यदि ((x-6)(x-13)=0), तो शून्य गुणनफल नियम से हल क्या होंगे?
If ((x-6)(x-13)=0), what are the solutions by zero product rule?
#quadratic
#zero-product
#method
A (x=6,13)
B (x=-6,-13)
C (x=0,19)
D (x=1,78)
Explanation opens after your attempt
Correct Answer
A. (x=6,13)
Step 1
Concept
((x-6)=0) or ((x-13)=0), so (x=6) or (x=13). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=6,13). ((x-6)=0) or ((x-13)=0), so (x=6) or (x=13). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
((x-6)=0) या ((x-13)=0), इसलिए (x=6) या (x=13) है। परीक्षा में हर गुणनखंड को अलग शून्य रखें।
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यदि ((x-5)(x-11)=0), तो शून्य गुणनफल नियम से हल क्या होंगे?
If ((x-5)(x-11)=0), what are the solutions by zero product rule?
#quadratic
#zero-product
#method
A (x=5,11)
B (x=-5,-11)
C (x=0,16)
D (x=1,55)
Explanation opens after your attempt
Correct Answer
A. (x=5,11)
Step 1
Concept
((x-5)=0) or ((x-11)=0), so (x=5) or (x=11). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=5,11). ((x-5)=0) or ((x-11)=0), so (x=5) or (x=11). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
((x-5)=0) या ((x-11)=0), इसलिए (x=5) या (x=11) है। परीक्षा में हर गुणनखंड को अलग शून्य रखें।
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यदि ((x-4)(x-9)=0), तो शून्य गुणनफल नियम से हल क्या होंगे?
If ((x-4)(x-9)=0), what are the solutions by zero product rule?
#quadratic
#zero-product
#method
A (x=4,9)
B (x=-4,-9)
C (x=0,13)
D (x=1,36)
Explanation opens after your attempt
Correct Answer
A. (x=4,9)
Step 1
Concept
((x-4)=0) or ((x-9)=0), so (x=4) or (x=9). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=4,9). ((x-4)=0) or ((x-9)=0), so (x=4) or (x=9). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
((x-4)=0) या ((x-9)=0), इसलिए (x=4) या (x=9) है। परीक्षा में हर गुणनखंड को अलग शून्य रखें।
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यदि ((x-3)(x-7)=0), तो शून्य गुणनफल नियम से हल क्या होंगे?
If ((x-3)(x-7)=0), what are the solutions by zero product rule?
#quadratic
#zero-product
#method
A (x=3,7)
B (x=-3,-7)
C (x=0,10)
D (x=1,21)
Explanation opens after your attempt
Correct Answer
A. (x=3,7)
Step 1
Concept
((x-3)=0) or ((x-7)=0), so (x=3) or (x=7). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=3,7). ((x-3)=0) or ((x-7)=0), so (x=3) or (x=7). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
((x-3)=0) या ((x-7)=0), इसलिए (x=3) या (x=7) है। परीक्षा में हर गुणनखंड को अलग शून्य रखें।
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यदि ((x-2)(x-5)=m) और (m=0), तो हल क्या होंगे?
If ((x-2)(x-5)=m) and (m=0), what are the solutions?
#quadratic
#zero-product
#parameter-case
A (x=2,5)
B (x=-2,-5)
C (x=0,7)
D (x=1,10)
Explanation opens after your attempt
Correct Answer
A. (x=2,5)
Step 1
Concept
Putting (m=0) gives ((x-2)(x-5)=0), so (x=2) or (x=5). In exams, apply zero product rule directly.
Step 2
Why this answer is correct
The correct answer is A. (x=2,5). Putting (m=0) gives ((x-2)(x-5)=0), so (x=2) or (x=5). In exams, apply zero product rule directly.
Step 3
Exam Tip
(m=0) रखने पर ((x-2)(x-5)=0), इसलिए (x=2) या (x=5) है। परीक्षा में शून्य गुणनफल नियम सीधे लगाएं।
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\(11x^2+12x+1=0\) के मूल क्या होंगे?
