100 results found for "equal-negative-roots" in Class 10.
\(x^2+12x+\lambda=0\) 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 \(x^2+bx+49=0\) की जड़ें समान और ऋणात्मक हैं, तो (b) का मान क्या होगा?
If \(x^2+bx+49=0\) has equal and negative roots, what is the value of (b)?
#quadratic-roots
#equal-negative-roots
#coefficient
A (-14)
B (14)
C (7)
D (-7)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(b^2-196=0\), so \(b=\pm14\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=14).
Step 2
Why this answer is correct
The correct answer is B. (14). For equal roots, \(b^2-196=0\), so \(b=\pm14\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=14).
Step 3
Exam Tip
समान जड़ों के लिए \(b^2-196=0\), इसलिए \(b=\pm14\)। समान जड़ \(-\frac{b}{2}\) ऋणात्मक होनी चाहिए, अतः (b=14)।
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यदि \(x^2+bx+25=0\) की जड़ें समान और ऋणात्मक हैं, तो (b) का मान क्या होगा?
If \(x^2+bx+25=0\) has equal and negative roots, what is the value of (b)?
#quadratic-roots
#equal-negative-roots
#coefficient
A (10)
B (-10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(b^2-100=0\), so \(b=\pm10\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=10).
Step 2
Why this answer is correct
The correct answer is A. (10). For equal roots, \(b^2-100=0\), so \(b=\pm10\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=10).
Step 3
Exam Tip
समान जड़ों के लिए \(b^2-100=0\), इसलिए \(b=\pm10\)। समान जड़ \(-\frac{b}{2}\) ऋणात्मक होनी चाहिए, अतः (b=10)।
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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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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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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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यदि \(x^2+kx+49=0\) के मूल एक दूसरे के बराबर और ऋणात्मक हैं तो (k) क्या होगा?
If roots of \(x^2+kx+49=0\) are equal and negative, what is (k)?
#roots
#equal_negative_roots
#parameter
A (14)
B (-14)
C (7)
D (-7)
Explanation opens after your attempt
Step 1
Concept
The equal negative roots are (-7) and (-7). Their sum is (-14), so (k=14).
Step 2
Why this answer is correct
The correct answer is A. (14). The equal negative roots are (-7) and (-7). Their sum is (-14), so (k=14).
Step 3
Exam Tip
बराबर ऋणात्मक मूल (-7) और (-7) हैं। योग (-14) है इसलिए (k=14) है।
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यदि समीकरण \(x^2+px+25=0\) के मूल बराबर हैं और दोनों ऋणात्मक हैं तो (p) का मान क्या होगा?
If the roots of \(x^2+px+25=0\) are equal and both negative, what is the value of (p)?
#roots
#equal_negative_roots
#parameter
A (10)
B (-10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The equal negative roots are (-5) and (-5) because the product is (25). The sum is (-10), so (p=10).
Step 2
Why this answer is correct
The correct answer is A. (10). The equal negative roots are (-5) and (-5) because the product is (25). The sum is (-10), so (p=10).
Step 3
Exam Tip
बराबर ऋणात्मक मूल (-5) और (-5) होंगे क्योंकि गुणनफल (25) है। योग (-10) है इसलिए (p=10) है।
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यदि \(x^2+kx+36=0\) के मूल एक दूसरे के बराबर और ऋणात्मक हैं तो (k) क्या होगा?
If roots of \(x^2+kx+36=0\) are equal and negative, what is (k)?
#roots
#equal_negative_roots
#parameter
A (12)
B (-12)
C (6)
D (-6)
Explanation opens after your attempt
Step 1
Concept
The equal negative roots are (-6) and (-6). Their sum is (-12), so (-k=-12) gives (k=12).
Step 2
Why this answer is correct
The correct answer is A. (12). The equal negative roots are (-6) and (-6). Their sum is (-12), so (-k=-12) gives (k=12).
Step 3
Exam Tip
बराबर ऋणात्मक मूल (-6) और (-6) हैं। योग (-12) है इसलिए (-k=-12) से (k=12) है।
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यदि समीकरण \(x^2+px+16=0\) के मूल बराबर हैं और दोनों ऋणात्मक हैं तो (p) का मान क्या होगा?
If the roots of \(x^2+px+16=0\) are equal and both negative, what is the value of (p)?
