\(8x^2-32x=0\) को हल करते समय कौनसी गलती नहीं करनी चाहिए?
Which mistake should be avoided while solving \(8x^2-32x=0\)?
#quadratic
#common-mistake
#zero-root
A (x=0) को छोड़ना / Missing (x=0)
B (8x) सामान्य गुणनखंड लेना / Taking (8x) common
C (8x(x-4)=0) लिखना / Writing (8x(x-4)=0)
D (x=4) को मूल मानना / Taking (x=4) as a root
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ना / Missing (x=0)
Step 1
Concept
(8x-2 -32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ना / Missing (x=0). (8x-2 -32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).
Step 3
Exam Tip
(8x-2 -32x=8x(x-4)), इसलिए (x=0) और (x=4) दोनों मूल हैं। परीक्षा में चर से भाग देने पर (x=0) छूट सकता है।
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\(6x^2-18x=0\) को हल करते समय कौनसी गलती नहीं करनी चाहिए?
Which mistake should be avoided while solving \(6x^2-18x=0\)?
#quadratic
#common-mistake
#zero-root
A (x=0) को छोड़ना / Missing (x=0)
B (6x) सामान्य गुणनखंड लेना / Taking (6x) common
C (6x(x-3)=0) लिखना / Writing (6x(x-3)=0)
D (x=3) को मूल मानना / Taking (x=3) as a root
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ना / Missing (x=0)
Step 1
Concept
(6x-2 -18x=6x(x-3)), so (x=0) and (x=3) are both roots. In exams, dividing by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ना / Missing (x=0). (6x-2 -18x=6x(x-3)), so (x=0) and (x=3) are both roots. In exams, dividing by the variable can miss (x=0).
Step 3
Exam Tip
(6x-2 -18x=6x(x-3)), इसलिए (x=0) और (x=3) दोनों मूल हैं। परीक्षा में चर से भाग देने पर (x=0) छूट सकता है।
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\(5x^2-20x=0\) को हल करते समय कौनसी सामान्य गलती हो सकती है?
What common mistake can occur while solving \(5x^2-20x=0\)?
#quadratic
#common-mistake
#zero-root
A (x=0) को छोड़ देना / Missing (x=0)
B (x=4) को लिखना / Writing (x=4)
C (5x) सामान्य गुणनखंड लेना / Taking (5x) common
D (5x(x-4)=0) लिखना / Writing (5x(x-4)=0)
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ देना / Missing (x=0)
Step 1
Concept
The correct form is (5x(x-4)=0), giving (x=0) and (x=4). In exams, dividing directly by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ देना / Missing (x=0). The correct form is (5x(x-4)=0), giving (x=0) and (x=4). In exams, dividing directly by the variable can miss (x=0).
Step 3
Exam Tip
सही रूप (5x(x-4)=0) है, जिससे (x=0) और (x=4) मिलते हैं। परीक्षा में चर से सीधे भाग देने से (x=0) छूट सकता है।
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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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\(x^2-5x=0\) 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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\(x^2-3x=0\) में (x=0) क्यों एक मूल है?
Why is (x=0) a root of \(x^2-3x=0\)?
#quadratic
#common-mistake
#zero-root
A क्योंकि (x(x-3)=0) / Because (x(x-3)=0)
B क्योंकि \(x^2=3\) / Because \(x^2=3\)
C क्योंकि (x-3=3) / Because (x-3=3)
D क्योंकि (x=3x) / Because (x=3x)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (x(x-3)=0) / Because (x(x-3)=0)
Step 1
Concept
(x-2 -3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x(x-3)=0) / Because (x(x-3)=0). (x-2 -3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).
Step 3
Exam Tip
(x-2 -3x=x(x-3)), इसलिए शून्य गुणनफल नियम से (x=0) मिलता है। परीक्षा में (x) से भाग देकर (x=0) न खोएं।
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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=0) समीकरण \(2x^2+mx+n=0\) का मूल है तो (n) का मान क्या होगा?
If (x=0) is a root of \(2x^2+mx+n=0\), what will be the value of (n)?
#roots
#zero_root
#condition
A (0)
B (2)
C (m)
D (1)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) leaves only (n). Therefore (n=0) must hold.
Step 2
Why this answer is correct
The correct answer is A. (0). Putting (x=0) leaves only (n). Therefore (n=0) must hold.
