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 \(x^2-5x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=17\), तो जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-5x+c=0\) and \(\alpha^2+\beta^2=17\), what are the roots?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (4) / (1) and (4)
B (2) और (3) / (2) and (3)
C (0) और (5) / (0) and (5)
D (-1) और (6) / (-1) and (6)
Explanation opens after your attempt
Correct Answer
A. (1) और (4) / (1) and (4)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 2
Why this answer is correct
The correct answer is A. (1) और (4) / (1) and (4). Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha^2+\beta^2=17\) है। \(25-2\alpha\beta=17\) से \(\alpha\beta=4\), इसलिए जड़ें (1) और (4) हैं।
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(x-2 -2x+\(a^2+2\)=0) की जड़ों की प्रकृति क्या है?
What is the nature of the roots of (x-2 -2x+\(a^2+2\)=0)?
#quadratic-roots
#nature-of-roots
#always-non-real
A दो वास्तविक भिन्न / Two real distinct
B समान वास्तविक / Equal real
C हर (a) के लिए वास्तविक नहीं / Not real for every (a)
D एक जड़ शून्य / One root is zero
Explanation opens after your attempt
Correct Answer
C. हर (a) के लिए वास्तविक नहीं / Not real for every (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is C. हर (a) के लिए वास्तविक नहीं / Not real for every (a). The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+2\)=-4\(a^2+1\)) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं हैं।
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यदि (3x-2 -(3h+1)x+h=0) की एक जड़ \(\frac{1}{3}\) है, तो दूसरी जड़ क्या है?
If one root of (3x-2 -(3h+1)x+h=0) is \(\frac{1}{3}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{3}\)
C (3h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{3}\) है। एक जड़ \(\frac{1}{3}\) है, इसलिए दूसरी जड़ (h) होगी।
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यदि \(x^2+px+q=0\) की जड़ें (-3) और (7) हैं, तो (p-q) का मान क्या है?
If the roots of \(x^2+px+q=0\) are (-3) and (7), what is the value of (p-q)?
#quadratic-roots
#coefficients
#forming-equation
A (13)
B (15)
C (17)
D (21)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (4), so (p=-4). The product is (-21), so (q=-21), hence (p-q=17).
Step 2
Why this answer is correct
The correct answer is C. (17). The sum of roots is (4), so (p=-4). The product is (-21), so (q=-21), hence (p-q=17).
Step 3
Exam Tip
जड़ों का योग (4) है, इसलिए (p=-4)। गुणनफल (-21) है, इसलिए (q=-21), अतः (p-q=17)।
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यदि \(x^2+8x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2 +3\alpha\beta) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2+8x+12=0\), what is (\(\alpha-\beta\)2 +3\alpha\beta)?
#quadratic-roots
#difference-expression
#sum-product
A (48)
B (50)
C (52)
D (56)
Explanation opens after your attempt
Step 1
Concept
Here (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta=64-48=16). Therefore (16+3(12)=52).
Step 2
Why this answer is correct
The correct answer is C. (52). Here (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta=64-48=16). Therefore (16+3(12)=52).
Step 3
Exam Tip
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta=64-48=16) है। इसलिए (16+3(12)=52)।
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यदि \(x^2-6x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2-5\alpha+\beta^2-5\beta\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-6x+5=0\), what is \(\alpha^2-5\alpha+\beta^2-5\beta\)?
#quadratic-roots
#root-expression
#sum-product
A (-6)
B (-4)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=6\) and \(\alpha\beta=5\). Since \(\alpha^2+\beta^2=26\), the value is (26-5\(\alpha+\beta\)=-4).
Step 2
Why this answer is correct
The correct answer is B. (-4). Here \(\alpha+\beta=6\) and \(\alpha\beta=5\). Since \(\alpha^2+\beta^2=26\), the value is (26-5\(\alpha+\beta\)=-4).