What will be the roots of \(11x^2+12x+1=0\)?
#quadratic
#fraction-roots
#zero-product
A \(x=-\frac{1}{11},-1\)
B \(x=\frac{1}{11},1\)
C (x=-11,-1)
D (x=11,1)
Explanation opens after your attempt
Correct Answer
A. \(x=-\frac{1}{11},-1\)
Step 1
Concept
((11x+1)(x+1)=0), so \(x=-\frac{1}{11}\) and (-1). In exams, ((11x+1)=0) gives \(-\frac{1}{11}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{1}{11},-1\). ((11x+1)(x+1)=0), so \(x=-\frac{1}{11}\) and (-1). In exams, ((11x+1)=0) gives \(-\frac{1}{11}\).
Step 3
Exam Tip
((11x+1)(x+1)=0), इसलिए \(x=-\frac{1}{11}\) और (-1) हैं। परीक्षा में ((11x+1)=0) से \(-\frac{1}{11}\) मिलता है।
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\(11x^2=77x\) को हल करने का सही तरीका क्या है?
What is the correct way to solve \(11x^2=77x\)?
#quadratic
#common-factor
#zero-product
A (11x(x-7)=0) लिखना / Write (11x(x-7)=0)
B (x=7) ही लिखना / Write only (x=7)
C (11x=77) लिखना / Write (11x=77)
D \(x^2=7x\) के बाद (x=7) ही लेना / After \(x^2=7x\), take only (x=7)
Explanation opens after your attempt
Correct Answer
A. (11x(x-7)=0) लिखना / Write (11x(x-7)=0)
Step 1
Concept
From \(11x^2-77x=0\), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (11x(x-7)=0) लिखना / Write (11x(x-7)=0). From \(11x^2-77x=0\), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(11x^2-77x=0\) से (11x(x-7)=0), इसलिए (x=0) और (x=7) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
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\(7x^2+8x+1=0\) के मूल क्या होंगे?
What will be the roots of \(7x^2+8x+1=0\)?
#quadratic
#fraction-roots
#zero-product
A \(x=-\frac{1}{7},-1\)
B \(x=\frac{1}{7},1\)
C (x=-7,-1)
D (x=7,1)
Explanation opens after your attempt
Correct Answer
A. \(x=-\frac{1}{7},-1\)
Step 1
Concept
((7x+1)(x+1)=0), so \(x=-\frac{1}{7}\) and (-1). In exams, ((7x+1)=0) gives \(-\frac{1}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{1}{7},-1\). ((7x+1)(x+1)=0), so \(x=-\frac{1}{7}\) and (-1). In exams, ((7x+1)=0) gives \(-\frac{1}{7}\).
Step 3
Exam Tip
((7x+1)(x+1)=0), इसलिए \(x=-\frac{1}{7}\) और (-1) हैं। परीक्षा में ((7x+1)=0) से \(-\frac{1}{7}\) मिलता है।
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\(9x^2=45x\) को हल करने का सही तरीका क्या है?
What is the correct way to solve \(9x^2=45x\)?
#quadratic
#common-factor
#zero-product
A (9x(x-5)=0) लिखना / Write (9x(x-5)=0)
B (x=5) ही लिखना / Write only (x=5)
C (9x=45) लिखना / Write (9x=45)
D \(x^2=5x\) के बाद (x=5) ही लेना / After \(x^2=5x\), take only (x=5)
Explanation opens after your attempt
Correct Answer
A. (9x(x-5)=0) लिखना / Write (9x(x-5)=0)
Step 1
Concept
From \(9x^2-45x=0\), (9x(x-5)=0), so (x=0) and (x=5). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (9x(x-5)=0) लिखना / Write (9x(x-5)=0). From \(9x^2-45x=0\), (9x(x-5)=0), so (x=0) and (x=5). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(9x^2-45x=0\) से (9x(x-5)=0), इसलिए (x=0) और (x=5) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
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\(5x^2+6x+1=0\) के मूल क्या होंगे?
What will be the roots of \(5x^2+6x+1=0\)?