Step 3
Exam Tip
(x=0) रखने पर केवल (n) बचता है। इसलिए (n=0) होना चाहिए।
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यदि (x=0) समीकरण \(ax^2+bx+c=0\) का मूल है और \(a\ne0\) है तो कौन सा निष्कर्ष सही है?
If (x=0) is a root of \(ax^2+bx+c=0\) and \(a\ne0\), which conclusion is correct?
#roots
#zero_root
#condition
A (c=0)
B (a=0)
C (b=0)
D \(c\ne0\)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) makes the equation (c=0). Therefore the constant term is zero for a zero root.
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) makes the equation (c=0). Therefore the constant term is zero for a zero root.
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य मूल के लिए अचर पद शून्य होता है।
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यदि (x=0) समीकरण \(ax^2+bx+c=0\) का मूल है और \(a\ne0\) है तो (c) के बारे में क्या सही है?
If (x=0) is a root of \(ax^2+bx+c=0\) and \(a\ne0\), what is true about (c)?
#roots
#zero_root
#condition
A (c=0)
B (c=a)
C (c=b)
D \(c\ne0\)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) makes the equation (c=0). Therefore the constant term is zero for a zero root.
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) makes the equation (c=0). Therefore the constant term is zero for a zero root.
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य मूल के लिए अचर पद शून्य होता है।
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यदि (0) समीकरण \(5x^2+kx+c=0\) का मूल है तो कौन सा पद शून्य होना चाहिए?
If (0) is a root of \(5x^2+kx+c=0\), which term must be zero?
#roots
#zero_root
#condition
A अचर पद (c) / Constant term (c)
B गुणांक (5) / Coefficient (5)
C मध्य गुणांक (k) / Middle coefficient (k)
D चर (x) / Variable (x)
Explanation opens after your attempt
Correct Answer
A. अचर पद (c) / Constant term (c)
Step 1
Concept
Putting (x=0) leaves only (c). Therefore for a zero root, (c=0) must hold.
Step 2
Why this answer is correct
The correct answer is A. अचर पद (c) / Constant term (c). Putting (x=0) leaves only (c). Therefore for a zero root, (c=0) must hold.
Step 3
Exam Tip
(x=0) रखने पर केवल (c) बचता है। इसलिए शून्य मूल के लिए (c=0) होना चाहिए।
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समीकरण \(x^2+6x=0\) के मूल कौन से हैं?
What are the roots of \(x^2+6x=0\)?
#roots
#zero_root
#common_factor
A (0) और (-6) / (0) and (-6)
B (0) और (6) / (0) and (6)
C (6) और (-6) / (6) and (-6)
D (1) और (-6) / (1) and (-6)
Explanation opens after your attempt
Correct Answer
A. (0) और (-6) / (0) and (-6)
Step 1
Concept
(x-2 +6x=x(x+6)). Therefore the roots are (0) and (-6).
Step 2
Why this answer is correct
The correct answer is A. (0) और (-6) / (0) and (-6). (x-2 +6x=x(x+6)). Therefore the roots are (0) and (-6).
Step 3
Exam Tip
(x-2 +6x=x(x+6)) है। इसलिए मूल (0) और (-6) हैं।
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किस समीकरण के मूल (0) और (8) हैं?
Which equation has roots (0) and (8)?
#roots
#equation_from_roots
#zero_root
A \(x^2-8x=0\)
B \(x^2+8x=0\)
C \(x^2-64=0\)
D \(x^2+64=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x=0\)
Step 1
Concept
With roots (0) and (8), the equation is (x(x-8)=0). Expanding it gives \(x^2-8x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x=0\). With roots (0) and (8), the equation is (x(x-8)=0). Expanding it gives \(x^2-8x=0\).
Step 3
Exam Tip
मूल (0) और (8) होने पर समीकरण (x(x-8)=0) होगा। इसे खोलने पर \(x^2-8x=0\) मिलता है।
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यदि (0) समीकरण \(2x^2+bx+c=0\) का मूल है तो (c) का मान कैसा होगा?
If (0) is a root of \(2x^2+bx+c=0\), what must be the value of (c)?
#roots
#zero_root
#condition
A (c=0)
B (c=2)
C (c=b)
D (c=-2)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) must give (c=0). Therefore the constant term is zero for a zero root.
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) must give (c=0). Therefore the constant term is zero for a zero root.
