100 results found for "one-root" in Class 10.
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 (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 (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 \(x^2-25x+q=0\) 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 \(x^2+ax+12=0\) 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 \(x^2+ax+a=0\) 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 \(x^2+ax+a=0\) 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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एक पौधे की जड़ को एक ओर पानी और दूसरी ओर प्रकाश मिले तो जड़ की दिशा समझाने में कौन से अनुवर्तन अधिक उपयोगी होंगे?
If a plant root gets water on one side and light on the other which tropisms will best explain its direction?
#hydrotropism
#phototropism
#root direction
A जलानुवर्तन और प्रकाशानुवर्तन / Hydrotropism and phototropism
B स्पर्शानुवर्तन और रसायनानुवर्तन / Thigmotropism and chemotropism
C केवल परावर्ती चाप / Only reflex arc
D केवल हार्मोन रक्त नियंत्रण / Only hormone blood control
Explanation opens after your attempt
Correct Answer
A. जलानुवर्तन और प्रकाशानुवर्तन / Hydrotropism and phototropism
Step 1
Concept
A root can grow towards water so hydrotropism is important.
Step 2
Why this answer is correct
A root can also grow away from light.
Step 3
Exam Tip
Therefore both stimuli must be considered to explain direction. चरण 1: जड़ पानी की ओर बढ़ सकती है इसलिए जलानुवर्तन महत्वपूर्ण है। चरण 2: जड़ प्रकाश से दूर भी बढ़ सकती है। चरण 3: इसलिए दिशा समझाने में दोनों उद्दीपन ध्यान में रखने होंगे।
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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 \(x^2-13x+42=0\) 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 \(x^2-11x+30=0\) 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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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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महासभा में एक देश एक मत का सिद्धांत किस विचार को दर्शाता है?
The one country one vote rule in the General Assembly reflects which idea?
#general assembly
#one country one vote
#equality
A सदस्य देशों की औपचारिक समानता / Formal equality of member states
B बड़ी शक्तियों का विशेष शासन / Special rule of great powers
C सैन्य शक्ति के आधार पर मत / Votes based on military power
D उपनिवेशी नियंत्रण / Colonial control
Explanation opens after your attempt
Correct Answer
A. सदस्य देशों की औपचारिक समानता / Formal equality of member states
Step 1
Concept
Each member has one vote in the General Assembly. Exam tip: connect it with the feature of a universal discussion forum.
Step 2
Why this answer is correct
The correct answer is A. सदस्य देशों की औपचारिक समानता / Formal equality of member states. Each member has one vote in the General Assembly. Exam tip: connect it with the feature of a universal discussion forum.
Step 3
Exam Tip
महासभा में हर सदस्य देश को एक मत मिलता है। परीक्षा में इसे सार्वभौमिक चर्चा मंच की विशेषता से जोड़ें।
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यदि \(x^2+px+64=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+64=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(12\sqrt{2}\)
B \(-12\sqrt{2}\)
C \(16\sqrt{2}\)
D (24)
Explanation opens after your attempt
Correct Answer
A. \(12\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=64\) से \(r=4\sqrt{2}\) और \(p=3r=12\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो (q) का मान क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (150)
B (250)
C (15)
D (25)
Explanation opens after your attempt
Step 1
Concept
The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (150). The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (15) है, इसलिए \(q=10\times15=150\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि \(x^2+px+49=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+49=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \( \frac{21\sqrt{2}}{2}\)
B \(-\frac{21\sqrt{2}}{2}\)
C \(14\sqrt{2}\)
D (21)
Explanation opens after your attempt
Correct Answer
A. \( \frac{21\sqrt{2}}{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{21\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=49\) और \(p=3r=\frac{21\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
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यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो (q) का मान क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (126)
B (207)
C (14)
D (23)
Explanation opens after your attempt
Step 1
Concept
The other root is (14), so \(q=9\times14=126\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (126). The other root is (14), so \(q=9\times14=126\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (14) है, इसलिए \(q=9\times14=126\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि \(x^2+px+36=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+36=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(9\sqrt{2}\)
B \(-9\sqrt{2}\)
C \(12\sqrt{2}\)
D (18)
Explanation opens after your attempt
Correct Answer
A. \(9\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=36\) से \(r=3\sqrt{2}\) और \(p=3r=9\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो (q) का मान क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (104)
B (168)
C (13)
D (21)
Explanation opens after your attempt
Step 1
Concept
The other root is (13), so \(q=8\times13=104\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (104). The other root is (13), so \(q=8\times13=104\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (13) है, इसलिए \(q=8\times13=104\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि \(x^2+px+25=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+25=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \( \frac{15\sqrt{2}}{2}\)
B \(-\frac{15\sqrt{2}}{2}\)
C \(10\sqrt{2}\)
D (15)
Explanation opens after your attempt
Correct Answer
A. \( \frac{15\sqrt{2}}{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{15\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
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यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो (q) का मान क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (54)
B (90)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
The other root is (9), so \(q=6\times9=54\). In exams, when (a=1), (c) equals the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (54). The other root is (9), so \(q=6\times9=54\). In exams, when (a=1), (c) equals the product of roots.