#quadratic
#fraction-roots
#zero-product
A \(x=-\frac{1}{5},-1\)
B \(x=\frac{1}{5},1\)
C (x=-5,-1)
D (x=5,1)
Explanation opens after your attempt
Correct Answer
A. \(x=-\frac{1}{5},-1\)
Step 1
Concept
((5x+1)(x+1)=0), so \(x=-\frac{1}{5}\) and (-1). In exams, ((5x+1)=0) gives \(-\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{1}{5},-1\). ((5x+1)(x+1)=0), so \(x=-\frac{1}{5}\) and (-1). In exams, ((5x+1)=0) gives \(-\frac{1}{5}\).
Step 3
Exam Tip
((5x+1)(x+1)=0), इसलिए \(x=-\frac{1}{5}\) और (-1) हैं। परीक्षा में ((5x+1)=0) से \(-\frac{1}{5}\) मिलता है।
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\(7x^2=28x\) को हल करने का सही तरीका क्या है?
What is the correct way to solve \(7x^2=28x\)?
#quadratic
#common-factor
#zero-product
A (7x(x-4)=0) लिखना / Write (7x(x-4)=0)
B (x=4) ही लिखना / Write only (x=4)
C (7x=28) लिखना / Write (7x=28)
D \(x^2=4x\) के बाद (x=4) ही लेना / After \(x^2=4x\), take only (x=4)
Explanation opens after your attempt
Correct Answer
A. (7x(x-4)=0) लिखना / Write (7x(x-4)=0)
Step 1
Concept
From \(7x^2-28x=0\), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (7x(x-4)=0) लिखना / Write (7x(x-4)=0). From \(7x^2-28x=0\), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(7x^2-28x=0\) से (7x(x-4)=0), इसलिए (x=0) और (x=4) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
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\(3x^2-11x+6=0\) के मूल क्या हैं?
What are the roots of \(3x^2-11x+6=0\)?
#quadratic
#fraction-roots
#zero-product
A \(x=3,\frac{2}{3}\)
B \(x=-3,-\frac{2}{3}\)
C (x=2,3)
D \(x=\frac{3}{2},3\)
Explanation opens after your attempt
Correct Answer
A. \(x=3,\frac{2}{3}\)
Step 1
Concept
((3x-2)(x-3)=0), so \(x=\frac{2}{3}\) and (x=3). In exams, solve (3x-2=0) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=3,\frac{2}{3}\). ((3x-2)(x-3)=0), so \(x=\frac{2}{3}\) and (x=3). In exams, solve (3x-2=0) carefully.
Step 3
Exam Tip
((3x-2)(x-3)=0), इसलिए \(x=\frac{2}{3}\) और (x=3) हैं। परीक्षा में (3x-2=0) को ध्यान से हल करें।
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\(9x^2-27x=0\) के मूल क्या हैं?
What are the roots of \(9x^2-27x=0\)?
#quadratic
#zero-product
#roots
A (x=0,3)
B (x=0,-3)
C (x=9,3)
D (x=-9,-3)
Explanation opens after your attempt
Correct Answer
A. (x=0,3)
Step 1
Concept
(9x(x-3)=0), so (x=0) or (x=3). In exams, apply zero product rule directly.
Step 2
Why this answer is correct
The correct answer is A. (x=0,3). (9x(x-3)=0), so (x=0) or (x=3). In exams, apply zero product rule directly.
Step 3
Exam Tip
(9x(x-3)=0), इसलिए (x=0) या (x=3) है। परीक्षा में शून्य गुणनफल नियम सीधे लगाएं।
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\(4x^2+28x=0\) के मूल क्या हैं?
What are the roots of \(4x^2+28x=0\)?
#quadratic
#zero-product
#roots
A (x=0,-7)
B (x=0,7)
C (x=4,-7)
D (x=-4,7)
Explanation opens after your attempt
Correct Answer
A. (x=0,-7)
Step 1
Concept
(4x(x+7)=0), so (x=0) or (x=-7). In exams, do not miss the root (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0,-7). (4x(x+7)=0), so (x=0) or (x=-7). In exams, do not miss the root (x=0).
