100 results found for "root checking" 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 (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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क्या (x=2) समीकरण \(x^2-4=0\) का मूल है?
Is (x=2) a root of \(x^2-4=0\)?
#quadratic-equations
#root-checking
#easy
A हाँ / Yes
B नहीं / No
C केवल (x=-2) मूल है / Only (x=-2) is a root
D निर्धारित नहीं हो सकता / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=2) gives \(2^2-4=0\). To check a root, substitute the value and see if the result is (0).
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=2) gives \(2^2-4=0\). To check a root, substitute the value and see if the result is (0).
Step 3
Exam Tip
(x=2) रखने पर \(2^2-4=0\) मिलता है। मूल जांचने के लिए मान रखकर परिणाम (0) देखें।
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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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क्या (x=-3) समीकरण \(2x^2+7x+3=0\) का मूल है?
Is (x=-3) a root of \(2x^2+7x+3=0\)?
#roots
#negative_root
#checking
A हाँ / Yes
B नहीं / No
C केवल (x=3) मूल है / Only (x=3) is a root
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=-3) gives (18-21+3=0). Watch the signs carefully when substituting a negative root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=-3) gives (18-21+3=0). Watch the signs carefully when substituting a negative root.
Step 3
Exam Tip
(x=-3) रखने पर (18-21+3=0) मिलता है। ऋणात्मक मूल रखते समय चिन्हों को ध्यान से देखें।
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क्या (x=5) समीकरण \(x^2-10x+25=0\) का मूल है?
Is (x=5) a root of \(x^2-10x+25=0\)?
#roots
#checking
#substitution
A हाँ / Yes
B नहीं / No
C केवल (x=-5) मूल है / Only (x=-5) is a root
D कोई वास्तविक मूल नहीं / No real root
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=5) gives (25-50+25=0). Therefore (5) is its root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=5) gives (25-50+25=0). Therefore (5) is its root.
Step 3
Exam Tip
(x=5) रखने पर (25-50+25=0) मिलता है। इसलिए (5) इसका मूल है।
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क्या (x=-2) समीकरण \(3x^2+2x-8=0\) का मूल है?
Is (x=-2) a root of \(3x^2+2x-8=0\)?
#roots
#negative_root
#checking
A हाँ / Yes
B नहीं / No
C केवल (x=2) मूल है / Only (x=2) is a root
D कोई वास्तविक मूल नहीं / No real root
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=-2) gives (12-4-8=0). Handle signs carefully when substituting a negative root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=-2) gives (12-4-8=0). Handle signs carefully when substituting a negative root.
Step 3
Exam Tip
(x=-2) रखने पर (12-4-8=0) मिलता है। ऋणात्मक मूल रखते समय चिन्हों को ध्यान से संभालें।
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क्या (x=4) समीकरण \(x^2-9x+20=0\) का मूल है?
Is (x=4) a root of \(x^2-9x+20=0\)?
#roots
#checking
#substitution
A हाँ / Yes
B नहीं / No
C केवल (x=5) मूल है / Only (x=5) is a root
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=4) gives (16-36+20=0). Therefore (4) is a root of this equation.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=4) gives (16-36+20=0). Therefore (4) is a root of this equation.
Step 3
Exam Tip
(x=4) रखने पर (16-36+20=0) मिलता है। इसलिए (4) इस समीकरण का मूल है।
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क्या (x=-1) समीकरण \(2x^2+5x+3=0\) का मूल है?
Is (x=-1) a root of \(2x^2+5x+3=0\)?
#roots
#negative_root
#checking
A हाँ / Yes
B नहीं / No
C केवल (x=1) मूल है / Only (x=1) is a root
D कोई वास्तविक मूल नहीं / No real root
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=-1) gives (2-5+3=0). Check signs carefully for a negative root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=-1) gives (2-5+3=0). Check signs carefully for a negative root.
Step 3
Exam Tip
(x=-1) रखने पर (2-5+3=0) मिलता है। ऋणात्मक मूल में चिन्हों की जांच ध्यान से करें।
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क्या (x=3) समीकरण \(x^2-7x+12=0\) का मूल है?
Is (x=3) a root of \(x^2-7x+12=0\)?
