100 results found for "conjugate-root" in Class 10.
किस विकल्प में अपरिमेय संख्या को परिमेय संख्या में बदलने के लिए सही संयुग्मी चुना गया है?
In which option is the correct conjugate chosen to rationalize an irrational denominator?
#rationalization
#conjugate
#common-error
A \(\frac{1}{5+\sqrt{2}}\) के लिए \(5-\sqrt{2}\) / For \(\frac{1}{5+\sqrt{2}}\) use \(5-\sqrt{2}\)
B \(\frac{1}{5+\sqrt{2}}\) के लिए \(5+\sqrt{2}\) / For \(\frac{1}{5+\sqrt{2}}\) use \(5+\sqrt{2}\)
C \(\frac{1}{5+\sqrt{2}}\) के लिए \(\sqrt{2}-5\) नहीं / For \(\frac{1}{5+\sqrt{2}}\) not \(\sqrt{2}-5\)
D \(\frac{1}{5+\sqrt{2}}\) के लिए \(5\sqrt{2}\) / For \(\frac{1}{5+\sqrt{2}}\) use \(5\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5+\sqrt{2}}\) के लिए \(5-\sqrt{2}\) / For \(\frac{1}{5+\sqrt{2}}\) use \(5-\sqrt{2}\)
Step 1
Concept
The conjugate of \(5+\sqrt{2}\) is \(5-\sqrt{2}\). In exams changing the middle sign is the key idea of a conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5+\sqrt{2}}\) के लिए \(5-\sqrt{2}\) / For \(\frac{1}{5+\sqrt{2}}\) use \(5-\sqrt{2}\). The conjugate of \(5+\sqrt{2}\) is \(5-\sqrt{2}\). In exams changing the middle sign is the key idea of a conjugate.
Step 3
Exam Tip
\(5+\sqrt{2}\) का संयुग्मी \(5-\sqrt{2}\) है। परीक्षा में बीच का चिन्ह बदलना ही संयुग्मी बनाने की मुख्य बात है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?
In which equation is (x=0) one root and the other root negative?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2+7x=0\)
B \(x^2-7x=0\)
C \(x^2+7=0\)
D \(x^2-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x=0\)
Step 1
Concept
(x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x=0\). (x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 3
Exam Tip
(x-2 +7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल धनात्मक है?
In which equation is (x=0) one root and the other root positive?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2-6x=0\)
B \(x^2+6x=0\)
C \(x^2+6=0\)
D \(x^2-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x=0\)
Step 1
Concept
(x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x=0\). (x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 3
Exam Tip
(x-2 -6x=x(x-6)), इसलिए मूल (0) और (6) हैं। दूसरा मूल धनात्मक है।
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यदि \(\alpha+\beta=18\) और \(\alpha\beta=74\), तो कौन सा संयुग्मी अपरिमेय युग्म संभव है?
If \(\alpha+\beta=18\) and \(\alpha\beta=74\), which conjugate irrational pair is possible?
#zeroes
#conjugates
#sum-product
A \(9+\sqrt{7}\) और \(9-\sqrt{7}\) / \(9+\sqrt{7}\) and \(9-\sqrt{7}\)
B \(8+\sqrt{7}\) और \(8-\sqrt{7}\) / \(8+\sqrt{7}\) and \(8-\sqrt{7}\)
C \(9+\sqrt{5}\) और \(9-\sqrt{5}\) / \(9+\sqrt{5}\) and \(9-\sqrt{5}\)
D \(7+\sqrt{9}\) और \(7-\sqrt{9}\) / \(7+\sqrt{9}\) and \(7-\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
A. \(9+\sqrt{7}\) और \(9-\sqrt{7}\) / \(9+\sqrt{7}\) and \(9-\sqrt{7}\)
Step 1
Concept
The sum of \(9+\sqrt{7}\) and \(9-\sqrt{7}\) is (18), and the product is (81-7=74). In exams check both sum and product of options.
Step 2
Why this answer is correct
The correct answer is A. \(9+\sqrt{7}\) और \(9-\sqrt{7}\) / \(9+\sqrt{7}\) and \(9-\sqrt{7}\). The sum of \(9+\sqrt{7}\) and \(9-\sqrt{7}\) is (18), and the product is (81-7=74). In exams check both sum and product of options.
Step 3
Exam Tip
\(9+\sqrt{7}\) और \(9-\sqrt{7}\) का योग (18) और गुणनफल (81-7=74) है। परीक्षा में विकल्पों का योग और गुणनफल दोनों जांचें।
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यदि \(\alpha+\beta=10\) और \(\alpha\beta=21\), तो कौन सा संयुग्मी अपरिमेय युग्म संभव है?
If \(\alpha+\beta=10\) and \(\alpha\beta=21\), which conjugate irrational pair is possible?
#zeroes
#reasoning
#option-audit
A \(5+\sqrt{4}\) और \(5-\sqrt{4}\) / \(5+\sqrt{4}\) and \(5-\sqrt{4}\)
B \(5+\sqrt{5}\) और \(5-\sqrt{5}\) / \(5+\sqrt{5}\) and \(5-\sqrt{5}\)
C \(6+\sqrt{5}\) और \(6-\sqrt{5}\) / \(6+\sqrt{5}\) and \(6-\sqrt{5}\)
D \(4+\sqrt{5}\) और \(4-\sqrt{5}\) / \(4+\sqrt{5}\) and \(4-\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(5+\sqrt{5}\) और \(5-\sqrt{5}\) / \(5+\sqrt{5}\) and \(5-\sqrt{5}\)
Step 1
Concept
The pair \(5+\sqrt{5}\) and \(5-\sqrt{5}\) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.
Step 2
Why this answer is correct
The correct answer is B. \(5+\sqrt{5}\) और \(5-\sqrt{5}\) / \(5+\sqrt{5}\) and \(5-\sqrt{5}\). The pair \(5+\sqrt{5}\) and \(5-\sqrt{5}\) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.
Step 3
Exam Tip
\(5+\sqrt{5}\) और \(5-\sqrt{5}\) का योग (10) और गुणनफल (25-5=20) है इसलिए यह भी नहीं है। सही युग्म (5+2) और (5-2) परिमेय होगा इसलिए दिए विकल्पों में कोई नहीं।
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किस विकल्प में दिया बहुपद परिमेय गुणांकों वाला है और उसके शून्यक अपरिमेय संयुग्मी हैं?
Which option gives a polynomial with rational coefficients and irrational conjugate zeroes?
#rational-coefficients
#irrational-conjugates
#mcq
A \(x^2-6x+7\)
B \(x^2-\sqrt{2}x+1\)
C \(x^2-4x+4\)
D \(x^2+1\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+7\)
Step 1
Concept
For \(x^2-6x+7\), (D=36-28=8). The coefficients are rational and the zeroes are \(3\pm\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+7\). For \(x^2-6x+7\), (D=36-28=8). The coefficients are rational and the zeroes are \(3\pm\sqrt{2}\).