Step 3
Exam Tip
(4x(x+7)=0), इसलिए (x=0) या (x=-7) है। परीक्षा में (x=0) वाला मूल न छोड़ें।
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\(2x^2-7x+5=0\) के मूल क्या हैं?
What are the roots of \(2x^2-7x+5=0\)?
#quadratic
#fraction-roots
#zero-product
A \(x=1,\frac{5}{2}\)
B \(x=-1,-\frac{5}{2}\)
C (x=2,5)
D \(x=\frac{1}{2},5\)
Explanation opens after your attempt
Correct Answer
A. \(x=1,\frac{5}{2}\)
Step 1
Concept
((2x-5)(x-1)=0), so \(x=\frac{5}{2}\) and (x=1). In exams, solve (2x-5=0) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=1,\frac{5}{2}\). ((2x-5)(x-1)=0), so \(x=\frac{5}{2}\) and (x=1). In exams, solve (2x-5=0) carefully.
Step 3
Exam Tip
((2x-5)(x-1)=0), इसलिए \(x=\frac{5}{2}\) और (x=1) हैं। परीक्षा में (2x-5=0) को ध्यान से हल करें।
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\(7x^2-14x=0\) के मूल क्या हैं?
What are the roots of \(7x^2-14x=0\)?
#quadratic
#zero-product
#roots
A (x=0,2)
B (x=0,-2)
C (x=7,2)
D (x=-7,-2)
Explanation opens after your attempt
Correct Answer
A. (x=0,2)
Step 1
Concept
(7x(x-2)=0), so (x=0) or (x=2). In exams, apply zero product rule directly.
Step 2
Why this answer is correct
The correct answer is A. (x=0,2). (7x(x-2)=0), so (x=0) or (x=2). In exams, apply zero product rule directly.
Step 3
Exam Tip
(7x(x-2)=0), इसलिए (x=0) या (x=2) है। परीक्षा में शून्य गुणनफल नियम सीधे लगाएं।
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\(5x^2+15x=0\) के मूल क्या हैं?
What are the roots of \(5x^2+15x=0\)?
#quadratic
#zero-product
#roots
A (x=0,-3)
B (x=0,3)
C (x=5,-3)
D (x=-5,3)
Explanation opens after your attempt
Correct Answer
A. (x=0,-3)
Step 1
Concept
(5x(x+3)=0), so (x=0) or (x=-3). In exams, do not miss the root (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0,-3). (5x(x+3)=0), so (x=0) or (x=-3). In exams, do not miss the root (x=0).
Step 3
Exam Tip
(5x(x+3)=0), इसलिए (x=0) या (x=-3) है। परीक्षा में (x=0) वाला मूल न छोड़ें।
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\(x^2-16x=0\) के मूल क्या हैं?
What are the roots of \(x^2-16x=0\)?
#quadratic
#roots
#zero-product
A (x=0,16)
B (x=0,-16)
C (x=1,16)
D (x=-1,16)
Explanation opens after your attempt
Correct Answer
A. (x=0,16)
Step 1
Concept
(x(x-16)=0), so (x=0) or (x=16). In exams, use zero product rule on each factor separately.
Step 2
Why this answer is correct
The correct answer is A. (x=0,16). (x(x-16)=0), so (x=0) or (x=16). In exams, use zero product rule on each factor separately.
Step 3
Exam Tip
(x(x-16)=0), इसलिए (x=0) या (x=16) है। परीक्षा में शून्य गुणनफल नियम से हर गुणनखंड को अलग रखें।
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\(x^2-16x=0\) को हल करने का सही पहला कदम कौनसा है?
What is the correct first step to solve \(x^2-16x=0\)?
