\(x^2+14x+49=0\) को किस रूप में लिखा जा सकता है?
In which form can \(x^2+14x+49=0\) be written?
#quadratic
#perfect-square
#factorisation
A ((x+7)2 =0)
B ((x-7)2 =0)
C ((x+14)2 =0)
D ((x-14)2 =0)
Explanation opens after your attempt
Correct Answer
A. ((x+7)2 =0)
Step 1
Concept
(x-2 +14x+49=(x+7)2 ), so it is a perfect square. In exams, match it with \(2\cdot7x=14x\).
Step 2
Why this answer is correct
The correct answer is A. ((x+7)2 =0). (x-2 +14x+49=(x+7)2 ), so it is a perfect square. In exams, match it with \(2\cdot7x=14x\).
Step 3
Exam Tip
(x-2 +14x+49=(x+7)2 ), इसलिए यह पूर्ण वर्ग है। परीक्षा में \(2\cdot7x=14x\) से मिलान करें।
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\(x^2-4x-21=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2-4x-21=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x-7)(x+3))
B ((x+7)(x-3))
C ((x-21)(x+1))
D ((x+7)(x+3))
Explanation opens after your attempt
Correct Answer
A. ((x-7)(x+3))
Step 1
Concept
Since (-7+3=-4) and \(-7\times3=-21\), ((x-7)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 2
Why this answer is correct
The correct answer is A. ((x-7)(x+3)). Since (-7+3=-4) and \(-7\times3=-21\), ((x-7)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 3
Exam Tip
(-7+3=-4) और \(-7\times3=-21\), इसलिए ((x-7)(x+3)) सही है। परीक्षा में मिश्रित चिन्ह वाले गुणनखंड ध्यान से चुनें।
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\(x^2+11x+30=0\) को गुणनखंड रूप में कैसे लिखा जाएगा?
How will \(x^2+11x+30=0\) be written in factorised form?
#quadratic
#factorisation
#signs
A ((x+5)(x+6)=0)
B ((x-5)(x-6)=0)
C ((x+3)(x+10)=0)
D ((x+2)(x+15)=0)
Explanation opens after your attempt
Correct Answer
A. ((x+5)(x+6)=0)
Step 1
Concept
(5+6=11) and \(5\times6=30\), so the correct factors are ((x+5)(x+6)). In exams, positive (c) and positive (b) give both positive signs.
Step 2
Why this answer is correct
The correct answer is A. ((x+5)(x+6)=0). (5+6=11) and \(5\times6=30\), so the correct factors are ((x+5)(x+6)). In exams, positive (c) and positive (b) give both positive signs.
Step 3
Exam Tip
(5+6=11) और \(5\times6=30\), इसलिए सही गुणनखंड ((x+5)(x+6)) हैं। परीक्षा में धनात्मक (c) और धनात्मक (b) पर दोनों चिन्ह धनात्मक होते हैं।
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गुणनखंड विधि से \(x^2-9x+20=0\) के मूल क्या होंगे?
Using factorisation method, what will be the roots of \(x^2-9x+20=0\)?
#quadratic
#factorisation
#roots
A (x=4,5)
B (x=-4,-5)
C (x=2,10)
D (x=1,20)
Explanation opens after your attempt
Correct Answer
A. (x=4,5)
Step 1
Concept
(x-2 -9x+20=(x-4)(x-5)), so the roots are (4) and (5). In exams, check both sum and product.
Step 2
Why this answer is correct
The correct answer is A. (x=4,5). (x-2 -9x+20=(x-4)(x-5)), so the roots are (4) and (5). In exams, check both sum and product.
Step 3
Exam Tip
(x-2 -9x+20=(x-4)(x-5)), इसलिए मूल (4) और (5) हैं। परीक्षा में योग और गुणनफल दोनों जांचें।
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\(x^2+x-6=0\) के लिए सही गुणनखंड कौनसे हैं?
What are the correct factors for \(x^2+x-6=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x+3)(x-2))
B ((x-3)(x+2))
C ((x+6)(x-1))
D ((x-6)(x+1))
Explanation opens after your attempt
Correct Answer
A. ((x+3)(x-2))
Step 1
Concept
(3+(-2)=1) and \(3\times(-2)=-6\), so ((x+3)(x-2)) is correct. In exams, match signs with the product.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)(x-2)). (3+(-2)=1) and \(3\times(-2)=-6\), so ((x+3)(x-2)) is correct. In exams, match signs with the product.