#roots
#equal_negative_roots
#parameter
A (8)
B (-8)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
For equal roots, the roots are (-4) and (-4) because the product is (16). The sum is (-8), so (-p=-8) gives (p=8).
Step 2
Why this answer is correct
The correct answer is A. (8). For equal roots, the roots are (-4) and (-4) because the product is (16). The sum is (-8), so (-p=-8) gives (p=8).
Step 3
Exam Tip
बराबर मूलों के लिए मूल (-4) और (-4) होंगे क्योंकि गुणनफल (16) है। योग (-8) है इसलिए (-p=-8) से (p=8) 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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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 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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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 \(x^2+px+64=0\) 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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यदि (x-2 -2(a+5)x+a-2 +18=0) की जड़ें समान हैं, तो (a) का मान क्या होगा?
If (x-2 -2(a+5)x+a-2 +18=0) has equal roots, what is the value of (a)?
#quadratic-roots
#equal-roots
#discriminant
A \(-\frac{7}{10}\)
B \(\frac{7}{10}\)
C \(-\frac{10}{7}\)
D \(\frac{10}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{7}{10}\)
Step 1
Concept
For equal roots, put (D=0). From (4(a+5)2 -4\(a^2+18\)=0), we get (10a+7=0), so \(a=-\frac{7}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{7}{10}\). For equal roots, put (D=0). From (4(a+5)2 -4\(a^2+18\)=0), we get (10a+7=0), so \(a=-\frac{7}{10}\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखते हैं। (4(a+5)2 -4\(a^2+18\)=0) से (10a+7=0), इसलिए \(a=-\frac{7}{10}\)।
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यदि (x-2 -(u+v+2)x+(u+1)(v+1)=0) की जड़ें समान हैं, तो सही कथन क्या है?
If (x-2 -(u+v+2)x+(u+1)(v+1)=0) has equal roots, which statement is correct?
#quadratic-roots
#equal-roots
#general-form
A (u+v=0)
B (uv=1)
C (u=v)
D (u=-v)
Explanation opens after your attempt
Step 1
Concept
The roots of this equation are (u+1) and (v+1). For equal roots, (u+1=v+1), so (u=v).
Step 2
Why this answer is correct
The correct answer is C. (u=v). The roots of this equation are (u+1) and (v+1). For equal roots, (u+1=v+1), so (u=v).
Step 3
Exam Tip
इस समीकरण की जड़ें (u+1) और (v+1) हैं। समान जड़ों के लिए (u+1=v+1), इसलिए (u=v)।
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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 (x-2 -2(a-4)x+a-2 -20=0) की जड़ें समान हैं, तो (a) का मान क्या होगा?
If (x-2 -2(a-4)x+a-2 -20=0) has equal roots, what is the value of (a)?
#quadratic-roots
#equal-roots
#discriminant
A \(\frac{7}{2}\)
B \(\frac{9}{2}\)
C \(\frac{11}{2}\)
D (5)
Explanation opens after your attempt
Correct Answer
B. \(\frac{9}{2}\)
Step 1
Concept
For equal roots, (D=0). From (4(a-4)2 -4\(a^2-20\)=0), we get \(a=\frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{9}{2}\). For equal roots, (D=0). From (4(a-4)2 -4\(a^2-20\)=0), we get \(a=\frac{9}{2}\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) होता है। (4(a-4)2 -4\(a^2-20\)=0) से \(a=\frac{9}{2}\) मिलता है।
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यदि (x-2 -(u+v)x+uv=0) की जड़ें बराबर हैं, तो (u) और (v) के बारे में सही कथन क्या है?
If the roots of (x-2 -(u+v)x+uv=0) are equal, what is the correct statement about (u) and (v)?
#quadratic-roots
#equal-roots
#general-form
A (u=v)
B (u=-v)
C (uv=0)
D (u+v=0)
Explanation opens after your attempt
Step 1
Concept
The roots of this equation are (u) and (v). For equal roots, it is necessary that (u=v).
Step 2
Why this answer is correct
The correct answer is A. (u=v). The roots of this equation are (u) and (v). For equal roots, it is necessary that (u=v).
Step 3
Exam Tip
इस समीकरण की जड़ें (u) और (v) हैं। जड़ें बराबर होने के लिए (u=v) होना जरूरी है।
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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 -2(k+2)x+k-2 +5=0) की जड़ें समान हैं, तो (k) का मान क्या होगा?