Step 3
Exam Tip
\(x^2-6x+7\) में (D=36-28=8) है। गुणांक परिमेय हैं और शून्यक \(3\pm\sqrt{2}\) होंगे।
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यदि (4x-2 -(4h+1)x+h=0) की एक जड़ \(\frac{1}{4}\) है, तो दूसरी जड़ क्या है?
If one root of (4x-2 -(4h+1)x+h=0) is \(\frac{1}{4}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{4}\)
C (4h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{4}\) है। एक जड़ \(\frac{1}{4}\) है, इसलिए दूसरी जड़ (h) होगी।
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यदि (x-2 -(m+9)x+9m=0) की एक जड़ (9) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+9)x+9m=0) is (9), what is the other root?
#quadratic-roots
#other-root
#parameter
A (m)
B (9m)
C (m+9)
D \(\frac{m}{9}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 3
Exam Tip
जड़ों का गुणनफल (9m) है। एक जड़ (9) है, इसलिए दूसरी जड़ \(\frac{9m}{9}=m\) होगी।
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यदि (2x-2 -(3p+2)x+p(p+2)=0) की एक जड़ (p) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(3p+2)x+p(p+2)=0) is (p), what will be the other root?
#quadratic-roots
#other-root
#parametric-equation
A \(\frac{p+2}{2}\)
B (p+2)
C \(\frac{p}{2}\)
D (2p+2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{p+2}{2}\)
Step 1
Concept
The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{p+2}{2}\). The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 3
Exam Tip
जड़ों का गुणनफल (\frac{p(p+2)}{2}) है। एक जड़ (p) होने पर दूसरी जड़ \(\frac{p+2}{2}\) होगी।
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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 (2x-2 -(h+1)x+h=0) की जड़ों में से एक हमेशा \(\frac{1}{2}\) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(h+1)x+h=0) is always \(\frac{1}{2}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{2}\)
C (2h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 2
Why this answer is correct
The correct answer is A. (h). The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 3
Exam Tip
गुणनफल \(\frac{h}{2}\) है। एक जड़ \(\frac{1}{2}\) है, इसलिए दूसरी जड़ \(\frac{h}{2}\div\frac{1}{2}=h\) होगी।
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यदि (x-2 -(m+2)x+3m=0) की एक जड़ (3) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+2)x+3m=0) is (3), what is the other root?
#quadratic-roots
#other-root
#parametric-equation
A (m)
B (3m)
C (m+2)
D \(\frac{m}{3}\)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 2
Why this answer is correct
The correct answer is A. (m). Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 3
Exam Tip
(x=3) रखने पर समीकरण हर (m) के लिए सही हो जाता है। गुणनफल (3m) है और एक जड़ (3), इसलिए दूसरी जड़ (m) है।
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यदि (x-2 -(m-2 )x+m-6 =0) की एक जड़ (3) है, तो दूसरी जड़ क्या होगी?
If one root of (x-2 -(m-2 )x+m-6 =0) is (3), what is the other root?
#quadratic-roots
#other-root
#error-check
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 2
Why this answer is correct
The correct answer is A. (1). Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 3
Exam Tip
(x=3) रखने पर (9-3(m-2 )+m-6 =0), इसलिए \(m=\frac{9}{2}\)। गुणनफल \(m-6=-\frac{3}{2}\) है, अतः दूसरी जड़ \(-\frac{1}{2}\) होगी, इसलिए कोई विकल्प सही नहीं है।
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यदि (x-2 -(m+7)x+7m=0) का एक मूल (7) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+7)x+7m=0) is (7), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (7m)
C (m+7)
D (m-7 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (7m) है और एक मूल (7) है। इसलिए दूसरा मूल \(\frac{7m}{7}=m\) है।
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यदि \(x^2+ax+54=0\) का एक मूल (6) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+54=0\) is (6), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
B दूसरा मूल (-9), (a=3) / other root (-9), (a=3)
C दूसरा मूल (9), (a=15) / other root (9), (a=15)
D दूसरा मूल (-6), (a=0) / other root (-6), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
Step 1
Concept
The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15). The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 3
Exam Tip
मूलों का गुणनफल (54) है, इसलिए दूसरा मूल (9) होगा। योग (15) है और (-a=15), इसलिए (a=-15)।
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यदि (x-2 -(m+6)x+6m=0) का एक मूल (6) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+6)x+6m=0) is (6), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (6m)
C (m+6)
D (m-6 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (6m) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{6m}{6}=m\) है।
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यदि \(x^2+ax+40=0\) का एक मूल (5) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+40=0\) is (5), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
B दूसरा मूल (-8), (a=3) / other root (-8), (a=3)
C दूसरा मूल (8), (a=13) / other root (8), (a=13)
D दूसरा मूल (-5), (a=0) / other root (-5), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
Step 1
Concept
The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13). The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 3
Exam Tip
मूलों का गुणनफल (40) है, इसलिए दूसरा मूल (8) होगा। योग (13) है और (-a=13), इसलिए (a=-13)।
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यदि (x-2 -(m+5)x+5m=0) का एक मूल (5) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+5)x+5m=0) is (5), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (5m)
C (m+5)
D (m-5 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (5m) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{5m}{5}=m\) है।
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यदि \(x^2+ax+24=0\) का एक मूल (4) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+24=0\) is (4), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
B दूसरा मूल (-6), (a=2) / other root (-6), (a=2)
C दूसरा मूल (6), (a=10) / other root (6), (a=10)
D दूसरा मूल (-4), (a=0) / other root (-4), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
Step 1
Concept
The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10). The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 3
Exam Tip
मूलों का गुणनफल (24) है, इसलिए दूसरा मूल (6) होगा। योग (10) है और (-a=10), इसलिए (a=-10)।
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यदि (x-2 -(m+4)x+4m=0) का एक मूल (4) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+4)x+4m=0) is (4), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (4m)
C (m+4)
D (m-4 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 3
Exam Tip
गुणनफल (4m) है और एक मूल (4) है। इसलिए दूसरा मूल \(\frac{4m}{4}=m\) है।
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यदि \(x^2+ax+18=0\) का एक मूल (2) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+18=0\) is (2), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
B दूसरा मूल (-9), (a=7) / other root (-9), (a=7)
C दूसरा मूल (9), (a=11) / other root (9), (a=11)
D दूसरा मूल (-2), (a=0) / other root (-2), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
Step 1
Concept
The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11). The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 3
Exam Tip
मूलों का गुणनफल (18) है, इसलिए दूसरा मूल (9) होगा। योग (11) है और (-a=11), इसलिए (a=-11)।
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यदि (x-2 -(m+2)x+2m=0) का एक मूल (2) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+2)x+2m=0) is (2), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (2m)
C (m+2)
D (m-2 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 3
Exam Tip
गुणनफल (2m) है और एक मूल (2) है। इसलिए दूसरा मूल \(\frac{2m}{2}=m\) है।
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यदि \(x^2+ax+12=0\) का एक मूल (3) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+12=0\) is (3), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
B दूसरा मूल (-4), (a=1) / other root (-4), (a=1)
C दूसरा मूल (4), (a=7) / other root (4), (a=7)
D दूसरा मूल (-3), (a=0) / other root (-3), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
Step 1
Concept
The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7). The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 3
Exam Tip
मूलों का गुणनफल (12) है, इसलिए दूसरा मूल (4) होगा। योग (7) है और (-a=7), इसलिए (a=-7)।
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यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (15)
B (10)
C (25)
D (-15)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (15). The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (25) है, इसलिए दूसरा मूल (25-10=15) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (14)
B (9)
C (23)
D (-14)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (14). The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (23) है, इसलिए दूसरा मूल (23-9=14) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (13)
B (8)
C (21)
D (-13)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (13). The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (21) है, इसलिए दूसरा मूल (21-8=13) होगा। परीक्षा में एक मूल दिया हो तो योग का उपयोग करें।
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यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (9)
B (6)
C (15)
D (-9)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (9). The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (15) है, इसलिए दूसरा मूल (15-6=9) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (5)
B (4)
C (9)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (5). The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (9) है, इसलिए दूसरा मूल (9-4=5) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (3)
B (2)
C (5)
D (-3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (3). The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 3
Exam Tip
मूलों का योग (5) है, इसलिए दूसरा मूल (5-2=3) होगा। परीक्षा में एक मूल दिया हो तो योग या गुणनफल का प्रयोग करें।
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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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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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यदि (x-2 +(k-3)x+k=0) की एक जड़ दूसरी जड़ की दुगुनी है, तो (k) का मान क्या होगा?