#roots
#checking
#substitution
A हाँ / Yes
B नहीं / No
C केवल (x=4) मूल है / Only (x=4) is a root
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=3) gives (9-21+12=0). Therefore (3) is a root of this equation.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=3) gives (9-21+12=0). Therefore (3) is a root of this equation.
Step 3
Exam Tip
(x=3) रखने पर (9-21+12=0) मिलता है। इसलिए (3) इस समीकरण का मूल है।
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निम्न में से कौन सा मान \(x^2-100=0\) का मूल नहीं है?
Which of the following values is not a root of \(x^2-100=0\)?
#roots
#not_a_root
#checking
A (10)
B (-10)
C (0)
D \(\sqrt{100}\)
Explanation opens after your attempt
Step 1
Concept
The roots of \(x^2-100=0\) are (10) and (-10). Substituting (0) does not make the equation true.
Step 2
Why this answer is correct
The correct answer is C. (0). The roots of \(x^2-100=0\) are (10) and (-10). Substituting (0) does not make the equation true.
Step 3
Exam Tip
\(x^2-100=0\) के मूल (10) और (-10) हैं। (0) रखने पर समीकरण सत्य नहीं होता।
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क्या (x=-1) समीकरण \(2x^2+x-1=0\) का मूल है?
Is (x=-1) a root of \(2x^2+x-1=0\)?
#roots
#negative_root
#checking
A हाँ / Yes
B नहीं / No
C केवल (x=1) पर / Only at (x=1)
D कोई वास्तविक मूल नहीं / No real root
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=-1) gives (2-1-1=0). Sign checking is important with a negative root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=-1) gives (2-1-1=0). Sign checking is important with a negative root.
Step 3
Exam Tip
(x=-1) रखने पर (2-1-1=0) मिलता है। ऋणात्मक मूल में चिन्हों की जांच जरूरी है।
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क्या (x=2) समीकरण \(x^2-4x+4=0\) का मूल है?
Is (x=2) a root of \(x^2-4x+4=0\)?
#roots
#checking
#substitution
A हाँ / Yes
B नहीं / No
C केवल (x=4) पर / Only at (x=4)
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=2) gives (4-8+4=0). Therefore (2) is a root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=2) gives (4-8+4=0). Therefore (2) is a root.
Step 3
Exam Tip
(x=2) रखने पर (4-8+4=0) मिलता है। इसलिए (2) इसका मूल है।
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निम्न में से कौन सा मान \(x^2-64=0\) का मूल नहीं है?
Which of the following values is not a root of \(x^2-64=0\)?
#roots
#not_a_root
#checking
A (8)
B (-8)
C (0)
D \(\sqrt{64}\)
Explanation opens after your attempt
Step 1
Concept
The roots of \(x^2-64=0\) are (8) and (-8). Substituting (0) does not make the equation true.
Step 2
Why this answer is correct
The correct answer is C. (0). The roots of \(x^2-64=0\) are (8) and (-8). Substituting (0) does not make the equation true.
Step 3
Exam Tip
\(x^2-64=0\) के मूल (8) और (-8) हैं। (0) रखने पर समीकरण सत्य नहीं होता।
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क्या (x=-2) समीकरण \(x^2+3x+2=0\) का मूल है?
Is (x=-2) a root of \(x^2+3x+2=0\)?
#roots
#negative_root
#checking
A हाँ / Yes
B नहीं / No
C केवल (x=2) पर / Only at (x=2)
D कोई वास्तविक मूल नहीं / No real root
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=-2) gives (4-6+2=0). Be careful with signs when substituting a negative value.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=-2) gives (4-6+2=0). Be careful with signs when substituting a negative value.
Step 3
Exam Tip
(x=-2) रखने पर (4-6+2=0) मिलता है। ऋणात्मक मान रखते समय चिन्हों पर ध्यान दें।
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क्या (x=1) समीकरण \(x^2-3x+2=0\) का मूल है?
Is (x=1) a root of \(x^2-3x+2=0\)?
#roots
#checking
#substitution
A हाँ / Yes
B नहीं / No
C केवल (x=2) पर / Only at (x=2)
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=1) gives (1-3+2=0). Therefore (1) is a root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=1) gives (1-3+2=0). Therefore (1) is a root.