Step 3
Exam Tip
(3+(-2)=1) और \(3\times(-2)=-6\), इसलिए ((x+3)(x-2)) सही है। परीक्षा में चिन्हों को गुणनफल से मिलाएं।
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\(2x^2-5x+3=0\) के गुणनखंड कौनसे हैं?
What are the factors of \(2x^2-5x+3=0\)?
#quadratic
#factorisation
#middle-term
A ((2x-3)(x-1))
B ((2x+3)(x+1))
C ((2x-1)(x-3))
D ((x-3)(x-2))
Explanation opens after your attempt
Correct Answer
A. ((2x-3)(x-1))
Step 1
Concept
(2x-2 -5x+3=(2x-3)(x-1)). In exams, multiply back to check the middle term (-5x).
Step 2
Why this answer is correct
The correct answer is A. ((2x-3)(x-1)). (2x-2 -5x+3=(2x-3)(x-1)). In exams, multiply back to check the middle term (-5x).
Step 3
Exam Tip
(2x-2 -5x+3=(2x-3)(x-1)) है। परीक्षा में गुणा करके मध्य पद (-5x) वापस जांचें।
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\(x^2-8x+15=0\) के गुणनखंड कौनसे होंगे?
What will be the factors of \(x^2-8x+15=0\)?
#quadratic
#factorisation
#negative-middle-term
A ((x-3)(x-5))
B ((x+3)(x+5))
C ((x-1)(x-15))
D ((x+1)(x+15))
Explanation opens after your attempt
Correct Answer
A. ((x-3)(x-5))
Step 1
Concept
(3+5=8) and \(3\times5=15\), so with signs we get ((x-3)(x-5)). In exams, for a negative middle term, look for both negative factors.
Step 2
Why this answer is correct
The correct answer is A. ((x-3)(x-5)). (3+5=8) and \(3\times5=15\), so with signs we get ((x-3)(x-5)). In exams, for a negative middle term, look for both negative factors.
Step 3
Exam Tip
(3+5=8) और \(3\times5=15\), इसलिए संकेतों के साथ ((x-3)(x-5)) मिलता है। परीक्षा में ऋणात्मक मध्य पद के लिए दोनों ऋणात्मक गुणनखंड देखें।
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\(4x^2-1=0\) को किस रूप में लिखा जा सकता है?
How can \(4x^2-1=0\) be written?
#quadratic
#difference-of-squares
#factorisation
A ((2x-1)(2x+1)=0)
B ((4x-1)(x+1)=0)
C ((2x-1)2 =0)
D ((x-1)(4x+1)=0)
Explanation opens after your attempt
Correct Answer
A. ((2x-1)(2x+1)=0)
Step 1
Concept
(4x-2 -1=(2x)2 -12 =(2x-1)(2x+1)). In exams, identify the squares first.
Step 2
Why this answer is correct
The correct answer is A. ((2x-1)(2x+1)=0). (4x-2 -1=(2x)2 -12 =(2x-1)(2x+1)). In exams, identify the squares first.
Step 3
Exam Tip
(4x-2 -1=(2x)2 -12 =(2x-1)(2x+1)) है। परीक्षा में पहले वर्गों को पहचानें।
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\(x^2+3x+2=0\) के मूल क्या होंगे?
What are the roots of \(x^2+3x+2=0\)?
#quadratic
#roots
#factorisation
A (x=-1,-2)
B (x=1,2)
C (x=-3,-2)
D (x=0,-2)
Explanation opens after your attempt
Correct Answer
A. (x=-1,-2)
Step 1
Concept
((x+1)(x+2)=0), so (x=-1) and (x=-2). In exams, from ((x+a)=0), write (x=-a).
Step 2
Why this answer is correct
The correct answer is A. (x=-1,-2). ((x+1)(x+2)=0), so (x=-1) and (x=-2). In exams, from ((x+a)=0), write (x=-a).
Step 3
Exam Tip
((x+1)(x+2)=0), इसलिए (x=-1) और (x=-2) हैं। परीक्षा में ((x+a)=0) से (x=-a) लिखें।
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\(x^2+3x+2=0\) को हल करने के लिए सबसे उपयुक्त आसान विधि कौनसी है?