If one root of (x-2 +(k-3)x+k=0) is twice the other root, what is the value of (k)?
#quadratic-roots
#roots-ratio
#parameter
A \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
B \(\frac{3+\sqrt{33}}{4}\) या \(\frac{3-\sqrt{33}}{4}\) / \(\frac{3+\sqrt{33}}{4}\) or \(\frac{3-\sqrt{33}}{4}\)
C (6) या (3) / (6) or (3)
D (9) या (2) / (9) or (2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
Step 1
Concept
Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\). Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 3
Exam Tip
जड़ें (r) और (2r) मानने पर (3r=3-k) और \(2r^2=k\) मिलता है। हल करने पर \(2k^2-21k+18=0\), इसलिए दिए गए दोनों मान मिलते हैं।
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यदि \(x^2-13x+42=0\) के मूलों में छोटा मूल \(\alpha\) और बड़ा मूल \(\beta\) है तो \(\beta-\alpha\) क्या है?
If the smaller root of \(x^2-13x+42=0\) is \(\alpha\) and the larger root is \(\beta\), what is \(\beta-\alpha\)?
#roots
#ordered_roots
#difference
A (1)
B (13)
C (42)
D (6)
Explanation opens after your attempt
Step 1
Concept
The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
Step 3
Exam Tip
समीकरण के मूल (6) और (7) हैं। इसलिए छोटा मूल (6) और बड़ा मूल (7) है तथा \(\beta-\alpha=1\) है।
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यदि \(x^2+ax+a=0\) का एक मूल (2) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (2), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{2}{3}\)
B \(\frac{2}{3}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{2}{3}\)
Step 1
Concept
Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{2}{3}\). Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 3
Exam Tip
(x=2) रखने पर (4+3a=0) से \(a=-\frac{4}{3}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{2}{3}\) होगा।
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यदि \(x^2-11x+30=0\) के मूलों में छोटा मूल \(\alpha\) और बड़ा मूल \(\beta\) है तो \(\beta-\alpha\) क्या है?
If the smaller root of \(x^2-11x+30=0\) is \(\alpha\) and the larger root is \(\beta\), what is \(\beta-\alpha\)?
#roots
#ordered_roots
#difference
A (1)
B (11)
C (30)
D (5)
Explanation opens after your attempt
Step 1
Concept
The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so \(\beta-\alpha=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so \(\beta-\alpha=1\).
Step 3
Exam Tip
समीकरण के मूल (5) और (6) हैं। इसलिए छोटा मूल (5) और बड़ा मूल (6) है तथा \(\beta-\alpha=1\) है।
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यदि \(x^2+ax+a=0\) का एक मूल (1) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (1), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{1}{2}\)
B \(\frac{1}{2}\)
C (2)
D (-2)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{1}{2}\)
Step 1
Concept
Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 3
Exam Tip
(x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा।
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समीकरण \(x^2-17x+70=0\) का एक मूल (7) है तो दूसरा मूल क्या है?
If one root of \(x^2-17x+70=0\) is (7), what is the other root?
#roots
#other_root
#product
A (10)
B (7)
C (17)
D (70)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 2
Why this answer is correct
The correct answer is A. (10). The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 3
Exam Tip
मूलों का गुणनफल (70) है और एक मूल (7) है। दूसरा मूल \(\frac{70}{7}=10\) होगा।
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समीकरण \(x^2-15x+54=0\) का एक मूल (6) है तो दूसरा मूल क्या है?
If one root of \(x^2-15x+54=0\) is (6), what is the other root?
#roots
#other_root
#product
A (9)
B (6)
C (15)
D (54)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 2
Why this answer is correct
The correct answer is A. (9). The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 3
Exam Tip
मूलों का गुणनफल (54) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{54}{6}=9\) है।
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समीकरण \(x^2-13x+40=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-13x+40=0\) is (5), what is the other root?
#roots
#other_root
#product
A (8)
B (5)
C (13)
D (40)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 2
Why this answer is correct
The correct answer is A. (8). The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 3
Exam Tip
मूलों का गुणनफल (40) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{40}{5}=8\) है।
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यदि मूलों का योग (12) है और एक मूल (5) है तो दूसरा मूल क्या है?
If the sum of roots is (12) and one root is (5), what is the other root?
#roots
#other_root
#sum
A (5)
B (7)
C (12)
D (17)
Explanation opens after your attempt
Step 1
Concept
The other root is (12-5=7). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is (12-5=7). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (12-5=7) है। योग में से दिया हुआ मूल घटाएं।
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यदि एक मूल (6) है और मूलों का गुणनफल (48) है तो दूसरा मूल क्या होगा?
If one root is (6) and the product of roots is (48), what is the other root?
#roots
#other_root
#product
A (6)
B (8)
C (42)
D (48)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (8). The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{48}{6}=8\) होगा। गुणनफल को दिए हुए मूल से भाग करें।
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समीकरण \(x^2-11x+30=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-11x+30=0\) is (5), what is the other root?
#roots
#other_root
#factorisation
A (5)
B (6)
C (11)
D (30)
Explanation opens after your attempt
Step 1
Concept
(x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -11x+30=(x-5)(x-6)) है। इसलिए दूसरा मूल (6) है।
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यदि मूलों का योग (8) है और एक मूल (3) है तो दूसरा मूल क्या है?