If (x-2 -2(k+2)x+k-2 +5=0) has equal roots, what is the value of (k)?
#quadratic-roots
#equal-roots
#discriminant
A \(\frac{1}{4}\)
B (4)
C -\(\frac{1}{4}\)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{4}\)
Step 1
Concept
For equal roots, put (D=0). Simplifying (4(k+2)2 -4\(k^2+5\)=0) gives (4k-1=0), so \(k=\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{4}\). For equal roots, put (D=0). Simplifying (4(k+2)2 -4\(k^2+5\)=0) gives (4k-1=0), so \(k=\frac{1}{4}\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। (4(k+2)2 -4\(k^2+5\)=0) से (4k+3=0) नहीं, बल्कि (4k-1=0) नहीं; सही सरलीकरण (4k-1=0) देता है, इसलिए \(k=\frac{1}{4}\)।
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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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\(x^2+kx+16=0\) 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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\(2x^2+\lambda x+8=0\) की जड़ें समान हों, तो \(\lambda\) के मान क्या होंगे?
If \(2x^2+\lambda x+8=0\) has equal roots, what are the values of \(\lambda\)?
#quadratic-roots
#equal-roots
#lambda
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, put (D=0). From \(\lambda^2-64=0\), we get \(\lambda=\pm8\).
Step 2
Why this answer is correct
The correct answer is A. (8) और (-8) / (8) and (-8). For equal roots, put (D=0). From \(\lambda^2-64=0\), we get \(\lambda=\pm8\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। \(\lambda^2-64=0\) से \(\lambda=\pm8\) 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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(kx-2 -2(k+1)x+(k+4)=0) की जड़ें समान हों, तो (k) का मान क्या है?
For (kx-2 -2(k+1)x+(k+4)=0) to have equal roots, what is the value of (k)?
#quadratic-roots
#equal-roots
#discriminant
A \(\frac{1}{2}\)
B \(-\frac{1}{2}\)
C (1)
D (2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
For equal roots, the discriminant (D=0). Since (D=4(1-2k)), we get \(k=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). For equal roots, the discriminant (D=0). Since (D=4(1-2k)), we get \(k=\frac{1}{2}\).
Step 3
Exam Tip
समान जड़ों के लिए विविक्तकर (D=0) होता है। (D=4(1-2k)) रखने पर \(k=\frac{1}{2}\) आता है।
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किस विकल्प में मूलों का योग ऋणात्मक और गुणनफल ऋणात्मक होगा?
In which option will the sum of roots be negative and the product of roots be negative?
#quadratic-equations
#sum-product
#signs
#expert
A \(x^2+5x-24=0\)
B \(x^2-5x-24=0\)
C \(x^2+5x+24=0\)
D \(x^2-5x+24=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x-24=0\)
Step 1
Concept
In the first option, the sum is \(-\frac{b}{a}=-5\) and the product is \(\frac{c}{a}=-24\). So both are negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x-24=0\). In the first option, the sum is \(-\frac{b}{a}=-5\) and the product is \(\frac{c}{a}=-24\). So both are negative.
Step 3
Exam Tip
पहले विकल्प में योग \(-\frac{b}{a}=-5\) और गुणनफल \(\frac{c}{a}=-24\) है। इसलिए दोनों ऋणात्मक हैं।
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समीकरण \(x^2+px+49=0\) के मूल समान और ऋणात्मक हैं। (p) का मान क्या होगा?
The roots of \(x^2+px+49=0\) are equal and negative. What is the value of (p)?
#quadratic-equations
#equal-negative-roots
#parameter
#expert
A (14)
B (-14)
C (7)
D (-7)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-196=0\) gives \(p=\pm14\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=14).
Step 2
Why this answer is correct
The correct answer is A. (14). For equal roots, \(p^2-196=0\) gives \(p=\pm14\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=14).
Step 3
Exam Tip
समान मूलों के लिए \(p^2-196=0\) से \(p=\pm14\) मिलता है। ऋणात्मक समान मूल के लिए \(-\frac{p}{2}<0\), इसलिए (p=14)।
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किस समीकरण में मूल बराबर और ऋणात्मक होंगे?
Which equation will have equal and negative roots?
