100 results found for "root-sign" 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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रेखा से संकेत चिह्न में स्पष्टता कैसे आती है?
How does line bring clarity in a sign symbol?
#sign design
#simple line
#clarity
A सरल सीमा और दिशा बनाकर / By making simple boundary and direction
B जटिलता बढ़ाकर / By increasing complexity
C कागज छिपाकर / By hiding paper
D फ्रेम हटाकर / By removing frame
Explanation opens after your attempt
Correct Answer
A. सरल सीमा और दिशा बनाकर / By making simple boundary and direction
Step 1
Concept
Simple line makes a sign quickly recognizable. Exam tip: keep simple line in sign design.
Step 2
Why this answer is correct
The correct answer is A. सरल सीमा और दिशा बनाकर / By making simple boundary and direction. Simple line makes a sign quickly recognizable. Exam tip: keep simple line in sign design.
Step 3
Exam Tip
सरल रेखा चिह्न को जल्दी पहचानने योग्य बनाती है। परीक्षा में sign design में simple line रखें।
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यदि एक चेतावनी संकेत बहुत गोल और कोमल आकारों से बना है तो किस कारण से उसका प्रभाव घट सकता है?
If a warning sign is made with very round and soft shapes why can its effect reduce?
#warning sign
#shape mood
#function
A गोल आकार चेतावनी की तीक्ष्णता को कम कर सकते हैं / Round shapes can reduce sharpness of warning
B गोल आकार हमेशा भय बढ़ाते हैं / Round shapes always increase fear
C आकार संकेत में उपयोगी नहीं होते / Shapes are not useful in signs
D कोमल आकार केवल मान बनाते हैं / Soft shapes only create value
Explanation opens after your attempt
Correct Answer
A. गोल आकार चेतावनी की तीक्ष्णता को कम कर सकते हैं / Round shapes can reduce sharpness of warning
Step 1
Concept
Clear and sharp shapes can be more effective in warning. Exam tip: choose shape according to function.
Step 2
Why this answer is correct
The correct answer is A. गोल आकार चेतावनी की तीक्ष्णता को कम कर सकते हैं / Round shapes can reduce sharpness of warning. Clear and sharp shapes can be more effective in warning. Exam tip: choose shape according to function.
Step 3
Exam Tip
चेतावनी में स्पष्ट और तीखे आकार अधिक प्रभावी हो सकते हैं। परीक्षा में कार्य के अनुसार आकार चुनें।
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यदि चेतावनी संकेत गोल और बहुत कोमल है तो क्या डिजाइन समस्या हो सकती है?
If a warning sign is round and very soft what design problem may occur?
#warning sign
#shape mood
#function
A चेतावनी का तीखापन कम हो सकता है / Sharpness of warning may reduce
B चेतावनी हमेशा बढ़ेगी / Warning will always increase
C रंग अपने आप गहरा होगा / Colour will become dark automatically
D स्थान पूर्ण होगा / Space will be complete
Explanation opens after your attempt
Correct Answer
A. चेतावनी का तीखापन कम हो सकता है / Sharpness of warning may reduce
Step 1
Concept
Sharp or clear shapes can make warning more effective. Exam tip: match shape mood and function.
Step 2
Why this answer is correct
The correct answer is A. चेतावनी का तीखापन कम हो सकता है / Sharpness of warning may reduce. Sharp or clear shapes can make warning more effective. Exam tip: match shape mood and function.
Step 3
Exam Tip
नुकीले या स्पष्ट आकार चेतावनी को अधिक प्रभावी बना सकते हैं। परीक्षा में shape mood और function मिलाएं।
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यदि किसी सूचना चिह्न में आकार सुंदर है पर तुरंत पहचान में नहीं आता तो डिजाइन की मुख्य कमी क्या है?
If shape in an information sign is beautiful but not quickly recognized what is the main design weakness?
#sign design
#clarity
#shape
A सौंदर्य अधिक है / Beauty is high
B पठनीयता और स्पष्टता कमजोर है / Readability and clarity are weak
C कागज मोटा है / Paper is thick
D रंग तटस्थ है / Colour is neutral
Explanation opens after your attempt
Correct Answer
B. पठनीयता और स्पष्टता कमजोर है / Readability and clarity are weak
Step 1
Concept
Quick recognition is necessary in a sign. Exam tip: prioritize clarity in functional design.
Step 2
Why this answer is correct
The correct answer is B. पठनीयता और स्पष्टता कमजोर है / Readability and clarity are weak. Quick recognition is necessary in a sign. Exam tip: prioritize clarity in functional design.