If the sum of roots is (8) and one root is (3), what is the other root?
#roots
#other_root
#sum
A (3)
B (5)
C (8)
D (11)
Explanation opens after your attempt
Step 1
Concept
The other root is (8-3=5). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (5). The other root is (8-3=5). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (8-3=5) है। योग में से दिया हुआ मूल घटाएं।
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यदि एक मूल (5) है और मूलों का गुणनफल (35) है तो दूसरा मूल क्या होगा?
If one root is (5) and the product of roots is (35), what is the other root?
#roots
#other_root
#product
A (5)
B (7)
C (30)
D (35)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{35}{5}=7\) होगा। गुणनफल में दिए हुए मूल से भाग करें।
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समीकरण \(x^2-9x+18=0\) का एक मूल (3) है तो दूसरा मूल क्या है?
If one root of \(x^2-9x+18=0\) is (3), what is the other root?
#roots
#other_root
#factorisation
A (3)
B (6)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
(x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -9x+18=(x-3)(x-6)) है। इसलिए दूसरा मूल (6) है।
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यदि मूलों का योग (5) है और एक मूल (2) है तो दूसरा मूल क्या है?
If the sum of roots is (5) and one root is (2) then what is the other root?
#roots
#other_root
#sum
A (2)
B (3)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
The other root is (5-2=3). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (3). The other root is (5-2=3). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (5-2=3) है। योग में से दिए हुए मूल को घटाएं।
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यदि एक मूल (3) है और मूलों का गुणनफल (12) है तो दूसरा मूल क्या होगा?
If one root is (3) and the product of roots is (12) then what is the other root?
#roots
#other_root
#product
A (3)
B (4)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 2
Why this answer is correct
The correct answer is B. (4). The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{12}{3}=4\) होगा। गुणनफल वाले प्रश्न में दिए मूल से भाग दें।
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समीकरण \(x^2-6x+8=0\) का एक मूल (4) है तो दूसरा मूल क्या है?
If one root of \(x^2-6x+8=0\) is (4) then what is the other root?
#roots
#other_root
#factorisation
A (1)
B (2)
C (-2)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 2
Why this answer is correct
The correct answer is B. (2). (x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 3
Exam Tip
(x-2 -6x+8=(x-4)(x-2)) इसलिए दूसरा मूल (2) है। दिए गए एक मूल से दूसरा गुणनखंड खोजें।
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किस समीकरण में (x=-2) मूल नहीं है?
In which equation is (x=-2) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2+2x=0\)
B \(x^2+5x+6=0\)
C \(2x^2+3x-2=0\)
D \(x^2-2x+4=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2-2x+4=0\)
Step 1
Concept
Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2-2x+4=0\). Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 3
Exam Tip
(x=-2) रखने पर \(4+4+4=12\neq0\) मिलता है। मूल न होने की जांच भी प्रतिस्थापन से करें।
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किस समीकरण में (x=1) मूल नहीं है?
In which equation is (x=1) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2-1=0\)
B \(x^2-3x+2=0\)
C \(2x^2+x-3=0\)
D \(x^2+x+1=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2+x+1=0\)
Step 1
Concept
Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2+x+1=0\). Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 3
Exam Tip
(x=1) रखने पर \(1+1+1=3\neq 0\) मिलता है। किसी विकल्प में मूल न होने की जांच भी प्रतिस्थापन से करें।
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यदि \(2+\sqrt{13}\) परिमेय गुणांकों वाले द्विघात बहुपद का एक शून्यक है, तो उस बहुपद में (x) का गुणांक किस रूप में हो सकता है?
If \(2+\sqrt{13}\) is one zero of a quadratic polynomial with rational coefficients, what can the coefficient of (x) be?
#rational-coefficients
#conjugate-root
#coefficient
A (-4)
B (4)
C \(-2-\sqrt{13}\)
D \(\sqrt{13}\)
Explanation opens after your attempt
Step 1
Concept
The other zero will be \(2-\sqrt{13}\), so the sum is (4). In a monic polynomial, the coefficient of (x) will be (-4).
Step 2
Why this answer is correct
The correct answer is A. (-4). The other zero will be \(2-\sqrt{13}\), so the sum is (4). In a monic polynomial, the coefficient of (x) will be (-4).
Step 3
Exam Tip
दूसरा शून्यक \(2-\sqrt{13}\) होगा, इसलिए योग (4) है। एकक बहुपद में (x) का गुणांक (-4) होगा।
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यदि (p(x)=x-2 +ax+7) का एक शून्यक \(\sqrt{7}\) है और (a) परिमेय है, तो (a) क्या होगा?
If one zero of (p(x)=x-2 +ax+7) is \(\sqrt{7}\) and (a) is rational, what is (a)?
#parameter-rational
#conjugate-root
#quadratic
A (0)
B \(-2\sqrt{7}\)
C \(2\sqrt{7}\)
D (7)
Explanation opens after your attempt
Step 1
Concept
The other zero will be \(-\sqrt{7}\), so the sum is (0) and (a=-0=0). With rational coefficients, take the conjugate zero.
Step 2
Why this answer is correct
The correct answer is A. (0). The other zero will be \(-\sqrt{7}\), so the sum is (0) and (a=-0=0). With rational coefficients, take the conjugate zero.
Step 3
Exam Tip
दूसरा शून्यक \(-\sqrt{7}\) होगा, इसलिए योग (0) और (a=-0=0) है। परिमेय गुणांक में संयुग्मी शून्यक लेना जरूरी है।
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कौन सा कथन हमेशा सही है यदि द्विघात बहुपद के परिमेय गुणांक और एक शून्यक \(\sqrt{13}\) है?
Which statement is always true if a quadratic polynomial has rational coefficients and one zero is \(\sqrt{13}\)?
#conjugate-rule
#rational-coefficients
#root
A दूसरा शून्यक \(-\sqrt{13}\) होगा / The other zero will be \(-\sqrt{13}\)
B दूसरा शून्यक \(\sqrt{13}\) ही होगा / The other zero must be \(\sqrt{13}\)
C दूसरा शून्यक (13) होगा / The other zero will be (13)
D दूसरा शून्यक \(\frac{1}{\sqrt{13}}\) होगा / The other zero will be \(\frac{1}{\sqrt{13}}\)
Explanation opens after your attempt
Correct Answer
A. दूसरा शून्यक \(-\sqrt{13}\) होगा / The other zero will be \(-\sqrt{13}\)
Step 1
Concept
For rational coefficients, the conjugate \(-\sqrt{13}\) of \(\sqrt{13}\) also appears when the linear coefficient is rational. This follows from \(a+\sqrt{b}\) and \(a-\sqrt{b}\).
Step 2
Why this answer is correct
The correct answer is A. दूसरा शून्यक \(-\sqrt{13}\) होगा / The other zero will be \(-\sqrt{13}\). For rational coefficients, the conjugate \(-\sqrt{13}\) of \(\sqrt{13}\) also appears when the linear coefficient is rational. This follows from \(a+\sqrt{b}\) and \(a-\sqrt{b}\).