Step 3
Exam Tip
\(\alpha+\beta=6\) और \(\alpha\beta=5\) है। \(\alpha^2+\beta^2=26\), इसलिए (26-5\(\alpha+\beta\)=-4)।
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\(x^2+10x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+10x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(\lambda<0\)
B \(0<\lambda<25\)
C \(\lambda=25\)
D \(\lambda>25\)
Explanation opens after your attempt
Correct Answer
B. \(0<\lambda<25\)
Step 1
Concept
For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).
Step 2
Why this answer is correct
The correct answer is B. \(0<\lambda<25\). For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-10) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(100-4\lambda>0\), इसलिए \(0<\lambda<25\)।
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यदि (x-2 -(2a+7)x+(a+3)(a+4)=0) की जड़ें लगातार पूर्णांक रूप में हैं, तो वे कौन-सी हैं?
If the roots of (x-2 -(2a+7)x+(a+3)(a+4)=0) are in consecutive integer form, which are they?
#quadratic-roots
#consecutive-roots
#parametric
A (a+2) और (a+5) / (a+2) and (a+5)
B (a+3) और (a+4) / (a+3) and (a+4)
C (a) और (a+7) / (a) and (a+7)
D (2a+3) और (4) / (2a+3) and (4)
Explanation opens after your attempt
Correct Answer
B. (a+3) और (a+4) / (a+3) and (a+4)
Step 1
Concept
The sum of roots is (2a+7) and the product is ((a+3)(a+4)). These match (a+3) and (a+4).
Step 2
Why this answer is correct
The correct answer is B. (a+3) और (a+4) / (a+3) and (a+4). The sum of roots is (2a+7) and the product is ((a+3)(a+4)). These match (a+3) and (a+4).
Step 3
Exam Tip
जड़ों का योग (2a+7) और गुणनफल ((a+3)(a+4)) है। ये (a+3) और (a+4) से मिलते हैं।
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यदि \(3x^2-13x+4=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-\beta\) का धनात्मक मान क्या है?
If \(\alpha,\beta\) are roots of \(3x^2-13x+4=0\), what is the positive value of \(\alpha-\beta\)?
#quadratic-roots
#difference-of-roots
#numerical
A \(\frac{7}{3}\)
B \(\frac{11}{3}\)
C \(\frac{13}{3}\)
D (3)
Explanation opens after your attempt
Correct Answer
B. \(\frac{11}{3}\)
Step 1
Concept
Use (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta). With \(\alpha+\beta=\frac{13}{3}\) and \(\alpha\beta=\frac{4}{3}\), the positive difference is \(\frac{11}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{11}{3}\). Use (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta). With \(\alpha+\beta=\frac{13}{3}\) and \(\alpha\beta=\frac{4}{3}\), the positive difference is \(\frac{11}{3}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta) लगाएँ। \(\alpha+\beta=\frac{13}{3}\) और \(\alpha\beta=\frac{4}{3}\), इसलिए धनात्मक अंतर \(\frac{11}{3}\) है।
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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 \(x^2-12x+m=0\) की दोनों जड़ें अभाज्य संख्याएँ हैं, तो (m) का मान क्या है?
If both roots of \(x^2-12x+m=0\) are prime numbers, what is the value of (m)?
#quadratic-roots
#prime-roots
#integer-roots
A (30)
B (35)
C (40)
D (45)
Explanation opens after your attempt
Step 1
Concept
The prime roots with sum (12) are (5) and (7). Their product is (35), so (m=35).
Step 2
Why this answer is correct
The correct answer is B. (35). The prime roots with sum (12) are (5) and (7). Their product is (35), so (m=35).
Step 3
Exam Tip
योग (12) वाली अभाज्य जड़ें (5) और (7) हैं। उनका गुणनफल (35) है, इसलिए (m=35)।
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यदि \(x^2-3x+1=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^4+\beta^4\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-3x+1=0\), what is \(\alpha^4+\beta^4\)?