Which easy method is most suitable to solve \(x^2+3x+2=0\)?
#quadratic
#method-selection
#factorisation
A गुणनखंड विधि / Factorisation method
B सिर्फ ग्राफ विधि / Only graph method
C प्रतिस्थापन विधि / Substitution method
D विलोपन विधि / Elimination method
Explanation opens after your attempt
Correct Answer
A. गुणनखंड विधि / Factorisation method
Step 1
Concept
It easily factors as ((x+1)(x+2)=0). In exams, factorisation is fast for questions with small coefficients.
Step 2
Why this answer is correct
The correct answer is A. गुणनखंड विधि / Factorisation method. It easily factors as ((x+1)(x+2)=0). In exams, factorisation is fast for questions with small coefficients.
Step 3
Exam Tip
यह ((x+1)(x+2)=0) में आसानी से टूटता है। परीक्षा में छोटे गुणांकों वाले प्रश्नों में गुणनखंड विधि तेज रहती है।
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\(3x^2-12x=0\) को गुणनखंड करके क्या मिलेगा?
What is obtained by factoring \(3x^2-12x=0\)?
#quadratic
#common-factor
#factorisation
A (3x(x-4)=0)
B (3(x-4)=0)
C (x(3x+12)=0)
D (3x(x+4)=0)
Explanation opens after your attempt
Correct Answer
A. (3x(x-4)=0)
Step 1
Concept
Taking common factor (3x) gives (3x(x-4)=0). In exams, take out the greatest common factor.
Step 2
Why this answer is correct
The correct answer is A. (3x(x-4)=0). Taking common factor (3x) gives (3x(x-4)=0). In exams, take out the greatest common factor.
Step 3
Exam Tip
सामान्य गुणनखंड (3x) निकालने पर (3x(x-4)=0) मिलता है। परीक्षा में सबसे बड़ा सामान्य गुणनखंड निकालें।
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\(x^2+2x-15=0\) में कौनसी संख्या जोड़ी गुणनखंड बनाने में मदद करेगी?
Which number pair helps in factoring \(x^2+2x-15=0\)?
#quadratic
#middle-term-splitting
#factorisation
A (5) और (-3) / (5) and (-3)
B (3) और (5) / (3) and (5)
C (-5) और (-3) / (-5) and (-3)
D (1) और (-15) / (1) and (-15)
Explanation opens after your attempt
Correct Answer
A. (5) और (-3) / (5) and (-3)
Step 1
Concept
(5+(-3)=2) and \(5\times(-3)=-15\), so this pair is correct. In exams, split the middle term using such a pair.
Step 2
Why this answer is correct
The correct answer is A. (5) और (-3) / (5) and (-3). (5+(-3)=2) and \(5\times(-3)=-15\), so this pair is correct. In exams, split the middle term using such a pair.
Step 3
Exam Tip
(5+(-3)=2) और \(5\times(-3)=-15\), इसलिए यह जोड़ी सही है। परीक्षा में मध्य पद को इसी जोड़ी से तोड़ें।
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\(x^2-7x+10=0\) के मूल क्या हैं?
What are the roots of \(x^2-7x+10=0\)?
#quadratic
#factorisation
#roots
A (x=2,5)
B (x=-2,-5)
C (x=1,10)
D (x=3,4)
Explanation opens after your attempt
Correct Answer
A. (x=2,5)
Step 1
Concept
(x-2 -7x+10=(x-2)(x-5)), so the roots are (2) and (5). In exams, check (2+5=7) and \(2\times5=10\).
Step 2
Why this answer is correct
The correct answer is A. (x=2,5). (x-2 -7x+10=(x-2)(x-5)), so the roots are (2) and (5). In exams, check (2+5=7) and \(2\times5=10\).
Step 3
Exam Tip
(x-2 -7x+10=(x-2)(x-5)), इसलिए मूल (2) और (5) हैं। परीक्षा में (2+5=7) और \(2\times5=10\) जांचें।
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\(x^2+4x+4=0\) किस रूप में लिखा जा सकता है?
In which form can \(x^2+4x+4=0\) be written?