Step 3
Exam Tip
चिह्न में तेजी से पहचान जरूरी है। परीक्षा में functional design में clarity को प्राथमिकता दें।
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यदि किसी चिह्न को दूर से पढ़ना है तो आकारों के बारे में सबसे सही निर्णय क्या होगा?
If a sign has to be read from far away what is the most correct decision about shapes?
#sign design
#shape
#clarity
A सरल बड़े और स्पष्ट आकार रखना / Keep simple large and clear shapes
B बहुत सूक्ष्म और जटिल आकार रखना / Keep very tiny and complex shapes
C सभी आकार समान पृष्ठभूमि में छिपा देना / Hide all shapes in same background
D केवल धुंधले जैविक आकार रखना / Keep only blurred organic shapes
Explanation opens after your attempt
Correct Answer
A. सरल बड़े और स्पष्ट आकार रखना / Keep simple large and clear shapes
Step 1
Concept
Simple shapes are quickly recognized from distance. Exam tip: remember simplicity for visibility.
Step 2
Why this answer is correct
The correct answer is A. सरल बड़े और स्पष्ट आकार रखना / Keep simple large and clear shapes. Simple shapes are quickly recognized from distance. Exam tip: remember simplicity for visibility.
Step 3
Exam Tip
सरल आकार दूर से जल्दी पहचाने जाते हैं। परीक्षा में visibility के लिए simplicity याद रखें।
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यदि \(3,8,13,18,\ldots\) को उलटे क्रम में लिखा जाए तो सार्व अंतर का चिह्न कैसा होगा?
If \(3,8,13,18,\ldots\) is written in reverse order, what happens to the sign of the common difference?
#ap
#reverse sequence
#sign of d
#hard
A धनात्मक ही रहेगा / It remains positive
B शून्य हो जाएगा / It becomes zero
C ऋणात्मक हो जाएगा / It becomes negative
D परिभाषित नहीं होगा / It becomes undefined
Explanation opens after your attempt
Correct Answer
C. ऋणात्मक हो जाएगा / It becomes negative
Step 1
Concept
The original (d=5), but in reverse order \(18,13,8,3,\ldots\), (d=-5). Reversing the order changes the sign.
Step 2
Why this answer is correct
The correct answer is C. ऋणात्मक हो जाएगा / It becomes negative. The original (d=5), but in reverse order \(18,13,8,3,\ldots\), (d=-5). Reversing the order changes the sign.
Step 3
Exam Tip
मूल (d=5) है, पर उलटे क्रम \(18,13,8,3,\ldots\) में (d=-5) होगा। क्रम उलटने पर चिह्न बदलता है।
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यदि \(2, 9, 16, 23,\ldots\) को उलटे क्रम में लिखा जाए, तो सार्व अंतर का चिह्न कैसा होगा?
If \(2, 9, 16, 23,\ldots\) is written in reverse order, what will happen to the sign of the common difference?
#ap
#reverse sequence
#sign of d
#hard
A वह धनात्मक ही रहेगा / It will remain positive
B वह शून्य हो जाएगा / It will become zero
C वह ऋणात्मक हो जाएगा / It will become negative
D वह परिभाषित नहीं होगा / It will be undefined
Explanation opens after your attempt
Correct Answer
C. वह ऋणात्मक हो जाएगा / It will become negative
Step 1
Concept
The original (d=7), and in reverse order \(23,16,9,2,\ldots\), (d=-7). Reversing the order changes the sign of the common difference.
Step 2
Why this answer is correct
The correct answer is C. वह ऋणात्मक हो जाएगा / It will become negative. The original (d=7), and in reverse order \(23,16,9,2,\ldots\), (d=-7). Reversing the order changes the sign of the common difference.
Step 3
Exam Tip
मूल (d=7) है और उलटा क्रम \(23,16,9,2,\ldots\) में (d=-7) होगा। क्रम उलटने पर सार्व अंतर का चिह्न बदल जाता है।
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समीकरण \(5x^2+2x+1=0\) में (D) का चिन्ह और मूलों की प्रकृति क्या है?
What is the sign of (D) and the nature of roots in \(5x^2+2x+1=0\)?
#quadratic-equations
#no-real-roots
#sign-of-discriminant
A (D<0), कोई वास्तविक मूल नहीं / (D<0), no real roots
B (D=0), समान वास्तविक मूल / (D=0), equal real roots
C (D>0), असमान वास्तविक मूल / (D>0), distinct real roots
D (D>0), समान मूल / (D>0), equal roots
Explanation opens after your attempt
Correct Answer
A. (D<0), कोई वास्तविक मूल नहीं / (D<0), no real roots
Step 1
Concept
Here (D=22 -4(5)(1)=-16). A negative discriminant means no real roots.