#quadratic-roots
#fourth-power
#root-expression
A (43)
B (45)
C (47)
D (49)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=3\) and \(\alpha\beta=1\). First \(\alpha^2+\beta^2=7\), then \(\alpha^4+\beta^4=7^2-2=47\).
Step 2
Why this answer is correct
The correct answer is C. (47). Here \(\alpha+\beta=3\) and \(\alpha\beta=1\). First \(\alpha^2+\beta^2=7\), then \(\alpha^4+\beta^4=7^2-2=47\).
Step 3
Exam Tip
\(\alpha+\beta=3\) और \(\alpha\beta=1\) है। पहले \(\alpha^2+\beta^2=7\), फिर \(\alpha^4+\beta^4=7^2-2=47\)।
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यदि \(x^2-6x+r=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2\beta+\alpha\beta^2=54\), तो (r) क्या है?
If \(\alpha,\beta\) are roots of \(x^2-6x+r=0\) and \(\alpha^2\beta+\alpha\beta^2=54\), what is (r)?
#quadratic-roots
#root-expression
#parameter
A (6)
B (8)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
We use (\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)). Here \(\alpha+\beta=6\), so (6r=54) and (r=9).
Step 2
Why this answer is correct
The correct answer is C. (9). We use (\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)). Here \(\alpha+\beta=6\), so (6r=54) and (r=9).
Step 3
Exam Tip
(\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)) होता है। यहाँ \(\alpha+\beta=6\), इसलिए (6r=54) और (r=9)।
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(x-2 -(u-v)x-uv=0) की जड़ें कौन-सी हैं?
What are the roots of (x-2 -(u-v)x-uv=0)?
#quadratic-roots
#general-roots
#sum-product
A (u) और (v) / (u) and (v)
B (u) और (-v) / (u) and (-v)
C (-u) और (v) / (-u) and (v)
D (u-v) और (uv) / (u-v) and (uv)
Explanation opens after your attempt
Correct Answer
B. (u) और (-v) / (u) and (-v)
Step 1
Concept
The sum of roots is (u-v) and the product is (-uv). These match (u) and (-v).
Step 2
Why this answer is correct
The correct answer is B. (u) और (-v) / (u) and (-v). The sum of roots is (u-v) and the product is (-uv). These match (u) and (-v).
Step 3
Exam Tip
जड़ों का योग (u-v) और गुणनफल (-uv) है। ये (u) और (-v) से मेल खाते हैं।
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यदि \(x^2-14x+q=0\) की जड़ें (3:4) के अनुपात में हैं, तो (q) का मान क्या होगा?
If the roots of \(x^2-14x+q=0\) are in the ratio (3:4), what is the value of (q)?
#quadratic-roots
#roots-ratio
#product
A (36)
B (42)
C (48)
D (56)
Explanation opens after your attempt
Step 1
Concept
Let the roots be (3r) and (4r). From (7r=14), (r=2), so the product is \(12r^2=48\).
Step 2
Why this answer is correct
The correct answer is C. (48). Let the roots be (3r) and (4r). From (7r=14), (r=2), so the product is \(12r^2=48\).
Step 3
Exam Tip
जड़ें (3r) और (4r) मानें। (7r=14) से (r=2), इसलिए गुणनफल \(12r^2=48\) है।
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यदि \(5x^2-4x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?
If the roots of \(5x^2-4x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?
#quadratic-roots
#non-real-roots
#discriminant
A \(\lambda<\frac{4}{5}\)
B \(\lambda=\frac{4}{5}\)
C \(\lambda>\frac{4}{5}\)
D \(\lambda\le\frac{4}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\lambda>\frac{4}{5}\)
Step 1
Concept
For non-real roots, (D<0) is required. From \(16-20\lambda<0\), we get \(\lambda>\frac{4}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(\lambda>\frac{4}{5}\). For non-real roots, (D<0) is required. From \(16-20\lambda<0\), we get \(\lambda>\frac{4}{5}\).