Step 3
Exam Tip
(x=1) रखने पर (1-3+2=0) मिलता है। इसलिए (1) इसका मूल है।
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निम्न में से कौन सा मान \(x^2-16=0\) का मूल नहीं है?
Which of the following values is not a root of \(x^2-16=0\)?
#roots
#not_a_root
#checking
A (4)
B (-4)
C (0)
D \(\sqrt{16}\)
Explanation opens after your attempt
Step 1
Concept
The roots of \(x^2-16=0\) are (4) and (-4). Substituting (0) does not satisfy the equation.
Step 2
Why this answer is correct
The correct answer is C. (0). The roots of \(x^2-16=0\) are (4) and (-4). Substituting (0) does not satisfy the equation.
Step 3
Exam Tip
\(x^2-16=0\) के मूल (4) और (-4) हैं। (0) रखने पर समीकरण संतुष्ट नहीं होता।
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यदि (3) समीकरण \(x^2+px-6=0\) का मूल है तो (p) का मान क्या है?
If (3) is a root of \(x^2+px-6=0\) then what is the value of (p)?
#roots
#parameter
#checking
A (1)
B (-1)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (9+3p-6=0) so (p=-1). For a parameter question substitute the root directly.
Step 2
Why this answer is correct
The correct answer is B. (-1). Putting (x=3) gives (9+3p-6=0) so (p=-1). For a parameter question substitute the root directly.
Step 3
Exam Tip
(x=3) रखने पर (9+3p-6=0) इसलिए (p=-1)। पैरामीटर के लिए मूल को सीधे रखें।
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क्या (x=0) समीकरण \(3x^2+2x=0\) का मूल है?
Is (x=0) a root of \(3x^2+2x=0\)?
#roots
#zero_root
#checking
A हाँ / Yes
B नहीं / No
C केवल जब (x=2) हो / Only when (x=2)
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=0) gives (3(0)2 +2(0)=0). Therefore (0) is a root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=0) gives (3(0)2 +2(0)=0). Therefore (0) is a root.
Step 3
Exam Tip
(x=0) रखने पर (3(0)2 +2(0)=0) मिलता है। इसलिए (0) मूल है।
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क्या (x=2) समीकरण \(x^2-5x+6=0\) का मूल है?
Is (x=2) a root of \(x^2-5x+6=0\)?
#roots
#checking
#numerical
A हाँ / Yes
B नहीं / No
C केवल (x=0) पर / Only at (x=0)
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=2) gives (4-10+6=0) so it is a root. In exams always check the final sum after substitution.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=2) gives (4-10+6=0) so it is a root. In exams always check the final sum after substitution.
Step 3
Exam Tip
(x=2) रखने पर (4-10+6=0) मिलता है इसलिए यह मूल है। परीक्षा में मान रखने के बाद अंतिम योग जरूर देखें।
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रेखा आधारित आकृति में संतुलन देखते समय केवल केंद्र क्यों पर्याप्त नहीं है?
Why is only the centre not enough while checking balance in a line-based figure?
#balance
#visual weight
#line
A क्योंकि दृश्य भार रेखा की मोटाई, दिशा और स्थान से भी बनता है / Because visual weight also comes from thickness, direction, and placement of line
B क्योंकि केंद्र कला में नहीं होता / Because centre does not exist in art
C क्योंकि रेखा का संतुलन से संबंध नहीं / Because line has no relation with balance
D क्योंकि मोटाई हमेशा समान होती है / Because thickness is always same
Explanation opens after your attempt
Correct Answer
A. क्योंकि दृश्य भार रेखा की मोटाई, दिशा और स्थान से भी बनता है / Because visual weight also comes from thickness, direction, and placement of line
Step 1
Concept
Balance comes from distribution of visual weight. Exam tip: see line qualities along with centre.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि दृश्य भार रेखा की मोटाई, दिशा और स्थान से भी बनता है / Because visual weight also comes from thickness, direction, and placement of line. Balance comes from distribution of visual weight. Exam tip: see line qualities along with centre.
Step 3
Exam Tip
संतुलन दृश्य भार के वितरण से बनता है। परीक्षा में केंद्र के साथ रेखा गुण भी देखें।
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रेखा अध्ययन में सबसे अच्छी जांच प्रक्रिया कौन सी है?