Step 2
Why this answer is correct
The correct answer is A. (D<0), कोई वास्तविक मूल नहीं / (D<0), no real roots. Here (D=22 -4(5)(1)=-16). A negative discriminant means no real roots.
Step 3
Exam Tip
यहाँ (D=22 -4(5)(1)=-16) है। ऋणात्मक विविक्तकर का अर्थ है कोई वास्तविक मूल नहीं।
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समीकरण \(2x^2-x+4=0\) में (D) का चिन्ह क्या है?
What is the sign of (D) in \(2x^2-x+4=0\)?
#quadratic equations
#D sign
#negative
A ऋणात्मक (D<0) / Negative (D<0)
B धनात्मक (D>0) / Positive (D>0)
C शून्य (D=0) / Zero (D=0)
D पूर्ण वर्ग (D=9) / Perfect square (D=9)
Explanation opens after your attempt
Correct Answer
A. ऋणात्मक (D<0) / Negative (D<0)
Step 1
Concept
Here (D=(-1)2 -4(2)(4)=-31). So (D) is negative.
Step 2
Why this answer is correct
The correct answer is A. ऋणात्मक (D<0) / Negative (D<0). Here (D=(-1)2 -4(2)(4)=-31). So (D) is negative.
Step 3
Exam Tip
यहाँ (D=(-1)2 -4(2)(4)=-31) है। इसलिए (D) ऋणात्मक है।
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समीकरण \(x^2+x+1=0\) के लिए (D) का चिन्ह क्या है?
What is the sign of (D) for the equation \(x^2+x+1=0\)?
#quadratic equations
#sign of D
#no real roots
A ऋणात्मक (D<0) / Negative (D<0)
B धनात्मक (D>0) / Positive (D>0)
C शून्य (D=0) / Zero (D=0)
D एक (D=1) / One (D=1)
Explanation opens after your attempt
Correct Answer
A. ऋणात्मक (D<0) / Negative (D<0)
Step 1
Concept
Here (D=(1)2 -4(1)(1)=-3). So (D) is negative.
Step 2
Why this answer is correct
The correct answer is A. ऋणात्मक (D<0) / Negative (D<0). Here (D=(1)2 -4(1)(1)=-3). So (D) is negative.
Step 3
Exam Tip
यहाँ (D=(1)2 -4(1)(1)=-3) है। इसलिए (D) ऋणात्मक है।
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समीकरण \(2x^2+4x+2=0\) के लिए विविक्तकर का चिन्ह क्या है?
What is the sign of the discriminant for \(2x^2+4x+2=0\)?
#quadratic equations
#sign of discriminant
#equal roots
A शून्य / zero
B धनात्मक / positive
C ऋणात्मक / negative
D अज्ञात / unknown
Explanation opens after your attempt
Correct Answer
A. शून्य / zero
Step 1
Concept
(D=42 -4(2)(2)=0). So this is the case of equal real roots.
Step 2
Why this answer is correct
The correct answer is A. शून्य / zero. (D=42 -4(2)(2)=0). So this is the case of equal real roots.
Step 3
Exam Tip
(D=42 -4(2)(2)=0) है। इसलिए यह समान वास्तविक मूलों की स्थिति है।
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समीकरण \(3x^2-2x+4=0\) के लिए विविक्तकर का चिन्ह क्या है?
What is the sign of the discriminant for \(3x^2-2x+4=0\)?
#quadratic equations
#sign of discriminant
#no real roots
A ऋणात्मक / negative
B धनात्मक / positive
C शून्य / zero
D निश्चित नहीं / not fixed
Explanation opens after your attempt
Correct Answer
A. ऋणात्मक / negative
Step 1
Concept
(D=(-2)2 -4(3)(4)=-44<0). Hence the discriminant is negative.
Step 2
Why this answer is correct
The correct answer is A. ऋणात्मक / negative. (D=(-2)2 -4(3)(4)=-44<0). Hence the discriminant is negative.
Step 3
Exam Tip
(D=(-2)2 -4(3)(4)=-44<0) है। अतः विविक्तकर ऋणात्मक है।
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यदि (p(x)=(x+9)(x-5)(x-12)) है, तो (5<x<12) में (p(x)) का चिह्न क्या होगा?
If (p(x)=(x+9)(x-5)(x-12)), what will be the sign of (p(x)) for (5<x<12)?
#sign interval
#factor form
#graph
A धनात्मक / Positive
B ऋणात्मक / Negative
C शून्य / Zero
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
B. ऋणात्मक / Negative
Step 1
Concept
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है, इसलिए गुणनफल ऋणात्मक है। टिप: प्रत्येक कारक का चिह्न अलग जांचें।
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यदि (p(x)=(x+6)(x-4)(x-10)) है, तो (4<x<10) में (p(x)) का चिह्न क्या होगा?
