यदि \(x^2-6x+8=0\) की जड़ें \(\alpha,\beta\) हैं, तो (\(\alpha-2\)\(\beta-2\)) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-6x+8=0\), what is (\(\alpha-2\)\(\beta-2\))?
#quadratic-roots
#root-expression
#sum-product
A (0)
B (2)
C (4)
D (8)
Explanation opens after your attempt
Step 1
Concept
(\(\alpha-2\)\(\beta-2\)=\alpha\beta-2\(\alpha+\beta\)+4). Since \(\alpha+\beta=6\) and \(\alpha\beta=8\), the value is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). (\(\alpha-2\)\(\beta-2\)=\alpha\beta-2\(\alpha+\beta\)+4). Since \(\alpha+\beta=6\) and \(\alpha\beta=8\), the value is (0).
Step 3
Exam Tip
(\(\alpha-2\)\(\beta-2\)=\alpha\beta-2\(\alpha+\beta\)+4) है। \(\alpha+\beta=6\) और \(\alpha\beta=8\), इसलिए मान (0) है।
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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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\(x^2+3x-18=0\) की जड़ों का सही युग्म कौन-सा है?
Which is the correct pair of roots of \(x^2+3x-18=0\)?
#quadratic-roots
#factorisation
#root-pair
A (3) और (-6) / (3) and (-6)
B (-3) और (6) / (-3) and (6)
C (2) और (-9) / (2) and (-9)
D (-2) और (9) / (-2) and (9)
Explanation opens after your attempt
Correct Answer
A. (3) और (-6) / (3) and (-6)
Step 1
Concept
The numbers with product (-18) and sum (-3) are (3) and (-6). Hence this is the correct pair of roots.
Step 2
Why this answer is correct
The correct answer is A. (3) और (-6) / (3) and (-6). The numbers with product (-18) and sum (-3) are (3) and (-6). Hence this is the correct pair of roots.
Step 3
Exam Tip
गुणनफल (-18) और योग (-3) वाली संख्याएँ (3) और (-6) हैं। इसलिए यही जड़ों का सही युग्म है।
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\(4x^2-4x+1=0\) की दोनों जड़ों का मान क्या है?
What is the value of both roots of \(4x^2-4x+1=0\)?
#quadratic-roots
#equal-roots
#perfect-square
A \(\frac{1}{2}\)
B \(-\frac{1}{2}\)
C (1)
D (-1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
(4x-2 -4x+1=(2x-1)2 ). Therefore the equal root is \(x=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). (4x-2 -4x+1=(2x-1)2 ). Therefore the equal root is \(x=\frac{1}{2}\).
Step 3
Exam Tip
(4x-2 -4x+1=(2x-1)2 ) है। इसलिए समान जड़ \(x=\frac{1}{2}\) है।
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यदि (x=0), \(ax^2+bx+c=0\) की जड़ है, तो कौन-सी शर्त निश्चित रूप से सही है?
If (x=0) is a root of \(ax^2+bx+c=0\), which condition must be true?
#quadratic-roots
#zero-root
#root-verification
A (c=0)
B (b=0)
C (a=0)
D (a+b=0)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) gives (c=0). Thus the direct condition for zero to be a root is (c=0).
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) gives (c=0). Thus the direct condition for zero to be a root is (c=0).
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य जड़ होने की सीधी शर्त (c=0) 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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यदि \(x^2-4x+2=0\) की जड़ें \(\alpha,\beta\) हैं, तो (\(\alpha+2\)\(\beta+2\)) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-4x+2=0\), what is (\(\alpha+2\)\(\beta+2\))?
#quadratic-roots
#root-expression
#sum-product
A (10)
B (12)
C (14)
D (16)
Explanation opens after your attempt
Step 1
Concept
(\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4). Since \(\alpha+\beta=4\) and \(\alpha\beta=2\), the value is (14).
Step 2
Why this answer is correct
The correct answer is C. (14). (\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4). Since \(\alpha+\beta=4\) and \(\alpha\beta=2\), the value is (14).
Step 3
Exam Tip
(\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4) है। \(\alpha+\beta=4\) और \(\alpha\beta=2\), इसलिए मान (14) है।
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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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\(x^2+2x+c=0\) की जड़ें वास्तविक हैं और जड़ों का गुणनफल उनके योग से कम है, तो (c) पर सही शर्त क्या है?
For \(x^2+2x+c=0\), the roots are real and their product is less than their sum. What is the correct condition on (c)?
#quadratic-roots
#product-sum
#inequality
A (c<-2)
B \(c\le1\)
C (c> -2)
D \(-2<c\le1\)
Explanation opens after your attempt
Step 1
Concept
The sum is (-2) and the product is (c), so (c<-2) is needed. This condition also satisfies (D=4-4c>0).
Step 2
Why this answer is correct
The correct answer is A. (c<-2). The sum is (-2) and the product is (c), so (c<-2) is needed. This condition also satisfies (D=4-4c>0).
Step 3
Exam Tip
योग (-2) और गुणनफल (c) है, इसलिए (c<-2) चाहिए। यह शर्त (D=4-4c>0) को भी पूरा करती है।
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\(x^2-10x+k=0\) की जड़ें भिन्न अभाज्य संख्याएँ हैं, तो (k) का मान क्या है?