#quadratic
#common-factor
#zero-product
A (x(x-16)=0) लिखना / Write (x(x-16)=0)
B \(x^2=16\) लिखना / Write \(x^2=16\)
C (x-16=0) ही लेना / Take only (x-16=0)
D (x=16x) लिखना / Write (x=16x)
Explanation opens after your attempt
Correct Answer
A. (x(x-16)=0) लिखना / Write (x(x-16)=0)
Step 1
Concept
The common factor in \(x^2-16x\) is (x), so we write (x(x-16)=0). In exams, do not divide by the variable and lose (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x(x-16)=0) लिखना / Write (x(x-16)=0). The common factor in \(x^2-16x\) is (x), so we write (x(x-16)=0). In exams, do not divide by the variable and lose (x=0).
Step 3
Exam Tip
\(x^2-16x\) में सामान्य गुणनखंड (x) है, इसलिए (x(x-16)=0) लिखते हैं। परीक्षा में चर से भाग देकर (x=0) को न छोड़ें।
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\(x^2+5x=0\) को हल करने का सही पहला कदम क्या है?
What is the correct first step to solve \(x^2+5x=0\)?
#quadratic
#common-factor
#zero-product
A (x(x+5)=0) लिखना / Write (x(x+5)=0)
B \(x^2=5x\) लिखना / Write \(x^2=5x\)
C (x+5=0) ही लेना / Take only (x+5=0)
D (x=5) मान लेना / Assume (x=5)
Explanation opens after your attempt
Correct Answer
A. (x(x+5)=0) लिखना / Write (x(x+5)=0)
Step 1
Concept
Taking common factor (x) gives (x(x+5)=0). In exams, take the common factor first.
Step 2
Why this answer is correct
The correct answer is A. (x(x+5)=0) लिखना / Write (x(x+5)=0). Taking common factor (x) gives (x(x+5)=0). In exams, take the common factor first.
Step 3
Exam Tip
सामान्य गुणनखंड (x) निकालने पर (x(x+5)=0) मिलता है। परीक्षा में सामान्य गुणनखंड पहले निकालें।
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(x-2 -2(a-3)x+a-2 -16=0) की जड़ों का गुणनफल (0) हो, तो (a) के मान क्या हैं?
If the product of roots of (x-2 -2(a-3)x+a-2 -16=0) is (0), what are the values of (a)?
#quadratic-roots
#zero-product
#parameter
A (4) और (-4) / (4) and (-4)
B (3) और (-3) / (3) and (-3)
C (8) और (-8) / (8) and (-8)
D कोई मान नहीं / No value
Explanation opens after your attempt
Correct Answer
A. (4) और (-4) / (4) and (-4)
Step 1
Concept
The product of roots is \(a^2-16\). Setting it equal to (0) gives \(a^2=16\), so (a=4) or (a=-4).
Step 2
Why this answer is correct
The correct answer is A. (4) और (-4) / (4) and (-4). The product of roots is \(a^2-16\). Setting it equal to (0) gives \(a^2=16\), so (a=4) or (a=-4).
Step 3
Exam Tip
जड़ों का गुणनफल \(a^2-16\) है। इसे (0) रखने पर \(a^2=16\), इसलिए (a=4) या (a=-4)।
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(x-2 -2(a-2)x+a-2 -9=0) की जड़ों का गुणनफल (0) हो, तो (a) के मान क्या हैं?
If the product of the roots of (x-2 -2(a-2)x+a-2 -9=0) is (0), what are the values of (a)?
#quadratic-roots
#zero-product
#parameter
A (0) और (9) / (0) and (9)
B (3) और (-3) / (3) and (-3)
C (2) और (-2) / (2) and (-2)
D कोई मान नहीं / No value
Explanation opens after your attempt
Correct Answer
B. (3) और (-3) / (3) and (-3)
Step 1
Concept
The product of roots is \(a^2-9\). Setting it equal to (0) gives \(a^2=9\), so (a=3) or (a=-3).
Step 2
Why this answer is correct
The correct answer is B. (3) और (-3) / (3) and (-3). The product of roots is \(a^2-9\). Setting it equal to (0) gives \(a^2=9\), so (a=3) or (a=-3).
Step 3
Exam Tip
जड़ों का गुणनफल \(a^2-9\) है। इसे (0) रखने पर \(a^2=9\), इसलिए (a=3) या (a=-3)।
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(x-2 -2(a+1)x+a-2 -1=0) की जड़ों का गुणनफल (0) हो, तो (a) का कौन-सा मान संभव है?