Step 3
Exam Tip
परिमेय गुणांकों के लिए \(\sqrt{13}\) का संयुग्मी \(-\sqrt{13}\) भी आता है, जब रैखिक गुणांक परिमेय हो। यह नियम \(a+\sqrt{b}\) और \(a-\sqrt{b}\) पर आधारित है।
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यदि \(2+\sqrt{3}\) किसी परिमेय गुणांकों वाले द्विघात बहुपद का शून्यक है, तो दूसरा शून्यक क्या होगा?
If \(2+\sqrt{3}\) is a zero of a quadratic polynomial with rational coefficients, what will the other zero be?
#conjugate-root
#quadratic
#irrational
A \(2-\sqrt{3}\)
B \(-2+\sqrt{3}\)
C \(\sqrt{3}-2\)
D \(2+\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(2-\sqrt{3}\)
Step 1
Concept
With rational coefficients, the conjugate of an irrational zero is also a zero. So \(2-\sqrt{3}\) will be the other zero.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{3}\). With rational coefficients, the conjugate of an irrational zero is also a zero. So \(2-\sqrt{3}\) will be the other zero.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय शून्यक का संयुग्मी भी शून्यक होता है। इसलिए \(2-\sqrt{3}\) दूसरा शून्यक होगा।
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समीकरण \(9x^2-6x+1=0\) में समान मूल का मान क्या है?
What is the equal root in \(9x^2-6x+1=0\)?
#quadratic-equations
#equal-root-value
#perfect-square
A \(x=\frac{1}{3}\)
B \(x=-\frac{1}{3}\)
C (x=3)
D (x=1)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{1}{3}\)
Step 1
Concept
Here (9x-2 -6x+1=(3x-1)2 ). Therefore the equal root is \(x=\frac{1}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{3}\). Here (9x-2 -6x+1=(3x-1)2 ). Therefore the equal root is \(x=\frac{1}{3}\).
Step 3
Exam Tip
यहाँ (9x-2 -6x+1=(3x-1)2 ) है। इसलिए समान मूल \(x=\frac{1}{3}\) है।
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समीकरण \(4x^2-20x+25=0\) में समान मूल का मान क्या है?
What is the equal root in \(4x^2-20x+25=0\)?
#quadratic-equations
#equal-root-value
#perfect-square
A \(x=\frac{5}{2}\)
B \(x=-\frac{5}{2}\)
C (x=5)
D \(x=\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{2}\)
Step 1
Concept
Here (D=(-20)2 -4(4)(25)=0), and ((2x-5)2 =0). So the equal root is \(x=\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2}\). Here (D=(-20)2 -4(4)(25)=0), and ((2x-5)2 =0). So the equal root is \(x=\frac{5}{2}\).
Step 3
Exam Tip
यहाँ (D=(-20)2 -4(4)(25)=0) और ((2x-5)2 =0) है। इसलिए समान मूल \(x=\frac{5}{2}\) है।
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समीकरण \(x^2-8x+16=0\) में समान मूल का मान क्या है?
What is the equal root of \(x^2-8x+16=0\)?
#quadratic-equations
#equal-root-value
#factorisation
A (4)
B (-4)
C (8)
D (16)
Explanation opens after your attempt
Step 1
Concept
The equation becomes ((x-4)2 =0). The equal root is (x=4).
Step 2
Why this answer is correct
The correct answer is A. (4). The equation becomes ((x-4)2 =0). The equal root is (x=4).
Step 3
Exam Tip
समीकरण ((x-4)2 =0) बनता है। समान मूल सीधे (x=4) है।
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\(x^2-2\sqrt{17}x+17=0\) का मूल क्या है?
What is the root of \(x^2-2\sqrt{17}x+17=0\)?
#quadratic
#repeated-root
#irrational
A \(x=\sqrt{17}\)
B \(x=-\sqrt{17}\)
C (x=17)
D (x=-17)
Explanation opens after your attempt
Correct Answer
A. \(x=\sqrt{17}\)
Step 1
Concept
(\(x-\sqrt{17}\)2 =0), so the repeated root is \(\sqrt{17}\). In exams, ((x-a)2 =0) gives (x=a).
Step 2
Why this answer is correct
The correct answer is A. \(x=\sqrt{17}\). (\(x-\sqrt{17}\)2 =0), so the repeated root is \(\sqrt{17}\). In exams, ((x-a)2 =0) gives (x=a).
Step 3
Exam Tip
(\(x-\sqrt{17}\)2 =0), इसलिए दोहराया हुआ मूल \(\sqrt{17}\) है। परीक्षा में ((x-a)2 =0) से (x=a) मिलता है।
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\(10x^2-41x+40=0\) और \(15x^2-47x+30=0\) में कौनसा मूल समान है?
Which root is common to \(10x^2-41x+40=0\) and \(15x^2-47x+30=0\)?
#quadratic
#common-root
#factorisation
A \(x=\frac{5}{2}\)
B \(x=\frac{8}{5}\)
C \(x=\frac{3}{2}\)
D (x=4)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{2}\)
Step 1
Concept
The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2}\). The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{5}{2},\frac{8}{5}\) और दूसरे के मूल \(\frac{5}{2},\frac{4}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(x^2+2\sqrt{13}x+13=0\) का मूल क्या है?
What is the root of \(x^2+2\sqrt{13}x+13=0\)?
#quadratic
#repeated-root
#irrational
A \(x=-\sqrt{13}\)
B \(x=\sqrt{13}\)
C (x=-13)
D (x=13)
Explanation opens after your attempt
Correct Answer
A. \(x=-\sqrt{13}\)
Step 1
Concept
(\(x+\sqrt{13}\)2 =0), so the repeated root is \(-\sqrt{13}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\sqrt{13}\). (\(x+\sqrt{13}\)2 =0), so the repeated root is \(-\sqrt{13}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
(\(x+\sqrt{13}\)2 =0), इसलिए दोहराया हुआ मूल \(-\sqrt{13}\) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(8x^2-30x+27=0\) और \(12x^2-31x+20=0\) में कौनसा मूल समान है?
Which root is common to \(8x^2-30x+27=0\) and \(12x^2-31x+20=0\)?
#quadratic
#common-root
#factorisation
A \(x=\frac{3}{2}\)
B \(x=\frac{9}{4}\)
C \(x=\frac{4}{3}\)
D \(x=\frac{5}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(\frac{3}{2},\frac{9}{4}\), and the second has roots \(\frac{3}{2},\frac{10}{9}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{3}{2},\frac{9}{4}\), and the second has roots \(\frac{3}{2},\frac{10}{9}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{3}{2},\frac{9}{4}\) और दूसरे के मूल \(\frac{3}{2},\frac{10}{9}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(x^2-2\sqrt{11}x+11=0\) का मूल क्या है?