Step 3
Exam Tip
दूसरा मूल (9) है, इसलिए \(q=6\times9=54\) होगा। परीक्षा में (a=1) हो तो (c) मूलों के गुणनफल के बराबर होता है।
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यदि \(x^2+px+16=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+16=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(6\sqrt{2}\)
B \(-6\sqrt{2}\)
C \(8\sqrt{2}\)
D (12)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=16\) से \(r=2\sqrt{2}\) और \(p=3r=6\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो (q) का मान क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (20)
B (36)
C (5)
D (9)
Explanation opens after your attempt
Step 1
Concept
The other root is (5), so \(q=4\times5=20\). In exams, (c) equals the product of roots when (a=1).
Step 2
Why this answer is correct
The correct answer is A. (20). The other root is (5), so \(q=4\times5=20\). In exams, (c) equals the product of roots when (a=1).
Step 3
Exam Tip
दूसरा मूल (5) है, इसलिए \(q=4\times5=20\) होगा। परीक्षा में (c) मूलों के गुणनफल के बराबर होता है जब (a=1)।
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यदि \(x^2+px+9=0\) का एक मूल दूसरे का दुगुना है और दोनों ऋणात्मक हैं, तो (p) का सही सरल मान क्या है?
If one root of \(x^2+px+9=0\) is double the other and both are negative, what is the correct simplified value of (p)?
#quadratic
#roots-relation
#simplification
A \( \frac{9\sqrt{2}}{2}\)
B \(3\sqrt{2}\)
C \(9\sqrt{2}\)
D \( \frac{3\sqrt{2}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{9\sqrt{2}}{2}\)
Step 1
Concept
\(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{9\sqrt{2}}{2}\). \(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
\(\frac{9}{\sqrt{2}}\) को सरल करने पर \(\frac{9\sqrt{2}}{2}\) मिलता है। परीक्षा में हर को परिमेय बनाना न भूलें।
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यदि \(x^2+px+9=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+9=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(3\sqrt{2}\)
B \(-3\sqrt{2}\)
C \(9\sqrt{2}\)
D (6)
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=9\) और योग (-3r) है, इसलिए \(p=3r=3\sqrt{\frac{9}{2}}\) नहीं बल्कि \(r=\frac{3}{\sqrt{2}}\), अतः \(p=\frac{9}{\sqrt{2}}\) होता है। परीक्षा में ऐसे प्रश्नों में मानकर समीकरण बनाएं।
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यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो (q) का मान क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (6)
B (10)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
The other root is (3), so \(q=2\times3=6\). In exams, connect (c) with product of roots.
Step 2
Why this answer is correct
The correct answer is A. (6). The other root is (3), so \(q=2\times3=6\). In exams, connect (c) with product of roots.
Step 3
Exam Tip
दूसरा मूल (3) है, इसलिए \(q=2\times3=6\) होगा। परीक्षा में (c) को मूलों के गुणनफल से जोड़ें।
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निम्न में से किस समीकरण का एक मूल (0) है?
Which of the following equations has one root (0)?
#roots
#zero_root
#identification
A \(x^2+5x=0\)
B \(x^2+5=0\)
C \(x^2-5x+6=0\)
D \(x^2+2x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x=0\)
Step 1
Concept
(x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x=0\). (x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 3
Exam Tip
(x-2 +5x=x(x+5)) इसलिए (x=0) एक मूल है। अचर पद (0) हो तो शून्य मूल की संभावना देखें।
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यदि \(x^2+mx+64=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+64=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#expert
A ऐसा संभव नहीं है / It is not possible
B (m=64)
C (m=-64)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (64), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (64), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (64) है, इसलिए ऐसा संभव नहीं है।
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यदि \(x^2+mx+49=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+49=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#expert
A ऐसा संभव नहीं है / It is not possible
B (m=49)
C (m=-49)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (49), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (49), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (49) है, इसलिए ऐसा संभव नहीं है।
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यदि \(x^2+mx+36=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+36=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#expert
A ऐसा संभव नहीं है / It is not possible
B (m=36)
C (m=-36)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (36), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (36), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (36) है, इसलिए ऐसा संभव नहीं है।
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यदि \(x^2+mx+25=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+25=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#hard
A ऐसा संभव नहीं है / It is not possible
B (m=25)
C (m=-25)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (25), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (25), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (25) है, इसलिए ऐसा संभव नहीं है।
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यदि \(x^2+mx+16=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+16=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#hard
A ऐसा संभव नहीं है / It is not possible
B (m=16)
C (m=-16)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (16) है, इसलिए ऐसा संभव नहीं है।
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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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