The roots of \(x^2-10x+k=0\) are distinct prime numbers. What is (k)?
#quadratic-roots
#prime-roots
#integer-roots
A (21)
B (25)
C (16)
D (10)
Explanation opens after your attempt
Step 1
Concept
The distinct prime roots with sum (10) are (3) and (7). Their product is (21), so (k=21).
Step 2
Why this answer is correct
The correct answer is A. (21). The distinct prime roots with sum (10) are (3) and (7). Their product is (21), so (k=21).
Step 3
Exam Tip
योग (10) वाली भिन्न अभाज्य जड़ें (3) और (7) हैं। उनका गुणनफल (21) है, इसलिए (k=21)।
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(x-2 -(a+3)x+3a=0) के बारे में कौन-सा कथन हमेशा सही है?
Which statement is always true about (x-2 -(a+3)x+3a=0)?
#quadratic-roots
#root-verification
#parameter
A (3) हमेशा एक जड़ है / (3) is always a root
B (a) कभी जड़ नहीं है / (a) is never a root
C दोनों जड़ें हमेशा समान हैं / Both roots are always equal
D कोई वास्तविक जड़ नहीं है / There is no real root
Explanation opens after your attempt
Correct Answer
A. (3) हमेशा एक जड़ है / (3) is always a root
Step 1
Concept
Putting (x=3) gives (9-3(a+3)+3a=0). Hence (3) is always one root, and the other root is (a).
Step 2
Why this answer is correct
The correct answer is A. (3) हमेशा एक जड़ है / (3) is always a root. Putting (x=3) gives (9-3(a+3)+3a=0). Hence (3) is always one root, and the other root is (a).
Step 3
Exam Tip
(x=3) रखने पर (9-3(a+3)+3a=0) मिलता है। इसलिए (3) हमेशा एक जड़ है और दूसरी जड़ (a) होती है।
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यदि \(x^2+mx+n=0\) की जड़ें \(\alpha,\beta\) हैं, \(\alpha-\beta=4\) और \(\alpha\beta=5\), तो \(m^2\) क्या है?
If \(\alpha,\beta\) are roots of \(x^2+mx+n=0\), \(\alpha-\beta=4\), and \(\alpha\beta=5\), what is \(m^2\)?
#quadratic-roots
#difference-of-roots
#coefficient
A (16)
B (25)
C (36)
D (49)
Explanation opens after your attempt
Step 1
Concept
Use (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta). Since (16=\(\alpha+\beta\)2 -20), we get (\(\alpha+\beta\)2 =36) and \(m^2=36\).
Step 2
Why this answer is correct
The correct answer is C. (36). Use (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta). Since (16=\(\alpha+\beta\)2 -20), we get (\(\alpha+\beta\)2 =36) and \(m^2=36\).
Step 3
Exam Tip
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta) लगाएँ। (16=\(\alpha+\beta\)2 -20), इसलिए (\(\alpha+\beta\)2 =36) और \(m^2=36\)।
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यदि किसी द्विघात समीकरण की जड़ें (3) और (-2) हैं और \(x^2\) का गुणांक (2) है, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are (3) and (-2), and the coefficient of \(x^2\) is (2), which is the equation?
#quadratic-roots
#forming-equation
#leading-coefficient
A \(2x^2-2x-12=0\)
B \(2x^2+2x-12=0\)
C \(2x^2-2x+12=0\)
D \(x^2-x-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-2x-12=0\)
Step 1
Concept
The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-12=0\). The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 3
Exam Tip
मॉनिक समीकरण \(x^2-x-6=0\) है। \(x^2\) का गुणांक (2) चाहिए, इसलिए पूरे समीकरण को (2) से गुणा करें।
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यदि \(x^2-5x+1=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2+\beta^2\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-5x+1=0\), what is \(\alpha^2+\beta^2\)?
#quadratic-roots
#square-sum
#sum-product
A (21)
B (22)
C (23)
D (24)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha\beta=1\). Thus \(\alpha^2+\beta^2=25-2=23\).
Step 2
Why this answer is correct
The correct answer is C. (23). Here \(\alpha+\beta=5\) and \(\alpha\beta=1\). Thus \(\alpha^2+\beta^2=25-2=23\).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=1\) है। इसलिए \(\alpha^2+\beta^2=25-2=23\)।
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(x-2 -2(k+1)x+k-2 =0) की जड़ें वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
For (x-2 -2(k+1)x+k-2 =0) to have real and distinct roots, what is the correct condition on (k)?
#quadratic-roots
#distinct-roots
#discriminant
A \(k>-\frac{1}{2}\)
B \(k\ge-\frac{1}{2}\)
C \(k<-\frac{1}{2}\)
D \(k\le-\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(k>-\frac{1}{2}\)
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (D=4(2k+1)), so \(k>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k>-\frac{1}{2}\). For real and distinct roots, (D>0) is needed. Here (D=4(2k+1)), so \(k>-\frac{1}{2}\).