If (p(x)=(x+6)(x-4)(x-10)), what will be the sign of (p(x)) for (4<x<10)?
#sign interval
#factor form
#graph
A धनात्मक / Positive
B ऋणात्मक / Negative
C शून्य / Zero
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
B. ऋणात्मक / Negative
Step 1
Concept
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है, इसलिए गुणनफल ऋणात्मक है। टिप: प्रत्येक कारक का चिह्न अलग जांचें।
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यदि (p(x)=(x+5)(x-3)(x-9)) है तो (3<x<9) में (p(x)) का चिह्न क्या होगा?
If (p(x)=(x+5)(x-3)(x-9)), what will be the sign of (p(x)) for (3<x<9)?
#sign interval
#factor form
#graph
A धनात्मक / Positive
B ऋणात्मक / Negative
C शून्य / Zero
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
B. ऋणात्मक / Negative
Step 1
Concept
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check factor signs separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check factor signs separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है इसलिए गुणनफल ऋणात्मक है। टिप: कारकों के चिह्न अलग-अलग जांचें।
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यदि (p(x)=(x+4)(x-2)(x-7)) है, तो (2<x<7) में (p(x)) का चिह्न क्या होगा?
If (p(x)=(x+4)(x-2)(x-7)), what will be the sign of (p(x)) for (2<x<7)?
#sign interval
#factor form
#graph
A धनात्मक / Positive
B ऋणात्मक / Negative
C शून्य / Zero
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
B. ऋणात्मक / Negative
Step 1
Concept
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check factor signs separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check factor signs separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है, इसलिए गुणनफल ऋणात्मक है। टिप: कारकों के चिह्न अलग-अलग जाँचें।
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यदि (p(x)=(x+2)(x-1)(x-6)) है, तो (1<x<6) में (p(x)) का चिह्न क्या होगा?
If (p(x)=(x+2)(x-1)(x-6)), what will be the sign of (p(x)) for (1<x<6)?
#sign interval
#factor form
#graph
A धनात्मक / Positive
B ऋणात्मक / Negative
C शून्य / Zero
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
B. ऋणात्मक / Negative
Step 1
Concept
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है, इसलिए गुणनफल ऋणात्मक है। टिप: प्रत्येक कारक का चिह्न अलग जांचें।
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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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समीकरण \(9x^2-6x+1=0\) में समान मूल का मान क्या है?
What is the equal root in \(9x^2-6x+1=0\)?
#quadratic-equations
#equal-root-value
#perfect-square
A \(x=\frac{1}{3}\)
B \(x=-\frac{1}{3}\)
C (x=3)
D (x=1)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{1}{3}\)
Step 1
Concept
Here (9x-2 -6x+1=(3x-1)2 ). Therefore the equal root is \(x=\frac{1}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{3}\). Here (9x-2 -6x+1=(3x-1)2 ). Therefore the equal root is \(x=\frac{1}{3}\).
Step 3
Exam Tip
यहाँ (9x-2 -6x+1=(3x-1)2 ) है। इसलिए समान मूल \(x=\frac{1}{3}\) है।
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समीकरण \(4x^2-20x+25=0\) में समान मूल का मान क्या है?
What is the equal root in \(4x^2-20x+25=0\)?
#quadratic-equations
#equal-root-value
#perfect-square
A \(x=\frac{5}{2}\)
B \(x=-\frac{5}{2}\)
C (x=5)
D \(x=\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{2}\)
Step 1
Concept
Here (D=(-20)2 -4(4)(25)=0), and ((2x-5)2 =0). So the equal root is \(x=\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2}\). Here (D=(-20)2 -4(4)(25)=0), and ((2x-5)2 =0). So the equal root is \(x=\frac{5}{2}\).
Step 3
Exam Tip
यहाँ (D=(-20)2 -4(4)(25)=0) और ((2x-5)2 =0) है। इसलिए समान मूल \(x=\frac{5}{2}\) है।
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समीकरण \(x^2-8x+16=0\) में समान मूल का मान क्या है?
What is the equal root of \(x^2-8x+16=0\)?
#quadratic-equations
#equal-root-value
#factorisation
A (4)
B (-4)
C (8)
D (16)
Explanation opens after your attempt
Step 1
Concept
The equation becomes ((x-4)2 =0). The equal root is (x=4).
Step 2
Why this answer is correct
The correct answer is A. (4). The equation becomes ((x-4)2 =0). The equal root is (x=4).
Step 3
Exam Tip
समीकरण ((x-4)2 =0) बनता है। समान मूल सीधे (x=4) है।
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\(x^2-2\sqrt{17}x+17=0\) का मूल क्या है?