If the product of the roots of (x-2 -2(a+1)x+a-2 -1=0) is (0), which value of (a) is possible?
#quadratic-roots
#zero-product
#parameter
A (1) या (-1) / (1) or (-1)
B (0) केवल / (0) only
C (2) या (-2) / (2) or (-2)
D कोई मान नहीं / No value
Explanation opens after your attempt
Correct Answer
A. (1) या (-1) / (1) or (-1)
Step 1
Concept
The product of roots is \(a^2-1\). Setting it equal to (0) gives \(a^2=1\), so (a=1) or (a=-1).
Step 2
Why this answer is correct
The correct answer is A. (1) या (-1) / (1) or (-1). The product of roots is \(a^2-1\). Setting it equal to (0) gives \(a^2=1\), so (a=1) or (a=-1).
Step 3
Exam Tip
जड़ों का गुणनफल \(a^2-1\) है। इसे (0) रखने पर \(a^2=1\), इसलिए (a=1) या (a=-1)।
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यदि (x+6) और (x-2) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x+6) and (x-2) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (-6) और (2) / (-6) and (2)
B (6) और (-2) / (6) and (-2)
C (6) और (2) / (6) and (2)
D (-6) और (-2) / (-6) and (-2)
Explanation opens after your attempt
Correct Answer
A. (-6) और (2) / (-6) and (2)
Step 1
Concept
From (x+6=0), (x=-6), and from (x-2=0), (x=2). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (-6) और (2) / (-6) and (2). From (x+6=0), (x=-6), and from (x-2=0), (x=2). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x+6=0) से (x=-6) और (x-2=0) से (x=2) मिलता है। गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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यदि (x-4) और (x+7) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x-4) and (x+7) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (4) और (-7) / (4) and (-7)
B (-4) और (7) / (-4) and (7)
C (4) और (7) / (4) and (7)
D (-4) और (-7) / (-4) and (-7)
Explanation opens after your attempt
Correct Answer
A. (4) और (-7) / (4) and (-7)
Step 1
Concept
From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (4) और (-7) / (4) and (-7). From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x-4=0) से (x=4) और (x+7=0) से (x=-7) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है।
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यदि किसी द्विघात समीकरण के मूलों का गुणनफल (0) है तो सही कथन कौन सा है?
If the product of roots of a quadratic equation is (0), which statement is correct?
#roots
#zero_product
#reasoning
A कम से कम एक मूल (0) है / At least one root is (0)
B दोनों मूल जरूर (1) हैं / Both roots must be (1)
C दोनों मूल ऋणात्मक हैं / Both roots are negative
D कोई वास्तविक मूल नहीं है / There is no real root
Explanation opens after your attempt
Correct Answer
A. कम से कम एक मूल (0) है / At least one root is (0)
Step 1
Concept
If \(\alpha\beta=0\), then \(\alpha=0\) or \(\beta=0\). If the product is zero, always check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. कम से कम एक मूल (0) है / At least one root is (0). If \(\alpha\beta=0\), then \(\alpha=0\) or \(\beta=0\). If the product is zero, always check for a zero root.
Step 3
Exam Tip
यदि \(\alpha\beta=0\) है तो \(\alpha=0\) या \(\beta=0\) होगा। गुणनफल शून्य हो तो शून्य मूल जरूर देखें।
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यदि (x+2) और (x-5) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x+2) and (x-5) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (-2) और (5) / (-2) and (5)
B (2) और (-5) / (2) and (-5)
C (2) और (5) / (2) and (5)
D (-2) और (-5) / (-2) and (-5)
Explanation opens after your attempt
Correct Answer
A. (-2) और (5) / (-2) and (5)
Step 1
Concept
From (x+2=0), (x=-2), and from (x-5=0), (x=5). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (-2) और (5) / (-2) and (5). From (x+2=0), (x=-2), and from (x-5=0), (x=5). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x+2=0) से (x=-2) और (x-5=0) से (x=5) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है।
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