Step 3
Exam Tip
वास्तविक और भिन्न जड़ों के लिए (D>0) चाहिए। यहाँ (D=4(2k+1)), इसलिए \(k>-\frac{1}{2}\)।
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यदि \(x^2+px+q=0\) 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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\(\frac{1}{2}\) और \(\frac{3}{4}\) जड़ों वाला द्विघात समीकरण कौन-सा है?
Which quadratic equation has roots \(\frac{1}{2}\) and \(\frac{3}{4}\)?
#quadratic-roots
#forming-equation
#fractional-roots
A \(8x^2-10x+3=0\)
B \(8x^2+10x+3=0\)
C \(4x^2-10x+3=0\)
D \(8x^2-6x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(8x^2-10x+3=0\)
Step 1
Concept
The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 2
Why this answer is correct
The correct answer is A. \(8x^2-10x+3=0\). The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 3
Exam Tip
जड़ों का योग \(\frac{5}{4}\) और गुणनफल \(\frac{3}{8}\) है। समीकरण \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) को (8) से गुणा करें।
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यदि \(x^2+ax+12=0\) की एक जड़ दूसरी जड़ से (3) अधिक है, तो (a) के संभव मान क्या हैं?
If one root of \(x^2+ax+12=0\) is (3) more than the other root, what are the possible values of (a)?
#quadratic-roots
#difference-of-roots
#parameter
A \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
B \(\sqrt{21}\) और \(-\sqrt{21}\) / \(\sqrt{21}\) and \(-\sqrt{21}\)
C (3) और (-3) / (3) and (-3)
D (12) और (-12) / (12) and (-12)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
Step 1
Concept
Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\). Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 3
Exam Tip
जड़ें (r) और (r+3) मानने पर (r(r+3)=12) मिलता है। इससे जड़ों का योग \(\pm\sqrt{57}\) होता है, इसलिए \(a=\mp\sqrt{57}\)।
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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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यदि \(2x^2+3x-5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(2x^2+3x-5=0\), what is \(\alpha^2\beta+\alpha\beta^2\)?
#quadratic-roots
#root-expression
#sum-product
A \(\frac{15}{4}\)
B \(-\frac{15}{4}\)
C \(\frac{5}{4}\)
D \(-\frac{5}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{15}{4}\)
Step 1
Concept
(\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)). Since \(\alpha\beta=-\frac{5}{2}\) and \(\alpha+\beta=-\frac{3}{2}\), the value is \(\frac{15}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{15}{4}\). (\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)). Since \(\alpha\beta=-\frac{5}{2}\) and \(\alpha+\beta=-\frac{3}{2}\), the value is \(\frac{15}{4}\).
Step 3
Exam Tip
(\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)) होता है। \(\alpha\beta=-\frac{5}{2}\) और \(\alpha+\beta=-\frac{3}{2}\), इसलिए मान \(\frac{15}{4}\) है।
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(x-2 -2x+(p+3)=0) की वास्तविक जड़ें न हों, इसके लिए (p) पर सही शर्त क्या है?
For (x-2 -2x+(p+3)=0) to have no real roots, what is the correct condition on (p)?
#quadratic-roots
#no-real-roots
#inequality
A (p>-2)
B \(p\ge -2\)
C (p<-2)
D \(p\le -2\)
Explanation opens after your attempt
Step 1
Concept
For no real roots, (D<0) is required. Here (D=-4(p+2)), so (p>-2).
Step 2
Why this answer is correct
The correct answer is A. (p>-2). For no real roots, (D<0) is required. Here (D=-4(p+2)), so (p>-2).
Step 3
Exam Tip
वास्तविक जड़ें न होने के लिए (D<0) चाहिए। यहाँ (D=-4(p+2)), इसलिए (p>-2) है।
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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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(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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यदि \(x^2-7x+10=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-7x+10=0\), what is \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\)?
#quadratic-roots
#rational-expression
#sum-product
A \(\frac{5}{4}\)
B \(\frac{4}{5}\)
C \(\frac{7}{4}\)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{4}\)
Step 1
Concept
The denominator (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1=4). The numerator is \(\alpha+\beta-2=5\), so the value is \(\frac{5}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{4}\). The denominator (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1=4). The numerator is \(\alpha+\beta-2=5\), so the value is \(\frac{5}{4}\).
Step 3
Exam Tip
हर (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1=4) है। ऊपर \(\alpha+\beta-2=5\), इसलिए मान \(\frac{5}{4}\) 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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\(x^2-4x+k=0\) 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 किसी द्विघात समीकरण की जड़ें \(2+\sqrt{5}\) और \(2-\sqrt{5}\) हैं, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are \(2+\sqrt{5}\) and \(2-\sqrt{5}\), which is the equation?
#quadratic-roots
#forming-equation
#surd-roots
A \(x^2-4x-1=0\)
B \(x^2+4x-1=0\)
C \(x^2-4x+1=0\)
D \(x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-1=0\)
Step 1
Concept
\(The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-4x-1=0). The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 3
Exam Tip
जड़ों का योग (4) और गुणनफल (-1) है। \(समीकरण (x^2-(\)योग)x+गुणनफल\(=0) से उत्तर मिलता है\)।
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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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