#quadratic-equations
#equal-negative-roots
#perfect-square
#hard
A \(x^2+16x+64=0\)
B \(x^2-16x+64=0\)
C \(x^2-64=0\)
D \(x^2+64=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+16x+64=0\)
Step 1
Concept
(x-2 +16x+64=(x+8)2 ), so both roots are (-8). Both roots are equal and negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+16x+64=0\). (x-2 +16x+64=(x+8)2 ), so both roots are (-8). Both roots are equal and negative.
Step 3
Exam Tip
(x-2 +16x+64=(x+8)2 ), इसलिए दोनों मूल (-8) हैं। दोनों मूल बराबर और ऋणात्मक हैं।
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समीकरण \(x^2+px+9=0\) के मूल समान और ऋणात्मक हैं। (p) का मान क्या होगा?
The roots of \(x^2+px+9=0\) are equal and negative. What is the value of (p)?
#quadratic-equations
#equal-negative-roots
#parameter
#hard
A (6)
B (-6)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-36=0\) gives \(p=\pm6\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal roots, \(p^2-36=0\) gives \(p=\pm6\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=6).
Step 3
Exam Tip
समान मूलों के लिए \(p^2-36=0\) से \(p=\pm6\) मिलता है। ऋणात्मक समान मूल के लिए \(-\frac{p}{2}<0\), इसलिए (p=6)।
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समीकरण \(x^2+kx+81=0\) में यदि मूल समान और ऋणात्मक हैं, तो (k) का संभव मान कौन-सा है?
In \(x^2+kx+81=0\), if the roots are equal and negative, which possible value of (k) is correct?
#quadratic-equations
#equal-negative-roots
#parameter
#medium
A (18)
B (-18)
C (9)
D (-9)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(k^2=324\), and the equal root is \(-\frac{k}{2}\). For a negative root, (k=18) is correct.
Step 2
Why this answer is correct
The correct answer is A. (18). For equal roots, \(k^2=324\), and the equal root is \(-\frac{k}{2}\). For a negative root, (k=18) is correct.
Step 3
Exam Tip
समान मूलों के लिए \(k^2=324\) और समान मूल \(-\frac{k}{2}\) होगा। ऋणात्मक मूल के लिए (k=18) सही है।
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समीकरण \(x^2+kx+25=0\) में यदि मूल समान और ऋणात्मक हैं, तो (k) का संभव मान कौन-सा है?
In \(x^2+kx+25=0\), if the roots are equal and negative, which possible value of (k) is correct?
#quadratic-equations
#equal-negative-roots
#parameter
#medium
A (10)
B (-10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
For equal roots, (D=0) gives \(k^2=100\), and for equal negative roots \(-\frac{k}{2}<0\) is needed. Hence (k=10) is correct.
Step 2
Why this answer is correct
The correct answer is A. (10). For equal roots, (D=0) gives \(k^2=100\), and for equal negative roots \(-\frac{k}{2}<0\) is needed. Hence (k=10) is correct.
Step 3
Exam Tip
समान मूलों के लिए (D=0) से \(k^2=100\), और ऋणात्मक समान मूल के लिए \(-\frac{k}{2}<0\) चाहिए। इसलिए (k=10) सही है।
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सामान्य द्विघात समीकरण \(ax^2+bx+c=0\) में जड़ें समान हों, तो सही संबंध कौन-सा है?
For the general quadratic equation \(ax^2+bx+c=0\), which relation is correct when the roots are equal?
#quadratic-roots
#equal-roots
#standard-condition
A \(b^2=4ac\)
B \(b^2>4ac\)
C \(b^2<4ac\)
D (a+b+c=0)
Explanation opens after your attempt
Correct Answer
A. \(b^2=4ac\)
Step 1
Concept
For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 2
Why this answer is correct
The correct answer is A. \(b^2=4ac\). For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 3
Exam Tip
समान वास्तविक जड़ों के लिए विविक्तकर \(D=b^2-4ac=0\) होता है। इसलिए \(b^2=4ac\) सही संबंध है।
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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 \(4x^2-3x+k=0\) 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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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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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 \(x^2+px+18=0\) के मूलों का अनुपात (1:2) है और दोनों ऋणात्मक हैं तो (p) क्या होगा?
If roots of \(x^2+px+18=0\) are in the ratio (1:2) and both are negative, what is (p)?
#roots
#ratio_of_roots
#negative_roots
A (9)
B (-9)
C (6)
D (-6)
Explanation opens after your attempt
Step 1
Concept
The roots are (-3) and (-6) because the product is (18). Their sum is (-9), so (p=9).