Step 3
Exam Tip
वास्तविक नहीं होने के लिए (D<0) चाहिए। \(16-20\lambda<0\) से \(\lambda>\frac{4}{5}\) मिलता है।
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यदि \(x^2-8x+n=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=40\), तो (n) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-8x+n=0\) and \(\alpha^2+\beta^2=40\), what is the value of (n)?
#quadratic-roots
#square-sum
#parameter
A (10)
B (12)
C (14)
D (16)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=8\) and \(\alpha\beta=n\). From \(\alpha^2+\beta^2=64-2n=40\), we get (n=12).
Step 2
Why this answer is correct
The correct answer is B. (12). Here \(\alpha+\beta=8\) and \(\alpha\beta=n\). From \(\alpha^2+\beta^2=64-2n=40\), we get (n=12).
Step 3
Exam Tip
\(\alpha+\beta=8\) और \(\alpha\beta=n\) है। \(\alpha^2+\beta^2=64-2n=40\) से (n=12) मिलता है।
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(x-2 -2(a-2)x+a-2 -9=0) की जड़ों का गुणनफल (0) हो, तो (a) के मान क्या हैं?
If the product of the roots of (x-2 -2(a-2)x+a-2 -9=0) is (0), what are the values of (a)?
#quadratic-roots
#zero-product
#parameter
A (0) और (9) / (0) and (9)
B (3) और (-3) / (3) and (-3)
C (2) और (-2) / (2) and (-2)
D कोई मान नहीं / No value
Explanation opens after your attempt
Correct Answer
B. (3) और (-3) / (3) and (-3)
Step 1
Concept
The product of roots is \(a^2-9\). Setting it equal to (0) gives \(a^2=9\), so (a=3) or (a=-3).
Step 2
Why this answer is correct
The correct answer is B. (3) और (-3) / (3) and (-3). The product of roots is \(a^2-9\). Setting it equal to (0) gives \(a^2=9\), so (a=3) or (a=-3).
Step 3
Exam Tip
जड़ों का गुणनफल \(a^2-9\) है। इसे (0) रखने पर \(a^2=9\), इसलिए (a=3) या (a=-3)।
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यदि \(x^2+ax+b=0\) की जड़ें \(\frac{1}{3+\sqrt{5}}\) और \(\frac{1}{3-\sqrt{5}}\) हैं, तो (a) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(\frac{1}{3+\sqrt{5}}\) and \(\frac{1}{3-\sqrt{5}}\), what is the value of (a)?
#quadratic-roots
#rationalisation
#surd-roots
A \(-\frac{3}{2}\)
B \(\frac{3}{2}\)
C (-3)
D (3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{3}{2}\)
Step 1
Concept
After rationalising, the sum of roots is \(\frac{3}{2}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\). After rationalising, the sum of roots is \(\frac{3}{2}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{3}{2}\).
Step 3
Exam Tip
रैशनलाइज करने पर जड़ों का योग \(\frac{3}{2}\) मिलता है। \(x^2+ax+b=0\) में जड़ों का योग (-a) होता है, इसलिए \(a=-\frac{3}{2}\)।
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यदि \(x^2-sx+1=0\) की जड़ें \(\tan\theta\) और \(\cot\theta\) हो सकती हैं, तो वास्तविक \(\theta\) के लिए (s) पर सही शर्त क्या है?
If the roots of \(x^2-sx+1=0\) can be \(\tan\theta\) and \(\cot\theta\), what is the correct condition on (s) for real \(\theta\)?
#quadratic-roots
#trigonometric-roots
#condition
A (-2<s<2)
B (s=0)
C \(s^2\ge4\)
D \(s^2<4\)
Explanation opens after your attempt
Correct Answer
C. \(s^2\ge4\)
Step 1
Concept
We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
Step 2
Why this answer is correct
The correct answer is C. \(s^2\ge4\). We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
Step 3
Exam Tip
\(\tan\theta\cdot\cot\theta=1\) और \(\tan\theta+\cot\theta=s\) है। वास्तविक मानों के लिए \(s^2-4\ge0\), इसलिए \(s^2\ge4\)।
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यदि \(x^2-11x+24=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-11x+24=0\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?