What is the root of \(x^2-2\sqrt{17}x+17=0\)?
#quadratic
#repeated-root
#irrational
A \(x=\sqrt{17}\)
B \(x=-\sqrt{17}\)
C (x=17)
D (x=-17)
Explanation opens after your attempt
Correct Answer
A. \(x=\sqrt{17}\)
Step 1
Concept
(\(x-\sqrt{17}\)2 =0), so the repeated root is \(\sqrt{17}\). In exams, ((x-a)2 =0) gives (x=a).
Step 2
Why this answer is correct
The correct answer is A. \(x=\sqrt{17}\). (\(x-\sqrt{17}\)2 =0), so the repeated root is \(\sqrt{17}\). In exams, ((x-a)2 =0) gives (x=a).
Step 3
Exam Tip
(\(x-\sqrt{17}\)2 =0), इसलिए दोहराया हुआ मूल \(\sqrt{17}\) है। परीक्षा में ((x-a)2 =0) से (x=a) मिलता है।
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\(10x^2-41x+40=0\) और \(15x^2-47x+30=0\) में कौनसा मूल समान है?
Which root is common to \(10x^2-41x+40=0\) and \(15x^2-47x+30=0\)?
#quadratic
#common-root
#factorisation
A \(x=\frac{5}{2}\)
B \(x=\frac{8}{5}\)
C \(x=\frac{3}{2}\)
D (x=4)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{2}\)
Step 1
Concept
The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2}\). The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{5}{2},\frac{8}{5}\) और दूसरे के मूल \(\frac{5}{2},\frac{4}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(x^2+2\sqrt{13}x+13=0\) का मूल क्या है?
What is the root of \(x^2+2\sqrt{13}x+13=0\)?
#quadratic
#repeated-root
#irrational
A \(x=-\sqrt{13}\)
B \(x=\sqrt{13}\)
C (x=-13)
D (x=13)
Explanation opens after your attempt
Correct Answer
A. \(x=-\sqrt{13}\)
Step 1
Concept
(\(x+\sqrt{13}\)2 =0), so the repeated root is \(-\sqrt{13}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\sqrt{13}\). (\(x+\sqrt{13}\)2 =0), so the repeated root is \(-\sqrt{13}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
(\(x+\sqrt{13}\)2 =0), इसलिए दोहराया हुआ मूल \(-\sqrt{13}\) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(8x^2-30x+27=0\) और \(12x^2-31x+20=0\) में कौनसा मूल समान है?
Which root is common to \(8x^2-30x+27=0\) and \(12x^2-31x+20=0\)?
#quadratic
#common-root
#factorisation
A \(x=\frac{3}{2}\)
B \(x=\frac{9}{4}\)
C \(x=\frac{4}{3}\)
D \(x=\frac{5}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(\frac{3}{2},\frac{9}{4}\), and the second has roots \(\frac{3}{2},\frac{10}{9}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{3}{2},\frac{9}{4}\), and the second has roots \(\frac{3}{2},\frac{10}{9}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{3}{2},\frac{9}{4}\) और दूसरे के मूल \(\frac{3}{2},\frac{10}{9}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(x^2-2\sqrt{11}x+11=0\) का मूल क्या है?
What is the root of \(x^2-2\sqrt{11}x+11=0\)?
#quadratic
#repeated-root
#irrational
A \(x=\sqrt{11}\)
B \(x=-\sqrt{11}\)
C (x=11)
D (x=-11)
Explanation opens after your attempt
Correct Answer
A. \(x=\sqrt{11}\)
Step 1
Concept
(\(x-\sqrt{11}\)2 =0), so the repeated root is \(\sqrt{11}\). In exams, ((x-a)2 =0) gives (x=a).
Step 2
Why this answer is correct
The correct answer is A. \(x=\sqrt{11}\). (\(x-\sqrt{11}\)2 =0), so the repeated root is \(\sqrt{11}\). In exams, ((x-a)2 =0) gives (x=a).
Step 3
Exam Tip
(\(x-\sqrt{11}\)2 =0), इसलिए दोहराया हुआ मूल \(\sqrt{11}\) है। परीक्षा में ((x-a)2 =0) से (x=a) मिलता है।
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\(6x^2-19x+15=0\) और \(10x^2-27x+18=0\) में कौनसा मूल समान है?
Which root is common to \(6x^2-19x+15=0\) and \(10x^2-27x+18=0\)?