What is the best checking process in line study?
#visual-art
#line
#hard
#class-10
A दिशा भार दूरी भाव और उद्देश्य को साथ देखना / Checking direction weight spacing mood and purpose together
B केवल रेखा गिनना / Only counting lines
C सिर्फ कागज देखना / Only looking at paper
D माध्यम का नाम याद करना / Memorizing only medium name
Explanation opens after your attempt
Correct Answer
A. दिशा भार दूरी भाव और उद्देश्य को साथ देखना / Checking direction weight spacing mood and purpose together
Step 1
Concept
The effect of line is formed by many qualities. In exams do not depend on only one quality while analyzing.
Step 2
Why this answer is correct
The correct answer is A. दिशा भार दूरी भाव और उद्देश्य को साथ देखना / Checking direction weight spacing mood and purpose together. The effect of line is formed by many qualities. In exams do not depend on only one quality while analyzing.
Step 3
Exam Tip
रेखा का प्रभाव कई गुणों से बनता है। परीक्षा में विश्लेषण करते समय केवल एक गुण पर निर्भर न रहें।
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किस विकल्प में रेखा से अनुपात जांचने की भूमिका सही है?
Which option correctly states the role of line in checking proportion?
#visual-art
#line
#hard
#class-10
A हल्की निर्माण रेखाएं भागों का अनुपात जांचती हैं / Light construction lines check proportion of parts
B मोटी अंतिम रेखा हमेशा अनुपात छिपाती है / Thick final line always hides proportion
C रेखा अनुपात से जुड़ी नहीं / Line is not related to proportion
D केवल रंग अनुपात बताता है / Only colour tells proportion
Explanation opens after your attempt
Correct Answer
A. हल्की निर्माण रेखाएं भागों का अनुपात जांचती हैं / Light construction lines check proportion of parts
Step 1
Concept
Construction lines help improve proportion and structure. In exams make a light plan before final line.
Step 2
Why this answer is correct
The correct answer is A. हल्की निर्माण रेखाएं भागों का अनुपात जांचती हैं / Light construction lines check proportion of parts. Construction lines help improve proportion and structure. In exams make a light plan before final line.
Step 3
Exam Tip
निर्माण रेखाएं अनुपात और ढांचे को सुधारने में सहायक होती हैं। परीक्षा में अंतिम रेखा से पहले हल्की योजना बनाएं।
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रंगीन चित्र को धूसर में जांचना किस निर्णय में सबसे अधिक मदद करता है?
Checking a coloured picture in grey helps most in which decision?
#grayscale
#value
#readability
A मान संरचना और पठनीयता / Value structure and readability
B कागज का प्रकार / Type of paper
C ब्रश की गुणवत्ता / Brush quality
D फ्रेम का आकार / Size of frame
Explanation opens after your attempt
Correct Answer
A. मान संरचना और पठनीयता / Value structure and readability
Step 1
Concept
Light-dark relation is clear in grey view. Exam tip: check value readability.
Step 2
Why this answer is correct
The correct answer is A. मान संरचना और पठनीयता / Value structure and readability. Light-dark relation is clear in grey view. Exam tip: check value readability.
Step 3
Exam Tip
धूसर रूप में हल्का गहरा संबंध साफ दिखता है। परीक्षा में value readability जांचें।
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कोरल सागर की लड़ाई को जापानी विस्तार रोकने की दिशा में क्यों महत्वपूर्ण माना जाता है?
Why is the Battle of the Coral Sea considered important in checking Japanese expansion?
#world history
#world wars
#coral sea
#japanese expansion
A इसने पोर्ट मोरेस्बी की दिशा में जापानी योजना को रोका / It stopped the Japanese plan toward Port Moresby
B इसने जर्मनी को सोवियत संघ से बचाया / It saved Germany from Soviet Union
C इसने राष्ट्र संघ बनाया / It created the League
D इसने फ्रांस को आत्मसमर्पण कराया / It made France surrender
Explanation opens after your attempt
Correct Answer
A. इसने पोर्ट मोरेस्बी की दिशा में जापानी योजना को रोका / It stopped the Japanese plan toward Port Moresby
Step 1
Concept
Coral Sea slowed Japanese expansion in the Pacific. For exams treat it as an important check before Midway.