#quadratic
#perfect-square
#factorisation
A ((x+2)2 =0)
B ((x-2)2 =0)
C ((x+4)2 =0)
D ((x-4)2 =0)
Explanation opens after your attempt
Correct Answer
A. ((x+2)2 =0)
Step 1
Concept
\(x^2+4x+4\) is a perfect square and equals ((x+2)2 ). In exams, recognize the pattern \(a^2+2ab+b^2\).
Step 2
Why this answer is correct
The correct answer is A. ((x+2)2 =0). \(x^2+4x+4\) is a perfect square and equals ((x+2)2 ). In exams, recognize the pattern \(a^2+2ab+b^2\).
Step 3
Exam Tip
\(x^2+4x+4\) एक पूर्ण वर्ग है और यह ((x+2)2 ) के बराबर है। परीक्षा में \(a^2+2ab+b^2\) पैटर्न पहचानें।
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\(x^2-2x-8=0\) के लिए सही गुणनखंड कौनसे हैं?
What are the correct factors for \(x^2-2x-8=0\)?
#quadratic
#factorisation
#signs
A ((x-4)(x+2))
B ((x+4)(x-2))
C ((x-8)(x+1))
D ((x+8)(x-1))
Explanation opens after your attempt
Correct Answer
A. ((x-4)(x+2))
Step 1
Concept
Since (-4+2=-2) and \(-4\times2=-8\), ((x-4)(x+2)) is correct. In exams, match both product and sum.
Step 2
Why this answer is correct
The correct answer is A. ((x-4)(x+2)). Since (-4+2=-2) and \(-4\times2=-8\), ((x-4)(x+2)) is correct. In exams, match both product and sum.
Step 3
Exam Tip
(-4+2=-2) और \(-4\times2=-8\), इसलिए ((x-4)(x+2)) सही है। परीक्षा में गुणनफल और योग दोनों मिलाएं।
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शून्य गुणनफल नियम से (x(x-4)=0) के हल क्या हैं?
Using zero product rule, what are the solutions of (x(x-4)=0)?
#quadratic
#zero-product-rule
#factorisation
A (x=0,4)
B (x=0,-4)
C (x=4,8)
D (x=1,4)
Explanation opens after your attempt
Correct Answer
A. (x=0,4)
Step 1
Concept
If (x(x-4)=0), then (x=0) or (x-4=0). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=0,4). If (x(x-4)=0), then (x=0) or (x-4=0). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
यदि (x(x-4)=0), तो (x=0) या (x-4=0) होगा। परीक्षा में हर गुणनखंड को अलग-अलग शून्य के बराबर रखें।
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गुणनखंड विधि में \(x^2+7x+12=0\) को कैसे लिखा जाएगा?
In factorisation method, how will \(x^2+7x+12=0\) be written?
#quadratic
#factorisation
#standard-form
A ((x+3)(x+4)=0)
B ((x-3)(x-4)=0)
C ((x+2)(x+6)=0)
D ((x-1)(x+12)=0)
Explanation opens after your attempt
Correct Answer
A. ((x+3)(x+4)=0)
Step 1
Concept
Because (3+4=7) and \(3\times4=12\), the correct factors are ((x+3)(x+4)). In exams, pay close attention to signs.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)(x+4)=0). Because (3+4=7) and \(3\times4=12\), the correct factors are ((x+3)(x+4)). In exams, pay close attention to signs.
Step 3
Exam Tip
क्योंकि (3+4=7) और \(3\times4=12\), इसलिए सही गुणनखंड ((x+3)(x+4)) हैं। परीक्षा में संकेतों पर विशेष ध्यान दें।
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गुणनखंड विधि से \(x^2-5x+6=0\) के मूल क्या होंगे?
Using factorisation method, what are the roots of \(x^2-5x+6=0\)?
#quadratic
#factorisation
#roots
A (x=2,3)
B (x=1,6)
C (x=-2,-3)
D (x=0,5)
Explanation opens after your attempt
Correct Answer
A. (x=2,3)
Step 1
Concept
(x-2 -5x+6=(x-2)(x-3)), so the roots are (2) and (3). In exams, first find two numbers whose product is (6) and sum is (-5).