#quadratic
#common-root
#factorisation
A \(x=\frac{3}{2}\)
B \(x=\frac{5}{3}\)
C \(x=\frac{6}{5}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(\frac{3}{2},\frac{5}{3}\), and the second has roots \(\frac{3}{2},\frac{6}{5}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{3}{2},\frac{5}{3}\), and the second has roots \(\frac{3}{2},\frac{6}{5}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{3}{2},\frac{5}{3}\) और दूसरे के मूल \(\frac{3}{2},\frac{6}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(x^2+2\sqrt{7}x+7=0\) का मूल क्या है?
What is the root of \(x^2+2\sqrt{7}x+7=0\)?
#quadratic
#repeated-root
#irrational
A \(x=-\sqrt{7}\)
B \(x=\sqrt{7}\)
C (x=-7)
D (x=7)
Explanation opens after your attempt
Correct Answer
A. \(x=-\sqrt{7}\)
Step 1
Concept
(\(x+\sqrt{7}\)2 =0), so the repeated root is \(-\sqrt{7}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\sqrt{7}\). (\(x+\sqrt{7}\)2 =0), so the repeated root is \(-\sqrt{7}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
(\(x+\sqrt{7}\)2 =0), इसलिए दोहराया हुआ मूल \(-\sqrt{7}\) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(4x^2-12x+5=0\) और \(6x^2-17x+12=0\) में कौनसा मूल समान है?
Which root is common to \(4x^2-12x+5=0\) and \(6x^2-17x+12=0\)?
#quadratic
#common-root
#audit
A \(x=\frac{3}{2}\)
B \(x=\frac{1}{2}\)
C \(x=\frac{4}{3}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(\frac{1}{2},\frac{5}{2}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so none of the listed values is common. In exams, solve both equations correctly before comparing.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{1}{2},\frac{5}{2}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so none of the listed values is common. In exams, solve both equations correctly before comparing.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{1}{2},\frac{5}{2}\) हैं और दूसरे के मूल \(\frac{3}{2},\frac{4}{3}\) हैं, इसलिए दिए विकल्पों में समान मूल नहीं है। परीक्षा में तुलना से पहले दोनों समीकरण सही हल करें।
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\(5x^2-16x+12=0\) और \(6x^2-17x+12=0\) में कौनसा मूल समान है?
Which root is common to \(5x^2-16x+12=0\) and \(6x^2-17x+12=0\)?
#quadratic
#common-root
#hard
A \(x=\frac{3}{2}\)
B \(x=\frac{4}{5}\)
C \(x=\frac{4}{3}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{6}{5},2\) नहीं बल्कि \(\frac{6}{5}\) और (2) हैं, इसलिए यह विकल्प नहीं है। सही जांच में दोनों समीकरणों के मूल क्रमशः \(2,\frac{6}{5}\) और \(\frac{3}{2},\frac{4}{3}\) हैं।
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\(x^2-2\sqrt{5}x+5=0\) का मूल क्या है?
What is the root of \(x^2-2\sqrt{5}x+5=0\)?
#quadratic
#repeated-root
#irrational
A \(x=\sqrt{5}\)
B \(x=-\sqrt{5}\)
C (x=5)
D (x=-5)
Explanation opens after your attempt
Correct Answer
A. \(x=\sqrt{5}\)
Step 1
Concept
(\(x-\sqrt{5}\)2 =0), so the repeated root is \(\sqrt{5}\). In exams, ((x-a)2 =0) gives (x=a).
Step 2
Why this answer is correct
The correct answer is A. \(x=\sqrt{5}\). (\(x-\sqrt{5}\)2 =0), so the repeated root is \(\sqrt{5}\). In exams, ((x-a)2 =0) gives (x=a).
Step 3
Exam Tip
(\(x-\sqrt{5}\)2 =0), इसलिए दोहराया हुआ मूल \(\sqrt{5}\) है। परीक्षा में ((x-a)2 =0) से (x=a) मिलता है।
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\(3x^2-10x+8=0\) और \(4x^2-12x+8=0\) में कौनसा मूल समान है?
Which root is common to \(3x^2-10x+8=0\) and \(4x^2-12x+8=0\)?
#quadratic
#common-root
#hard
A (x=2)
B \(x=\frac{4}{3}\)
C (x=1)
D \(x=\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are \(2,\frac{4}{3}\), and the roots of the second are (2,1). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are \(2,\frac{4}{3}\), and the roots of the second are (2,1). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(2,\frac{4}{3}\) और दूसरे के मूल (2,1) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग-अलग हल करें।
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\(x^2+2\sqrt{3}x+3=0\) का मूल क्या है?
What is the root of \(x^2+2\sqrt{3}x+3=0\)?