Step 2
Why this answer is correct
The correct answer is A. इसने पोर्ट मोरेस्बी की दिशा में जापानी योजना को रोका / It stopped the Japanese plan toward Port Moresby. Coral Sea slowed Japanese expansion in the Pacific. For exams treat it as an important check before Midway.
Step 3
Exam Tip
कोरल सागर ने प्रशांत में जापानी विस्तार को धीमा किया। परीक्षा में इसे मिडवे से पहले की महत्वपूर्ण रोक मानें।
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पुलकेशिन द्वितीय द्वारा हर्षवर्धन को रोकना किस बात का संकेत था?
What did Pulakeshin II checking Harshavardhana indicate?
#indian history famous rulers pulakeshin ii
A मुगल सत्ता की शुरुआत / Beginning of Mughal power
B दक्षिण की चालुक्य शक्ति की मजबूती / Strength of Chalukya power in the south
C पाल विश्वविद्यालयों की स्थापना / Establishment of Pala universities
D सिख राज्य का विस्तार / Expansion of the Sikh state
Explanation opens after your attempt
Correct Answer
B. दक्षिण की चालुक्य शक्ति की मजबूती / Strength of Chalukya power in the south
Step 1
Concept
Pulakeshin II checked Harsha's southern advance near the Narmada. Link it with the conflict between northern and southern powers.
Step 2
Why this answer is correct
The correct answer is B. दक्षिण की चालुक्य शक्ति की मजबूती / Strength of Chalukya power in the south. Pulakeshin II checked Harsha's southern advance near the Narmada. Link it with the conflict between northern and southern powers.
Step 3
Exam Tip
पुलकेशिन द्वितीय ने नर्मदा के पास हर्ष की दक्षिण बढ़त रोकी। इसे उत्तर और दक्षिण शक्तियों के संघर्ष से जोड़ें।
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ऐहोल अभिलेख में हर्ष को रोकने का उल्लेख किस राजनीतिक परिस्थिति को दिखाता है?
The mention of checking Harsha in the Aihole inscription shows which political situation?
#ancient-history
#aihole
#pulakeshin-ii
A दक्कन और उत्तर भारत के बीच शक्ति संतुलन / Balance of power between Deccan and North India
B सिंधु नगरों का पतन / Decline of Indus cities
C मौर्य धम्म नीति / Mauryan Dhamma policy
D पाल तिब्बत संपर्क / Pala Tibet contact
Explanation opens after your attempt
Correct Answer
A. दक्कन और उत्तर भारत के बीच शक्ति संतुलन / Balance of power between Deccan and North India
Step 1
Concept
Pulakeshin II projected Deccan power against North Indian politics. For exams, remember Chalukya-Harsha relations.
Step 2
Why this answer is correct
The correct answer is A. दक्कन और उत्तर भारत के बीच शक्ति संतुलन / Balance of power between Deccan and North India. Pulakeshin II projected Deccan power against North Indian politics. For exams, remember Chalukya-Harsha relations.
Step 3
Exam Tip
पुलकेशिन द्वितीय ने दक्कन की शक्ति को उत्तर भारतीय राजनीति के सामने रखा। परीक्षा में चालुक्य हर्ष संबंध याद रखें।
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द्रव्यमान संरक्षण की विश्वसनीय जाँच के लिए सबसे उपयुक्त व्यवस्था कौन सी है?
Which arrangement is most suitable for reliable checking of conservation of mass?
#science
#class10
#hard
#conservation-of-mass
#experiment
A खुले पात्र में गैस निकलने देना / Allowing gas to escape in an open container
B बंद पात्र में पहले और बाद का कुल द्रव्यमान मापना / Measuring total mass before and after in a closed container
C केवल रंग बदलना देखना / Observing only colour change
D केवल गंध महसूस करना / Sensing only smell
Explanation opens after your attempt
Correct Answer
B. बंद पात्र में पहले और बाद का कुल द्रव्यमान मापना / Measuring total mass before and after in a closed container
Step 1
Concept
Matter should not escape while checking conservation of mass.
Step 2
Why this answer is correct
A closed container keeps gas and matter inside.