What is the root of \(x^2-2\sqrt{11}x+11=0\)?
#quadratic
#repeated-root
#irrational
A \(x=\sqrt{11}\)
B \(x=-\sqrt{11}\)
C (x=11)
D (x=-11)
Explanation opens after your attempt
Correct Answer
A. \(x=\sqrt{11}\)
Step 1
Concept
(\(x-\sqrt{11}\)2 =0), so the repeated root is \(\sqrt{11}\). In exams, ((x-a)2 =0) gives (x=a).
Step 2
Why this answer is correct
The correct answer is A. \(x=\sqrt{11}\). (\(x-\sqrt{11}\)2 =0), so the repeated root is \(\sqrt{11}\). In exams, ((x-a)2 =0) gives (x=a).
Step 3
Exam Tip
(\(x-\sqrt{11}\)2 =0), इसलिए दोहराया हुआ मूल \(\sqrt{11}\) है। परीक्षा में ((x-a)2 =0) से (x=a) मिलता है।
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\(6x^2-19x+15=0\) और \(10x^2-27x+18=0\) में कौनसा मूल समान है?
Which root is common to \(6x^2-19x+15=0\) and \(10x^2-27x+18=0\)?
#quadratic
#common-root
#factorisation
A \(x=\frac{3}{2}\)
B \(x=\frac{5}{3}\)
C \(x=\frac{6}{5}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(\frac{3}{2},\frac{5}{3}\), and the second has roots \(\frac{3}{2},\frac{6}{5}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{3}{2},\frac{5}{3}\), and the second has roots \(\frac{3}{2},\frac{6}{5}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{3}{2},\frac{5}{3}\) और दूसरे के मूल \(\frac{3}{2},\frac{6}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(x^2+2\sqrt{7}x+7=0\) का मूल क्या है?
What is the root of \(x^2+2\sqrt{7}x+7=0\)?
#quadratic
#repeated-root
#irrational
A \(x=-\sqrt{7}\)
B \(x=\sqrt{7}\)
C (x=-7)
D (x=7)
Explanation opens after your attempt
Correct Answer
A. \(x=-\sqrt{7}\)
Step 1
Concept
(\(x+\sqrt{7}\)2 =0), so the repeated root is \(-\sqrt{7}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\sqrt{7}\). (\(x+\sqrt{7}\)2 =0), so the repeated root is \(-\sqrt{7}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
(\(x+\sqrt{7}\)2 =0), इसलिए दोहराया हुआ मूल \(-\sqrt{7}\) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(4x^2-12x+5=0\) और \(6x^2-17x+12=0\) में कौनसा मूल समान है?
Which root is common to \(4x^2-12x+5=0\) and \(6x^2-17x+12=0\)?
#quadratic
#common-root
#audit
A \(x=\frac{3}{2}\)
B \(x=\frac{1}{2}\)
C \(x=\frac{4}{3}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(\frac{1}{2},\frac{5}{2}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so none of the listed values is common. In exams, solve both equations correctly before comparing.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{1}{2},\frac{5}{2}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so none of the listed values is common. In exams, solve both equations correctly before comparing.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{1}{2},\frac{5}{2}\) हैं और दूसरे के मूल \(\frac{3}{2},\frac{4}{3}\) हैं, इसलिए दिए विकल्पों में समान मूल नहीं है। परीक्षा में तुलना से पहले दोनों समीकरण सही हल करें।
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\(5x^2-16x+12=0\) और \(6x^2-17x+12=0\) में कौनसा मूल समान है?
Which root is common to \(5x^2-16x+12=0\) and \(6x^2-17x+12=0\)?
#quadratic
#common-root
#hard
A \(x=\frac{3}{2}\)
B \(x=\frac{4}{5}\)
C \(x=\frac{4}{3}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{6}{5},2\) नहीं बल्कि \(\frac{6}{5}\) और (2) हैं, इसलिए यह विकल्प नहीं है। सही जांच में दोनों समीकरणों के मूल क्रमशः \(2,\frac{6}{5}\) और \(\frac{3}{2},\frac{4}{3}\) हैं।
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\(x^2-2\sqrt{5}x+5=0\) का मूल क्या है?
What is the root of \(x^2-2\sqrt{5}x+5=0\)?
#quadratic
#repeated-root
#irrational
A \(x=\sqrt{5}\)
B \(x=-\sqrt{5}\)
C (x=5)
D (x=-5)
Explanation opens after your attempt
Correct Answer
A. \(x=\sqrt{5}\)
Step 1
Concept
(\(x-\sqrt{5}\)2 =0), so the repeated root is \(\sqrt{5}\). In exams, ((x-a)2 =0) gives (x=a).
Step 2
Why this answer is correct
The correct answer is A. \(x=\sqrt{5}\). (\(x-\sqrt{5}\)2 =0), so the repeated root is \(\sqrt{5}\). In exams, ((x-a)2 =0) gives (x=a).
Step 3
Exam Tip
(\(x-\sqrt{5}\)2 =0), इसलिए दोहराया हुआ मूल \(\sqrt{5}\) है। परीक्षा में ((x-a)2 =0) से (x=a) मिलता है।
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\(3x^2-10x+8=0\) और \(4x^2-12x+8=0\) में कौनसा मूल समान है?
Which root is common to \(3x^2-10x+8=0\) and \(4x^2-12x+8=0\)?
#quadratic
#common-root
#hard
A (x=2)
B \(x=\frac{4}{3}\)
C (x=1)
D \(x=\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are \(2,\frac{4}{3}\), and the roots of the second are (2,1). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are \(2,\frac{4}{3}\), and the roots of the second are (2,1). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(2,\frac{4}{3}\) और दूसरे के मूल (2,1) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग-अलग हल करें।
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\(x^2+2\sqrt{3}x+3=0\) का मूल क्या है?
What is the root of \(x^2+2\sqrt{3}x+3=0\)?
#quadratic
#repeated-root
#irrational
A \(x=-\sqrt{3}\)
B \(x=\sqrt{3}\)
C (x=-3)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=-\sqrt{3}\)
Step 1
Concept
(\(x+\sqrt{3}\)2 =0), so the repeated root is \(-\sqrt{3}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\sqrt{3}\). (\(x+\sqrt{3}\)2 =0), so the repeated root is \(-\sqrt{3}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
(\(x+\sqrt{3}\)2 =0), इसलिए दोहराया हुआ मूल \(-\sqrt{3}\) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(2x^2-5x+2=0\) और \(3x^2-8x+4=0\) में कौनसा मूल समान है?
Which root is common to \(2x^2-5x+2=0\) and \(3x^2-8x+4=0\)?