#quadratic-roots
#ratio-expression
#sum-product
A \(\frac{61}{24}\)
B \(\frac{67}{24}\)
C \(\frac{73}{24}\)
D \(\frac{79}{24}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{73}{24}\)
Step 1
Concept
We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=73\) and \(\alpha\beta=24\), so the value is \(\frac{73}{24}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{73}{24}\). We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=73\) and \(\alpha\beta=24\), so the value is \(\frac{73}{24}\).
Step 3
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\) है। यहाँ \(\alpha^2+\beta^2=73\) और \(\alpha\beta=24\), इसलिए मान \(\frac{73}{24}\) है।
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(x-2 -2(a+1)x+a-2 +2a=0) की जड़ें कौन-सी हैं?
What are the roots of (x-2 -2(a+1)x+a-2 +2a=0)?
#quadratic-roots
#parametric-roots
#sum-product
A (a) और (a+2) / (a) and (a+2)
B (a+1) और (a+1) / (a+1) and (a+1)
C (2a) और (2) / (2a) and (2)
D (a-2) और (a+2) / (a-2) and (a+2)
Explanation opens after your attempt
Correct Answer
A. (a) और (a+2) / (a) and (a+2)
Step 1
Concept
The sum of roots is (2a+2) and the product is \(a^2+2a\). These match (a) and (a+2).
Step 2
Why this answer is correct
The correct answer is A. (a) और (a+2) / (a) and (a+2). The sum of roots is (2a+2) and the product is \(a^2+2a\). These match (a) and (a+2).
Step 3
Exam Tip
जड़ों का योग (2a+2) और गुणनफल \(a^2+2a\) है। ये (a) और (a+2) से मिलते हैं।
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यदि (3x-2 -(4t+2)x+t(t+2)=0) की जड़ें (t) और \(\frac{t+2}{3}\) बताई गई हैं, तो यह कथन कब सत्य है?
If the roots of (3x-2 -(4t+2)x+t(t+2)=0) are said to be (t) and \(\frac{t+2}{3}\), when is this statement true?
#quadratic-roots
#parametric-roots
#verification
A केवल (t=0) पर / Only at (t=0)
B केवल (t=1) पर / Only at (t=1)
C हर (t) पर / For every (t)
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
C. हर (t) पर / For every (t)
Step 1
Concept
The sum of these two roots is \(\frac{4t+2}{3}\), and the product is (\frac{t(t+2)}{3}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
Step 2
Why this answer is correct
The correct answer is C. हर (t) पर / For every (t). The sum of these two roots is \(\frac{4t+2}{3}\), and the product is (\frac{t(t+2)}{3}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
Step 3
Exam Tip
इन दोनों जड़ों का योग \(\frac{4t+2}{3}\) और गुणनफल (\frac{t(t+2)}{3}) है। ये दिए गए समीकरण के \(-\frac{b}{a}\) और \(\frac{c}{a}\) से मेल खाते हैं।
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यदि \(x^2-3x-10=0\) की जड़ें \(\alpha,\beta\) हैं, तो (\(\alpha-4\)\(\beta-4\)) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-3x-10=0\), what is (\(\alpha-4\)\(\beta-4\))?
#quadratic-roots
#root-expression
#sum-product
A (-10)
B (-8)
C (-6)
D (6)
Explanation opens after your attempt
Step 1
Concept
We use (\(\alpha-4\)\(\beta-4\)=\alpha\beta-4\(\alpha+\beta\)+16). Since \(\alpha+\beta=3\) and \(\alpha\beta=-10\), the value is (-6).