Step 2
Why this answer is correct
The correct answer is A. (9). The roots are (-3) and (-6) because the product is (18). Their sum is (-9), so (p=9).
Step 3
Exam Tip
मूल (-3) और (-6) होंगे क्योंकि गुणनफल (18) है। उनका योग (-9) है इसलिए (p=9) है।
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यदि \(x^2+px+12=0\) के मूलों का अनुपात (1:3) है और दोनों ऋणात्मक हैं तो (p) क्या होगा?
If roots of \(x^2+px+12=0\) are in the ratio (1:3) and both are negative, what is (p)?
#roots
#ratio_of_roots
#negative_roots
A (8)
B (-8)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
The roots are (-2) and (-6) because the product is (12). Their sum is (-8), so (-p=-8) gives (p=8).
Step 2
Why this answer is correct
The correct answer is A. (8). The roots are (-2) and (-6) because the product is (12). Their sum is (-8), so (-p=-8) gives (p=8).
Step 3
Exam Tip
मूल (-2) और (-6) होंगे क्योंकि गुणनफल (12) है। उनका योग (-8) है इसलिए (-p=-8) से (p=8) है।
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यदि किसी द्विघात समीकरण के मूलों का गुणनफल ऋणात्मक है तो सही कथन कौन सा है?
If the product of roots of a quadratic equation is negative, which statement is correct?
#roots
#sign_of_roots
#reasoning
A एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative
B दोनों मूल धनात्मक हैं / Both roots are positive
C दोनों मूल ऋणात्मक हैं / Both roots are negative
D दोनों मूल बराबर हैं / Both roots are equal
Explanation opens after your attempt
Correct Answer
A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative
Step 1
Concept
A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
Step 2
Why this answer is correct
The correct answer is A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative. A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी होता है जब मूलों के चिन्ह विपरीत हों। इसलिए एक मूल धनात्मक और दूसरा ऋणात्मक होगा।
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यदि किसी द्विघात का विविक्तकर \(D=49-16h^2\) है, तो समान मूलों के लिए (h) के मान क्या होंगे?
If a quadratic has discriminant \(D=49-16h^2\), what are the values of (h) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A \(h=\frac{7}{4}\) या \(h=-\frac{7}{4}\) / \(h=\frac{7}{4}\) or \(h=-\frac{7}{4}\)
B (h=7) या (h=-7) / (h=7) or (h=-7)
C (h=4) या (h=-4) / (h=4) or (h=-4)
D (h=0)
Explanation opens after your attempt
Correct Answer
A. \(h=\frac{7}{4}\) या \(h=-\frac{7}{4}\) / \(h=\frac{7}{4}\) or \(h=-\frac{7}{4}\)
Step 1
Concept
For equal roots \(49-16h^2=0\) is needed. This gives \(h^2=\frac{49}{16}\), hence \(h=\pm\frac{7}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(h=\frac{7}{4}\) या \(h=-\frac{7}{4}\) / \(h=\frac{7}{4}\) or \(h=-\frac{7}{4}\). For equal roots \(49-16h^2=0\) is needed. This gives \(h^2=\frac{49}{16}\), hence \(h=\pm\frac{7}{4}\).
Step 3
Exam Tip
समान मूलों के लिए \(49-16h^2=0\) चाहिए। इससे \(h^2=\frac{49}{16}\), अतः \(h=\pm\frac{7}{4}\)।
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यदि (D=(m-8 )2 ) है, तो समान मूलों के लिए (m) का मान क्या होगा?
If (D=(m-8 )2 ), what is the value of (m) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (m=8)
B (m=-8)
C (m=0)
D हर (m) / Every (m)
Explanation opens after your attempt
Step 1
Concept
Equal roots need (D=0). From ((m-8 )2 =0), (m=8).
Step 2
Why this answer is correct
The correct answer is A. (m=8). Equal roots need (D=0). From ((m-8 )2 =0), (m=8).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। ((m-8 )2 =0) से (m=8)।
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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 \(x^2-2\theta x+3\theta=0\) 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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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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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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यदि किसी द्विघात का विविक्तकर \(D=36-9h^2\) है, तो समान मूलों के लिए (h) के मान क्या होंगे?