Step 3
Exam Tip
Therefore total mass before and after gives a correct comparison. चरण 1: द्रव्यमान संरक्षण जाँचते समय पदार्थ बाहर नहीं जाना चाहिए। चरण 2: बंद पात्र गैस और पदार्थ को अंदर रखता है। चरण 3: इसलिए पहले और बाद का कुल द्रव्यमान सही तुलना देता है।
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द्रव्यमान संरक्षण की विश्वसनीय जाँच के लिए सबसे उपयुक्त व्यवस्था कौन सी है?
Which arrangement is most suitable for reliable checking of conservation of mass?
#science
#class10
#hard
#conservation-of-mass
#experiment
A खुले पात्र में गैस निकलने देना / Allowing gas to escape in an open container
B बंद पात्र में पहले और बाद का कुल द्रव्यमान मापना / Measuring total mass before and after in a closed container
C केवल रंग बदलना देखना / Observing only colour change
D केवल गंध महसूस करना / Sensing only smell
Explanation opens after your attempt
Correct Answer
B. बंद पात्र में पहले और बाद का कुल द्रव्यमान मापना / Measuring total mass before and after in a closed container
Step 1
Concept
Matter should not escape while checking conservation of mass.
Step 2
Why this answer is correct
A closed container keeps gas and matter inside.
Step 3
Exam Tip
Therefore total mass before and after gives a correct comparison. चरण 1: द्रव्यमान संरक्षण जाँचते समय पदार्थ बाहर नहीं जाना चाहिए। चरण 2: बंद पात्र गैस और पदार्थ को अंदर रखता है। चरण 3: इसलिए पहले और बाद का कुल द्रव्यमान सही तुलना देता है।
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यदि खुले पात्र में गैस निकल जाए तो द्रव्यमान संरक्षण की जाँच में क्या सावधानी रखनी चाहिए?
If gas escapes from an open container what care should be taken while checking conservation of mass?
#science
#class10
#hard
#closed-container
#mass
A केवल रंग देखना चाहिए / Only colour should be observed
B बंद पात्र में कुल द्रव्यमान मापना चाहिए / Total mass should be measured in a closed container
C गैस को अनदेखा करना चाहिए / Gas should be ignored
D समीकरण का तीर बदलना चाहिए / The arrow of equation should be changed
Explanation opens after your attempt
Correct Answer
B. बंद पात्र में कुल द्रव्यमान मापना चाहिए / Total mass should be measured in a closed container
Step 1
Concept
An open container allows gas to escape.
Step 2
Why this answer is correct
The escaped gas will not be included in mass measurement.
Step 3
Exam Tip
Therefore a closed container is needed for correct checking. चरण 1: खुला पात्र गैस को बाहर जाने देता है। चरण 2: बाहर गई गैस का द्रव्यमान माप में शामिल नहीं होगा। चरण 3: इसलिए सही जाँच के लिए बंद पात्र जरूरी है।
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किस स्थिति में द्रव्यमान संरक्षण की जांच सबसे विश्वसनीय होगी?
In which situation will checking conservation of mass be most reliable?
#science
#class10
#hard
#conservation-of-mass
#experiment
A बंद पात्र में अभिक्रिया से पहले और बाद का कुल द्रव्यमान मापना / Measuring total mass before and after reaction in a closed container
B खुले पात्र में गैस निकलने देना / Allowing gas to escape in open container
C केवल रंग बदलना देखना / Observing only colour change
D केवल गंध महसूस करना / Sensing only smell
Explanation opens after your attempt
Correct Answer
A. बंद पात्र में अभिक्रिया से पहले और बाद का कुल द्रव्यमान मापना / Measuring total mass before and after reaction in a closed container
Step 1
Concept
Matter should not escape while checking conservation of mass.
Step 2
Why this answer is correct
A closed container keeps gas and matter inside.
Step 3
Exam Tip
Therefore mass before and after gives a correct comparison. चरण 1: द्रव्यमान संरक्षण जांचने के लिए पदार्थ बाहर नहीं जाना चाहिए। चरण 2: बंद पात्र गैस और पदार्थ को अंदर रखता है। चरण 3: इसलिए पहले और बाद का द्रव्यमान सही तुलना देता 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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