Step 2
Why this answer is correct
The correct answer is C. (-6). We use (\(\alpha-4\)\(\beta-4\)=\alpha\beta-4\(\alpha+\beta\)+16). Since \(\alpha+\beta=3\) and \(\alpha\beta=-10\), the value is (-6).
Step 3
Exam Tip
(\(\alpha-4\)\(\beta-4\)=\alpha\beta-4\(\alpha+\beta\)+16) है। \(\alpha+\beta=3\) और \(\alpha\beta=-10\), इसलिए मान (-6) है।
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यदि \(x^2-5x+c=0\) की दोनों जड़ें वास्तविक और धनात्मक हैं, तो (c) पर सही शर्त क्या है?
If both roots of \(x^2-5x+c=0\) are real and positive, what is the correct condition on (c)?
#quadratic-roots
#positive-roots
#condition
A (c<0)
B \(0<c\le\frac{25}{4}\)
C \(c>\frac{25}{4}\)
D (c=0)
Explanation opens after your attempt
Correct Answer
B. \(0<c\le\frac{25}{4}\)
Step 1
Concept
The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(0<c\le\frac{25}{4}\). The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
Step 3
Exam Tip
योग (5) धनात्मक है और दोनों जड़ों के लिए गुणनफल (c>0) चाहिए। वास्तविक जड़ों के लिए \(25-4c\ge0\), इसलिए \(0<c\le\frac{25}{4}\)।
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\(x^2-2qx+q^2-16=0\) की जड़ों का धनात्मक अंतर क्या है?
What is the positive difference between the roots of \(x^2-2qx+q^2-16=0\)?
#quadratic-roots
#root-pattern
#difference-of-roots
A (4)
B (6)
C (8)
D (16)
Explanation opens after your attempt
Step 1
Concept
The equation is ((x-q)2 -16=0). Thus the roots are (q+4) and (q-4), whose difference is (8).
Step 2
Why this answer is correct
The correct answer is C. (8). The equation is ((x-q)2 -16=0). Thus the roots are (q+4) and (q-4), whose difference is (8).
Step 3
Exam Tip
समीकरण ((x-q)2 -16=0) है। इसलिए जड़ें (q+4) और (q-4) हैं, जिनका अंतर (8) है।
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यदि \(x^2-4x-2=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2+4\beta+\alpha\beta\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-4x-2=0\), what is \(\alpha^2+4\beta+\alpha\beta\)?
#quadratic-roots
#root-property
#expression
A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
Since \(\alpha\) is a root, \(\alpha^2=4\alpha+2\), and \(\alpha\beta=-2\). The expression becomes (4\alpha+2+4\beta-2=4\(\alpha+\beta\)=16).
Step 2
Why this answer is correct
The correct answer is C. (16). Since \(\alpha\) is a root, \(\alpha^2=4\alpha+2\), and \(\alpha\beta=-2\). The expression becomes (4\alpha+2+4\beta-2=4\(\alpha+\beta\)=16).
Step 3
Exam Tip
क्योंकि \(\alpha\) जड़ है, \(\alpha^2=4\alpha+2\) होगा और \(\alpha\beta=-2\) है। व्यंजक (4\alpha+2+4\beta-2=4\(\alpha+\beta\)=16) बनता है।
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यदि \(x^2+ax+b=0\) की जड़ें \(3+\sqrt{2}\) और \(3-\sqrt{2}\) हैं, तो (a+b) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(3+\sqrt{2}\) and \(3-\sqrt{2}\), what is (a+b)?
#quadratic-roots
#surd-roots
#coefficients
A (-1)
B (1)
C (7)
D (13)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (6), so (a=-6). The product is (7), so (b=7), hence (a+b=1).
Step 2
Why this answer is correct
The correct answer is B. (1). The sum of roots is (6), so (a=-6). The product is (7), so (b=7), hence (a+b=1).
Step 3
Exam Tip
जड़ों का योग (6) है, इसलिए (a=-6)। गुणनफल (7) है, इसलिए (b=7), अतः (a+b=1)।
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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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