#quadratic
#common-root
#hard
A (x=2)
B \(x=\frac{1}{2}\)
C \(x=\frac{2}{3}\)
D (x=1)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are \(2,\frac{1}{2}\), and the roots of the second are \(2,\frac{2}{3}\). In exams, solve both equations separately for common root.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are \(2,\frac{1}{2}\), and the roots of the second are \(2,\frac{2}{3}\). In exams, solve both equations separately for common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(2,\frac{1}{2}\) और दूसरे के मूल \(2,\frac{2}{3}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(7x^2=175\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(7x^2=175\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm5\)
B (x=5)
C (x=-5)
D \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\)
Step 1
Concept
First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\). First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=25\) मिलता है, इसलिए \(x=\pm5\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(49x^2-42x+9=0\) का मूल क्या है?
What is the root of \(49x^2-42x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{7}\)
B \(x=-\frac{3}{7}\)
C \(x=\frac{7}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{7}\)
Step 1
Concept
((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{7}\). ((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((7x-3)2 =0), इसलिए (7x-3=0) और \(x=\frac{3}{7}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(x^2-11x+30=0\) और \(x^2-13x+42=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-11x+30=0\) and \(x^2-13x+42=0\)?
#quadratic
#common-root
#factorisation
A (x=6)
B (x=5)
C (x=7)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (5,6) और दूसरे के मूल (6,7) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
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\(36x^2-60x+25=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(36x^2-60x+25=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{5}{6}\)
B \(x=-\frac{5}{6}\)
C \(x=\frac{6}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{6}\)
Step 1
Concept
((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{6}\). ((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((6x-5)2 =0), इसलिए (6x-5=0) और \(x=\frac{5}{6}\) है। परीक्षा में दोहराए हुए मूल को सही भिन्न में लिखें।
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\(5x^2=80\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(5x^2=80\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm16\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=16\) मिलता है, इसलिए \(x=\pm4\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(25x^2-20x+4=0\) का मूल क्या है?
What is the root of \(25x^2-20x+4=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{2}{5}\)
B \(x=-\frac{2}{5}\)
C \(x=\frac{5}{2}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{2}{5}\)
Step 1
Concept
((5x-2)2 =0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{2}{5}\). ((5x-2)2 =0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((5x-2)2 =0), इसलिए (5x-2=0) और \(x=\frac{2}{5}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(x^2-7x+12=0\) और \(x^2-9x+20=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-7x+12=0\) and \(x^2-9x+20=0\)?
#quadratic
#common-root
#factorisation
A (x=4)
B (x=3)
C (x=5)
D (x=2)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=4). The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (3,4) और दूसरे के मूल (4,5) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
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\(16x^2-24x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(16x^2-24x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{4}\)
B \(x=-\frac{3}{4}\)
C \(x=\frac{4}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{4}\)
Step 1
Concept
((4x-3)2 =0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{4}\). ((4x-3)2 =0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((4x-3)2 =0), इसलिए (4x-3=0) और \(x=\frac{3}{4}\) है। परीक्षा में दोहराए हुए मूल को भी सही भिन्न में लिखें।
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\(3x^2=12\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2=12\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm2\)
B (x=2)
C (x=-2)
D \(x=\pm4\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm2\)
Step 1
Concept
First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2\). First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=4\) मिलता है, इसलिए \(x=\pm2\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(9x^2-30x+25=0\) का मूल क्या है?
What is the root of \(9x^2-30x+25=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{5}{3}\)
B \(x=-\frac{5}{3}\)
C \(x=\frac{3}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{3}\)
Step 1
Concept
((3x-5)2 =0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{3}\). ((3x-5)2 =0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((3x-5)2 =0), इसलिए (3x-5=0) और \(x=\frac{5}{3}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(x^2-5x+6=0\) और \(x^2-6x+8=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-5x+6=0\) and \(x^2-6x+8=0\)?
#quadratic
#common-root
#factorisation
A (x=2)
B (x=3)
C (x=4)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.
Step 3
Exam Tip
पहले समीकरण के मूल (2,3) और दूसरे के मूल (2,4) हैं। परीक्षा में दोनों समीकरण अलग-अलग हल करके समान मूल देखें।
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\(4x^2-12x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(4x^2-12x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{2}\)
B \(x=-\frac{3}{2}\)
C (x=3)
D \(x=\frac{2}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
((2x-3)2 =0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). ((2x-3)2 =0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.
Step 3
Exam Tip
((2x-3)2 =0), इसलिए (2x-3=0) और \(x=\frac{3}{2}\) है। परीक्षा में दोहराए हुए मूल को भी सही मान से लिखें।
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\(x^2-11x=0\) में (x=0) मूल क्यों है?
Why is (x=0) a root of \(x^2-11x=0\)?
#quadratic
#zero-root
#common-mistake
A क्योंकि (x(x-11)=0) / Because (x(x-11)=0)
B क्योंकि \(x^2=11\) / Because \(x^2=11\)
C क्योंकि (x=11x) / Because (x=11x)
D क्योंकि (x-11=11) / Because (x-11=11)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (x(x-11)=0) / Because (x(x-11)=0)
Step 1
Concept
(x-2 -11x=x(x-11)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x(x-11)=0) / Because (x(x-11)=0). (x-2 -11x=x(x-11)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 3
Exam Tip
(x-2 -11x=x(x-11)), इसलिए शून्य गुणनफल नियम से (x=0) मिलता है। परीक्षा में चर से भाग देकर यह मूल न खोएं।
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\(x^2=169\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=169\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm13\)
B (x=13)
C (x=-13)
D \(x=\pm169\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm13\)
Step 1
Concept
\(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm13\). \(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{169}=\pm13\) होता है। परीक्षा में केवल (13) लिखना अधूरा उत्तर है।
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\(x^2+18x+81=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+18x+81=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=-9)
B (x=9)
C (x=-18)
D (x=18)
Explanation opens after your attempt
Step 1
Concept
((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-9). ((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 3
Exam Tip
((x+9)2 =0), इसलिए दोहराया हुआ मूल (-9) है। परीक्षा में पूर्ण वर्ग समीकरण में दोनों मूल समान होते हैं।
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वर्गमूल विधि से \(x^2=144\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=144\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm12\)
B (x=12)
C (x=-12)
D \(x=\pm72\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm12\)
Step 1
Concept
\(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm12\). \(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{144}=\pm12\) होता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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\(x^2-5x=0\) में (x=0) मूल क्यों है?
Why is (x=0) a root of \(x^2-5x=0\)?