#quadratic
#repeated-root
#irrational
A \(x=-\sqrt{3}\)
B \(x=\sqrt{3}\)
C (x=-3)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=-\sqrt{3}\)
Step 1
Concept
(\(x+\sqrt{3}\)2 =0), so the repeated root is \(-\sqrt{3}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\sqrt{3}\). (\(x+\sqrt{3}\)2 =0), so the repeated root is \(-\sqrt{3}\). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
(\(x+\sqrt{3}\)2 =0), इसलिए दोहराया हुआ मूल \(-\sqrt{3}\) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(2x^2-5x+2=0\) और \(3x^2-8x+4=0\) में कौनसा मूल समान है?
Which root is common to \(2x^2-5x+2=0\) and \(3x^2-8x+4=0\)?
#quadratic
#common-root
#hard
A (x=2)
B \(x=\frac{1}{2}\)
C \(x=\frac{2}{3}\)
D (x=1)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are \(2,\frac{1}{2}\), and the roots of the second are \(2,\frac{2}{3}\). In exams, solve both equations separately for common root.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are \(2,\frac{1}{2}\), and the roots of the second are \(2,\frac{2}{3}\). In exams, solve both equations separately for common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(2,\frac{1}{2}\) और दूसरे के मूल \(2,\frac{2}{3}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
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\(7x^2=175\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(7x^2=175\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm5\)
B (x=5)
C (x=-5)
D \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\)
Step 1
Concept
First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\). First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=25\) मिलता है, इसलिए \(x=\pm5\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(49x^2-42x+9=0\) का मूल क्या है?
What is the root of \(49x^2-42x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{7}\)
B \(x=-\frac{3}{7}\)
C \(x=\frac{7}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{7}\)
Step 1
Concept
((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{7}\). ((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((7x-3)2 =0), इसलिए (7x-3=0) और \(x=\frac{3}{7}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(x^2-11x+30=0\) और \(x^2-13x+42=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-11x+30=0\) and \(x^2-13x+42=0\)?
#quadratic
#common-root
#factorisation
A (x=6)
B (x=5)
C (x=7)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (5,6) और दूसरे के मूल (6,7) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
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\(36x^2-60x+25=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(36x^2-60x+25=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{5}{6}\)
B \(x=-\frac{5}{6}\)
C \(x=\frac{6}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{6}\)
Step 1
Concept
((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{6}\). ((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((6x-5)2 =0), इसलिए (6x-5=0) और \(x=\frac{5}{6}\) है। परीक्षा में दोहराए हुए मूल को सही भिन्न में लिखें।
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\(5x^2=80\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(5x^2=80\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm16\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=16\) मिलता है, इसलिए \(x=\pm4\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(25x^2-20x+4=0\) का मूल क्या है?
What is the root of \(25x^2-20x+4=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{2}{5}\)
B \(x=-\frac{2}{5}\)
C \(x=\frac{5}{2}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{2}{5}\)
Step 1
Concept
((5x-2)2 =0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{2}{5}\). ((5x-2)2 =0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((5x-2)2 =0), इसलिए (5x-2=0) और \(x=\frac{2}{5}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(x^2-7x+12=0\) और \(x^2-9x+20=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-7x+12=0\) and \(x^2-9x+20=0\)?
#quadratic
#common-root
#factorisation
A (x=4)
B (x=3)
C (x=5)
D (x=2)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=4). The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (3,4) और दूसरे के मूल (4,5) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
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\(16x^2-24x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(16x^2-24x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{4}\)
B \(x=-\frac{3}{4}\)
C \(x=\frac{4}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{4}\)
Step 1
Concept
((4x-3)2 =0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{4}\). ((4x-3)2 =0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((4x-3)2 =0), इसलिए (4x-3=0) और \(x=\frac{3}{4}\) है। परीक्षा में दोहराए हुए मूल को भी सही भिन्न में लिखें।
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\(3x^2=12\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2=12\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm2\)
B (x=2)
C (x=-2)
D \(x=\pm4\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm2\)
Step 1
Concept
First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2\). First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=4\) मिलता है, इसलिए \(x=\pm2\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(9x^2-30x+25=0\) का मूल क्या है?
What is the root of \(9x^2-30x+25=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{5}{3}\)
B \(x=-\frac{5}{3}\)
C \(x=\frac{3}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{3}\)
Step 1
Concept
((3x-5)2 =0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{3}\). ((3x-5)2 =0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((3x-5)2 =0), इसलिए (3x-5=0) और \(x=\frac{5}{3}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(x^2-5x+6=0\) और \(x^2-6x+8=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-5x+6=0\) and \(x^2-6x+8=0\)?
#quadratic
#common-root
#factorisation
A (x=2)
B (x=3)
C (x=4)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.