Step 3
Exam Tip
(x=0) रखने पर (c=0) मिलना चाहिए। इसलिए शून्य मूल के लिए अचर पद शून्य होता है।
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समीकरण \(x^2-5x=0\) के मूल कौन से हैं?
What are the roots of \(x^2-5x=0\)?
#roots
#zero_root
#common_factor
A (0) और (5) / (0) and (5)
B (1) और (5) / (1) and (5)
C (-5) और (5) / (-5) and (5)
D (0) और (-5) / (0) and (-5)
Explanation opens after your attempt
Correct Answer
A. (0) और (5) / (0) and (5)
Step 1
Concept
(x-2 -5x=x(x-5)). Therefore the roots are (0) and (5).
Step 2
Why this answer is correct
The correct answer is A. (0) और (5) / (0) and (5). (x-2 -5x=x(x-5)). Therefore the roots are (0) and (5).
Step 3
Exam Tip
(x-2 -5x=x(x-5)) है। इसलिए मूल (0) और (5) हैं।
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किस समीकरण के मूल (0) और (-6) हैं?
Which equation has roots (0) and (-6)?
#roots
#equation_from_roots
#zero_root
A \(x^2+6x=0\)
B \(x^2-6x=0\)
C \(x^2+36=0\)
D \(x^2-36=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+6x=0\)
Step 1
Concept
With roots (0) and (-6), the equation is (x(x+6)=0). Expanding it gives \(x^2+6x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+6x=0\). With roots (0) and (-6), the equation is (x(x+6)=0). Expanding it gives \(x^2+6x=0\).
Step 3
Exam Tip
मूल (0) और (-6) होने पर समीकरण (x(x+6)=0) होगा। इसे खोलने पर \(x^2+6x=0\) मिलता है।
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निम्न में से किस समीकरण का एक मूल (0) है?
Which of the following equations has one root (0)?
#roots
#zero_root
#identification
A \(x^2+5x=0\)
B \(x^2+5=0\)
C \(x^2-5x+6=0\)
D \(x^2+2x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x=0\)
Step 1
Concept
(x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x=0\). (x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 3
Exam Tip
(x-2 +5x=x(x+5)) इसलिए (x=0) एक मूल है। अचर पद (0) हो तो शून्य मूल की संभावना देखें।
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किस शर्त पर (0) समीकरण \(ax^2+bx+c=0\) का मूल होगा?
Under which condition will (0) be a root of \(ax^2+bx+c=0\)?
#roots
#zero_root
#condition
A (c=0)
B (a=0)
C (b=0)
D (a=b)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) makes the equation (c=0). So for a zero root the constant term must be zero.
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) makes the equation (c=0). So for a zero root the constant term must be zero.
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य मूल के लिए अचर पद शून्य होना चाहिए।
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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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किस समीकरण के मूल (0) और (5) हैं?
Which equation has roots (0) and (5)?
#roots
#equation_from_roots
#zero_root
A \(x^2-5x=0\)
B \(x^2+5x=0\)
C \(x^2-25=0\)
D \(x^2+25=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-5x=0\)
Step 1
Concept
With roots (0) and (5) the equation is (x(x-5)=0) that is \(x^2-5x=0\). Remember the form (x-r) while forming an equation from roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-5x=0\). With roots (0) and (5) the equation is (x(x-5)=0) that is \(x^2-5x=0\). Remember the form (x-r) while forming an equation from roots.
Step 3
Exam Tip
मूल (0) और (5) होने पर समीकरण (x(x-5)=0) अर्थात \(x^2-5x=0\) होगा। मूलों से समीकरण बनाते समय (x-r) रूप याद रखें।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?
In which equation is (x=0) one root and the other root negative?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2+7x=0\)
B \(x^2-7x=0\)
C \(x^2+7=0\)
D \(x^2-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x=0\)
Step 1
Concept
(x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x=0\). (x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 3
Exam Tip
(x-2 +7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल धनात्मक है?
In which equation is (x=0) one root and the other root positive?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2-6x=0\)
B \(x^2+6x=0\)
C \(x^2+6=0\)
D \(x^2-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x=0\)
Step 1
Concept
(x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x=0\). (x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 3
Exam Tip
(x-2 -6x=x(x-6)), इसलिए मूल (0) और (6) हैं। दूसरा मूल धनात्मक है।
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