If a quadratic has discriminant \(D=36-9h^2\), what are the values of (h) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (h=2) या (h=-2) / (h=2) or (h=-2)
B (h=6) या (h=-6) / (h=6) or (h=-6)
C (h=3) या (h=-3) / (h=3) or (h=-3)
D (h=0)
Explanation opens after your attempt
Correct Answer
A. (h=2) या (h=-2) / (h=2) or (h=-2)
Step 1
Concept
For equal roots \(36-9h^2=0\) is needed. This gives \(h^2=4\), hence \(h=\pm2\).
Step 2
Why this answer is correct
The correct answer is A. (h=2) या (h=-2) / (h=2) or (h=-2). For equal roots \(36-9h^2=0\) is needed. This gives \(h^2=4\), hence \(h=\pm2\).
Step 3
Exam Tip
समान मूलों के लिए \(36-9h^2=0\) चाहिए। इससे \(h^2=4\), अतः \(h=\pm2\)।
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यदि (D=(w+2)2 ) है, तो समान मूलों के लिए (w) का मान क्या होगा?
If (D=(w+2)2 ), what is the value of (w) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (w=-2)
B (w=2)
C (w=0)
D हर (w) / Every (w)
Explanation opens after your attempt
Step 1
Concept
Equal roots need (D=0). From ((w+2)2 =0), (w=-2).
Step 2
Why this answer is correct
The correct answer is A. (w=-2). Equal roots need (D=0). From ((w+2)2 =0), (w=-2).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। ((w+2)2 =0) से (w=-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 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 \(x^2-2\mu x+2\mu=0\) 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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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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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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यदि किसी द्विघात का विविक्तकर \(D=25-4q^2\) है, तो समान मूलों के लिए (q) के मान क्या होंगे?
If a quadratic has discriminant \(D=25-4q^2\), what are the values of (q) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A \(q=\frac{5}{2}\) या \(q=-\frac{5}{2}\) / \(q=\frac{5}{2}\) or \(q=-\frac{5}{2}\)
B (q=5) या (q=-5) / (q=5) or (q=-5)
C (q=2) या (q=-2) / (q=2) or (q=-2)
D (q=0)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{5}{2}\) या \(q=-\frac{5}{2}\) / \(q=\frac{5}{2}\) or \(q=-\frac{5}{2}\)
Step 1
Concept
For equal roots \(25-4q^2=0\) is needed. This gives \(q^2=\frac{25}{4}\), hence \(q=\pm\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{5}{2}\) या \(q=-\frac{5}{2}\) / \(q=\frac{5}{2}\) or \(q=-\frac{5}{2}\). For equal roots \(25-4q^2=0\) is needed. This gives \(q^2=\frac{25}{4}\), hence \(q=\pm\frac{5}{2}\).
Step 3
Exam Tip
समान मूलों के लिए \(25-4q^2=0\) होना चाहिए। इससे \(q^2=\frac{25}{4}\), अतः \(q=\pm\frac{5}{2}\)।
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यदि (D=(s-2 )2 ) है, तो समान मूलों के लिए (s) का मान क्या होगा?
If (D=(s-2 )2 ), what is the value of (s) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (s=2)
B (s=-2)
C (s=0)
D हर (s) / Every (s)
Explanation opens after your attempt
Step 1
Concept
Equal roots need (D=0). From ((s-2 )2 =0), we get (s=2).
Step 2
Why this answer is correct
The correct answer is A. (s=2). Equal roots need (D=0). From ((s-2 )2 =0), we get (s=2).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। ((s-2 )2 =0) से (s=2) मिलता है।
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निम्न में से किस समीकरण के मूल समान हैं?
Which of the following equations has equal roots?
#quadratic-equations
#choose-equation
#equal-roots
A \(7x^2-10\sqrt{7}x+25=0\)
B \(7x^2-9\sqrt{7}x+25=0\)
C \(7x^2-8\sqrt{7}x+25=0\)
D \(7x^2-6\sqrt{7}x+25=0\)
Explanation opens after your attempt
Correct Answer
A. \(7x^2-10\sqrt{7}x+25=0\)
Step 1
Concept
In option (A), (D=\(-10\sqrt{7}\)2 -4(7)(25)=0). Equal roots need (D=0).
Step 2
Why this answer is correct
The correct answer is A. \(7x^2-10\sqrt{7}x+25=0\). In option (A), (D=\(-10\sqrt{7}\)2 -4(7)(25)=0). Equal roots need (D=0).