#quadratic
#zero-root
#common-mistake
A क्योंकि (x(x-5)=0) / Because (x(x-5)=0)
B क्योंकि \(x^2=5\) / Because \(x^2=5\)
C क्योंकि (x=5x) / Because (x=5x)
D क्योंकि (x-5=5) / Because (x-5=5)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (x(x-5)=0) / Because (x(x-5)=0)
Step 1
Concept
(x-2 -5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x(x-5)=0) / Because (x(x-5)=0). (x-2 -5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 3
Exam Tip
(x-2 -5x=x(x-5)), इसलिए शून्य गुणनफल नियम से (x=0) मिलता है। परीक्षा में चर से भाग देकर यह मूल न खोएं।
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\(x^2=121\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=121\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm11\)
B (x=11)
C (x=-11)
D \(x=\pm121\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm11\)
Step 1
Concept
\(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm11\). \(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{121}=\pm11\) होता है। परीक्षा में केवल (11) लिखना अधूरा उत्तर है।
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\(x^2+14x+49=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+14x+49=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=-7)
B (x=7)
C (x=-14)
D (x=14)
Explanation opens after your attempt
Step 1
Concept
((x+7)2 =0), so the repeated root is (-7). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. (x=-7). ((x+7)2 =0), so the repeated root is (-7). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
((x+7)2 =0), इसलिए दोहराया हुआ मूल (-7) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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वर्गमूल विधि से \(x^2=64\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=64\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm8\)
B (x=8)
C (x=-8)
D \(x=\pm32\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm8\)
Step 1
Concept
\(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm8\). \(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{64}=\pm8\) होता है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(x^2-12x+36=0\) का मूल क्या होगा?
What will be the root of \(x^2-12x+36=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=6)
B (x=-6)
C (x=12)
D (x=-12)
Explanation opens after your attempt
Step 1
Concept
((x-6)2 =0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). ((x-6)2 =0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.
Step 3
Exam Tip
((x-6)2 =0), इसलिए दोनों समान मूल (x=6) हैं। परीक्षा में पूर्ण वर्ग समीकरण में दोहराया हुआ मूल मिलता है।
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किस समीकरण को वर्गमूल विधि से सीधे हल किया जा सकता है?
Which equation can be solved directly by square root method?
#quadratic
#square-root-method
#method-selection
A ((x-2)2 =9)
B \(x^2+5x+6=0\)
C \(x^2+2x+1=0\)
D \(2x^2+3x+1=0\)
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 =9)
Step 1
Concept
\(In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 2
Why this answer is correct
\(The correct answer is A. ((x-2)^2=9). In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 3
Exam Tip
((x-2)2 =9) में सीधे वर्गमूल लिया जा सकता है। परीक्षा में ((expression\()^2=k) रूप को पहचानें\)।
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\(x^2-3x=0\) में (x=0) क्यों एक मूल है?
Why is (x=0) a root of \(x^2-3x=0\)?
#quadratic
#common-mistake
#zero-root
A क्योंकि (x(x-3)=0) / Because (x(x-3)=0)
B क्योंकि \(x^2=3\) / Because \(x^2=3\)
C क्योंकि (x-3=3) / Because (x-3=3)
D क्योंकि (x=3x) / Because (x=3x)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (x(x-3)=0) / Because (x(x-3)=0)
Step 1
Concept
(x-2 -3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x(x-3)=0) / Because (x(x-3)=0). (x-2 -3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).
Step 3
Exam Tip
(x-2 -3x=x(x-3)), इसलिए शून्य गुणनफल नियम से (x=0) मिलता है। परीक्षा में (x) से भाग देकर (x=0) न खोएं।
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\(x^2=49\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=49\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm7\)
B (x=7)
C (x=-7)
D \(x=\pm49\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm7\)
Step 1
Concept
\(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm7\). \(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 3
Exam Tip
\(x=\pm\sqrt{49}=\pm7\) होता है। परीक्षा में केवल धनात्मक मूल लिखना सामान्य गलती है।
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\(x^2-6x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2-6x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=3)
B (x=-3)
C (x=6)
D (x=9)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+9=(x-3)2 ), so the repeated root is (3). In exams, a perfect square gives equal roots.
Step 2
Why this answer is correct
The correct answer is A. (x=3). (x-2 -6x+9=(x-3)2 ), so the repeated root is (3). In exams, a perfect square gives equal roots.
Step 3
Exam Tip
(x-2 -6x+9=(x-3)2 ), इसलिए दोहराया हुआ मूल (3) है। परीक्षा में पूर्ण वर्ग में दोनों मूल समान होते हैं।
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यदि \(x^2=16\), तो वर्गमूल विधि से (x) का मान क्या होगा?
If \(x^2=16\), what is the value of (x) by square root method?
#quadratic
#square-root-method
#roots
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm8\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x^2=16\) से \(x=\pm\sqrt{16}=\pm4\) मिलता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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यदि (x=2), \(kx^2-6x+4=0\) की जड़ है, तो (k) का मान क्या है?
If (x=2) is a root of \(kx^2-6x+4=0\), what is (k)?
#quadratic-roots
#root-verification
#parameter
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=2), we get (4k-12+4=0). Hence (4k=8) and (k=2).
Step 2
Why this answer is correct
The correct answer is B. (2). Putting (x=2), we get (4k-12+4=0). Hence (4k=8) and (k=2).
Step 3
Exam Tip
(x=2) रखने पर (4k-12+4=0) मिलता है। इसलिए (4k=8) और (k=2)।
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यदि (x-2 +(6k-3)x+5k=0) में (x=-1) मूल है, तो (k) का मान क्या है?
If (x=-1) is a root of (x-2 +(6k-3)x+5k=0), what is the value of (k)?
#quadratic-equations
#root-substitution
#parameter
#expert
A (2)
B (1)
C (0)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Putting (x=-1) gives (1-(6k-3)+5k=0). Thus (4-k=0), so (k=4).
Step 2
Why this answer is correct
The correct answer is A. (2). Putting (x=-1) gives (1-(6k-3)+5k=0). Thus (4-k=0), so (k=4).
Step 3
Exam Tip
(x=-1) रखने पर (1-(6k-3)+5k=0) मिलता है। इससे (4-k=0), इसलिए (k=4) होगा।
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यदि (x=3) समीकरण \(kx^2-8x+5=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?
If (x=3) is not a root of \(kx^2-8x+5=0\), what condition must (k) satisfy?
#quadratic-equations
#not-root
#parameter
#expert
A \(k\neq \frac{19}{9}\)
B \(k=\frac{19}{9}\)
C \(k\neq8\)
D (k=8)
Explanation opens after your attempt
Correct Answer
A. \(k\neq \frac{19}{9}\)
Step 1
Concept
Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq \frac{19}{9}\). Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).
Step 3
Exam Tip
(x=3) रखने पर बायां पक्ष (9k-24+5=9k-19) होता है। मूल न होने के लिए \(9k-19\neq0\), इसलिए \(k\neq\frac{19}{9}\)।
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यदि (x=-6) समीकरण \(2x^2+px-18=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-6) is a root of \(2x^2+px-18=0\), what is the value of (p)?
#quadratic-equations
#root-substitution
#parameter
#expert
A (9)
B (6)
C (-9)
D (-6)
Explanation opens after your attempt
Step 1
Concept
Putting (x=-6) gives (72-6p-18=0). Hence (p=9).
Step 2
Why this answer is correct
The correct answer is A. (9). Putting (x=-6) gives (72-6p-18=0). Hence (p=9).
Step 3
Exam Tip
(x=-6) रखने पर (72-6p-18=0) मिलता है। इससे (p=9) है।
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