)x+q=0) में \(q\neq-2\) हो, तो समान मूलों के लिए (q) का मान क्या है?
In ((q+2)x-2 -2(q-1)x+q=0), with \(q\neq-2\), what is the value of (q) for equal roots?
#quadratic-equations
#equal-roots
#parameter
A \(\frac{1}{2}\)
B (2)
C \(-\frac{1}{2}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). Setting (D=0) gives \(q=\frac{1}{4}\), so calculate carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). Setting (D=0) gives \(q=\frac{1}{4}\), so calculate carefully.
Step 3
Exam Tip
यहाँ (D=4(q-1)2 -4q(q+2)=4(1-4q)) है। (D=0) से \(q=\frac{1}{4}\) नहीं, सही गणना 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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समीकरण (x-2 -(s+4)x+4s=0) में समान मूलों के लिए (s) का मान क्या होगा?
What is the value of (s) for equal roots in (x-2 -(s+4)x+4s=0)?
#quadratic-equations
#equal-roots
#parameter
A (s=4)
B (s=-4)
C (s=0)
D (s=8)
Explanation opens after your attempt
Step 1
Concept
Here (D=(s+4)2 -16s=(s-4 )2 ). For equal roots (D=0), so (s=4).
Step 2
Why this answer is correct
The correct answer is A. (s=4). Here (D=(s+4)2 -16s=(s-4 )2 ). For equal roots (D=0), so (s=4).
Step 3
Exam Tip
यहाँ (D=(s+4)2 -16s=(s-4 )2 ) है। समान मूलों के लिए (D=0), इसलिए (s=4)।
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समीकरण (x-2 -(r+5)x+5r=0) के दो वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two real and distinct roots of (x-2 -(r+5)x+5r=0)?
#quadratic-equations
#parameter
#distinct-roots
A \(r\neq5\)
B (r=5)
C (r>5) मात्र / Only (r>5)
D (r<5) मात्र / Only (r<5)
Explanation opens after your attempt
Correct Answer
A. \(r\neq5\)
Step 1
Concept
Here (D=(r+5)2 -20r=(r-5)2 ). For two distinct roots (D>0), so \(r\neq5\).
Step 2
Why this answer is correct
The correct answer is A. \(r\neq5\). Here (D=(r+5)2 -20r=(r-5)2 ). For two distinct roots (D>0), so \(r\neq5\).
Step 3
Exam Tip
यहाँ (D=(r+5)2 -20r=(r-5)2 ) है। दो असमान मूलों के लिए (D>0), इसलिए \(r\neq5\)।
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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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