Step 3
Exam Tip
विकल्प (A) में (D=\(-10\sqrt{7}\)2 -4(7)(25)=0) है। समान मूलों के लिए (D=0) चाहिए।
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समीकरण (3x-2 -2(2k+1)x+(k+1)2 =0) में समान मूलों के लिए (k) के मान क्या होंगे?
For equal roots in (3x-2 -2(2k+1)x+(k+1)2 =0), what are the values of (k)?
#quadratic-equations
#equal-roots
#parameter
A (k=1) या (k=-2) / (k=1) or (k=-2)
B (k=2) या (k=-1) / (k=2) or (k=-1)
C (k=0) या (k=-3) / (k=0) or (k=-3)
D (k=3) या (k=-1) / (k=3) or (k=-1)
Explanation opens after your attempt
Correct Answer
A. (k=1) या (k=-2) / (k=1) or (k=-2)
Step 1
Concept
Here (D=4(2k+1)2 -12(k+1)2 =4(k-1)(k+2)). From (D=0), (k=1) or (k=-2).
Step 2
Why this answer is correct
The correct answer is A. (k=1) या (k=-2) / (k=1) or (k=-2). Here (D=4(2k+1)2 -12(k+1)2 =4(k-1)(k+2)). From (D=0), (k=1) or (k=-2).
Step 3
Exam Tip
यहाँ (D=4(2k+1)2 -12(k+1)2 =4(k-1)(k+2)) है। (D=0) से (k=1) या (k=-2)।
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यदि (x-2 +2(m-2 )x+\(m^2-3m+4\)=0) के मूल समान हों, तो (m) का मान क्या होगा?
If (x-2 +2(m-2 )x+\(m^2-3m+4\)=0) has equal roots, what is the value of (m)?
#quadratic-equations
#equal-roots
#parameter
A (0)
B (1)
C (2)
D (4)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(m-2 )2 -4\(m^2-3m+4\)=-4m). From (D=0), (m=0).
Step 2
Why this answer is correct
The correct answer is A. (0). Here (D=4(m-2 )2 -4\(m^2-3m+4\)=-4m). From (D=0), (m=0).
Step 3
Exam Tip
यहाँ (D=4(m-2 )2 -4\(m^2-3m+4\)=-4m) है। (D=0) से (m=0) मिलता है।
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समीकरण \(x^2-2\lambda x+\lambda=0\) के समान मूलों के लिए \(\lambda\) के मान कौन से हैं?
What are the values of \(\lambda\) for equal roots of \(x^2-2\lambda x+\lambda=0\)?
#quadratic-equations
#equal-roots
#parameter
A \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\)
B केवल \(\lambda=1\) / Only \(\lambda=1\)
C केवल \(\lambda=0\) / Only \(\lambda=0\)
D \(\lambda=-1\) या \(\lambda=1\) / \(\lambda=-1\) or \(\lambda=1\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\)
Step 1
Concept
For equal roots (D=4\lambda\(\lambda-1\)=0). Therefore \(\lambda=0\) or \(\lambda=1\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\). For equal roots (D=4\lambda\(\lambda-1\)=0). Therefore \(\lambda=0\) or \(\lambda=1\).
Step 3
Exam Tip
समान मूलों के लिए (D=4\lambda\(\lambda-1\)=0) है। इसलिए \(\lambda=0\) या \(\lambda=1\)।
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यदि समीकरण (kx-2 -2(k+1)x+(k+3)=0) में \(k\neq0\) हो, तो समान मूलों के लिए (k) का मान क्या होगा?
If \(k\neq0\) in (kx-2 -2(k+1)x+(k+3)=0), what is the value of (k) for equal roots?
#quadratic-equations
#parameter
#equal-roots
A (k=1)
B (k=-1)
C (k=3)
D (k=-3)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=4(k+1)2 -4k(k+3)=0). Simplifying gives (1-k=0), so (k=1).
Step 2
Why this answer is correct
The correct answer is A. (k=1). For equal roots (D=4(k+1)2 -4k(k+3)=0). Simplifying gives (1-k=0), so (k=1).
Step 3
Exam Tip
समान मूलों के लिए (D=4(k+1)2 -4k(k+3)=0) है। सरल करने पर (1-k=0), इसलिए (k=1)।
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