Step 3
Exam Tip
पहले समीकरण के मूल (2,3) और दूसरे के मूल (2,4) हैं। परीक्षा में दोनों समीकरण अलग-अलग हल करके समान मूल देखें।
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\(4x^2-12x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(4x^2-12x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{2}\)
B \(x=-\frac{3}{2}\)
C (x=3)
D \(x=\frac{2}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
((2x-3)2 =0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). ((2x-3)2 =0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.
Step 3
Exam Tip
((2x-3)2 =0), इसलिए (2x-3=0) और \(x=\frac{3}{2}\) है। परीक्षा में दोहराए हुए मूल को भी सही मान से लिखें।
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\(x^2-11x=0\) में (x=0) मूल क्यों है?
Why is (x=0) a root of \(x^2-11x=0\)?
#quadratic
#zero-root
#common-mistake
A क्योंकि (x(x-11)=0) / Because (x(x-11)=0)
B क्योंकि \(x^2=11\) / Because \(x^2=11\)
C क्योंकि (x=11x) / Because (x=11x)
D क्योंकि (x-11=11) / Because (x-11=11)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (x(x-11)=0) / Because (x(x-11)=0)
Step 1
Concept
(x-2 -11x=x(x-11)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x(x-11)=0) / Because (x(x-11)=0). (x-2 -11x=x(x-11)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 3
Exam Tip
(x-2 -11x=x(x-11)), इसलिए शून्य गुणनफल नियम से (x=0) मिलता है। परीक्षा में चर से भाग देकर यह मूल न खोएं।
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\(x^2=169\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=169\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm13\)
B (x=13)
C (x=-13)
D \(x=\pm169\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm13\)
Step 1
Concept
\(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm13\). \(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{169}=\pm13\) होता है। परीक्षा में केवल (13) लिखना अधूरा उत्तर है।
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\(x^2+18x+81=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+18x+81=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=-9)
B (x=9)
C (x=-18)
D (x=18)
Explanation opens after your attempt
Step 1
Concept
((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-9). ((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 3
Exam Tip
((x+9)2 =0), इसलिए दोहराया हुआ मूल (-9) है। परीक्षा में पूर्ण वर्ग समीकरण में दोनों मूल समान होते हैं।
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वर्गमूल विधि से \(x^2=144\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=144\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm12\)
B (x=12)
C (x=-12)
D \(x=\pm72\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm12\)
Step 1
Concept
\(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm12\). \(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{144}=\pm12\) होता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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\(x^2-5x=0\) में (x=0) मूल क्यों है?
Why is (x=0) a root of \(x^2-5x=0\)?
#quadratic
#zero-root
#common-mistake
A क्योंकि (x(x-5)=0) / Because (x(x-5)=0)
B क्योंकि \(x^2=5\) / Because \(x^2=5\)
C क्योंकि (x=5x) / Because (x=5x)
D क्योंकि (x-5=5) / Because (x-5=5)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (x(x-5)=0) / Because (x(x-5)=0)
Step 1
Concept
(x-2 -5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x(x-5)=0) / Because (x(x-5)=0). (x-2 -5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.
Step 3
Exam Tip
(x-2 -5x=x(x-5)), इसलिए शून्य गुणनफल नियम से (x=0) मिलता है। परीक्षा में चर से भाग देकर यह मूल न खोएं।
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\(x^2=121\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=121\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm11\)
B (x=11)
C (x=-11)
D \(x=\pm121\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm11\)
Step 1
Concept
\(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm11\). \(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{121}=\pm11\) होता है। परीक्षा में केवल (11) लिखना अधूरा उत्तर है।
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\(x^2+14x+49=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+14x+49=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=-7)
B (x=7)
C (x=-14)
D (x=14)
Explanation opens after your attempt
Step 1
Concept
((x+7)2 =0), so the repeated root is (-7). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. (x=-7). ((x+7)2 =0), so the repeated root is (-7). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
((x+7)2 =0), इसलिए दोहराया हुआ मूल (-7) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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वर्गमूल विधि से \(x^2=64\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=64\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm8\)
B (x=8)
C (x=-8)
D \(x=\pm32\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm8\)
Step 1
Concept
\(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm8\). \(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{64}=\pm8\) होता है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(x^2-12x+36=0\) का मूल क्या होगा?
What will be the root of \(x^2-12x+36=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=6)
B (x=-6)
C (x=12)
D (x=-12)
Explanation opens after your attempt
Step 1
Concept
((x-6)2 =0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). ((x-6)2 =0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.
Step 3
Exam Tip
((x-6)2 =0), इसलिए दोनों समान मूल (x=6) हैं। परीक्षा में पूर्ण वर्ग समीकरण में दोहराया हुआ मूल मिलता है।
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