Step 2
Why this answer is correct
The correct answer is A. (x=2,3). (x-2 -5x+6=(x-2)(x-3)), so the roots are (2) and (3). In exams, first find two numbers whose product is (6) and sum is (-5).
Step 3
Exam Tip
(x-2 -5x+6=(x-2)(x-3)), इसलिए मूल (2) और (3) हैं। परीक्षा में पहले ऐसे दो संख्याएं खोजें जिनका गुणनफल (6) और योग (-5) हो।
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\(x^2+3x-18=0\) की जड़ों का सही युग्म कौन-सा है?
Which is the correct pair of roots of \(x^2+3x-18=0\)?
#quadratic-roots
#factorisation
#root-pair
A (3) और (-6) / (3) and (-6)
B (-3) और (6) / (-3) and (6)
C (2) और (-9) / (2) and (-9)
D (-2) और (9) / (-2) and (9)
Explanation opens after your attempt
Correct Answer
A. (3) और (-6) / (3) and (-6)
Step 1
Concept
The numbers with product (-18) and sum (-3) are (3) and (-6). Hence this is the correct pair of roots.
Step 2
Why this answer is correct
The correct answer is A. (3) और (-6) / (3) and (-6). The numbers with product (-18) and sum (-3) are (3) and (-6). Hence this is the correct pair of roots.
Step 3
Exam Tip
गुणनफल (-18) और योग (-3) वाली संख्याएँ (3) और (-6) हैं। इसलिए यही जड़ों का सही युग्म है।
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यदि \(\alpha\) और \(\beta\) समीकरण \(x^2+4x-21=0\) के मूल हैं तो \(\left|\alpha-\beta\right|\) का मान क्या है?
If \(\alpha\) and \(\beta\) are roots of \(x^2+4x-21=0\), what is the value of \(\left|\alpha-\beta\right|\)?
#roots
#absolute_difference
#factorisation
A (10)
B (4)
C (7)
D (21)
Explanation opens after your attempt
Step 1
Concept
The roots are (3) and (-7). Therefore (\left|\alpha-\beta\right|=\left|3-(-7)\right|=10).
Step 2
Why this answer is correct
The correct answer is A. (10). The roots are (3) and (-7). Therefore (\left|\alpha-\beta\right|=\left|3-(-7)\right|=10).
Step 3
Exam Tip
समीकरण के मूल (3) और (-7) हैं। इसलिए (\left|\alpha-\beta\right|=\left|3-(-7)\right|=10) है।
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यदि \(\alpha\) और \(\beta\) समीकरण \(x^2+3x-18=0\) के मूल हैं तो \(\left|\alpha-\beta\right|\) का मान क्या है?
If \(\alpha\) and \(\beta\) are roots of \(x^2+3x-18=0\), what is the value of \(\left|\alpha-\beta\right|\)?
#roots
#absolute_difference
#factorisation
A (9)
B (3)
C (6)
D (18)
Explanation opens after your attempt
Step 1
Concept
The roots are (3) and (-6). Therefore (\left|\alpha-\beta\right|=\left|3-(-6)\right|=9).
Step 2
Why this answer is correct
The correct answer is A. (9). The roots are (3) and (-6). Therefore (\left|\alpha-\beta\right|=\left|3-(-6)\right|=9).
Step 3
Exam Tip
समीकरण के मूल (3) और (-6) हैं। इसलिए (\left|\alpha-\beta\right|=\left|3-(-6)\right|=9) है।
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समीकरण \(10x^2-17x+3=0\) के मूल कौन से हैं?
What are the roots of \(10x^2-17x+3=0\)?
#roots
#factorisation
#fraction_roots
A \(\frac{3}{2}\) और \(\frac{1}{5}\) / \(\frac{3}{2}\) and \(\frac{1}{5}\)
B -\(\frac{3}{2}\) और \(-\frac{1}{5}\) / \(-\frac{3}{2}\) and \(-\frac{1}{5}\)
C (2) और (3) / (2) and (3)
D \(\frac{5}{3}\) और \(\frac{1}{2}\) / \(\frac{5}{3}\) and \(\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{2}\) और \(\frac{1}{5}\) / \(\frac{3}{2}\) and \(\frac{1}{5}\)
Step 1
Concept
(10x-2 -17x+3=(2x-3)(5x-1)). Therefore the roots are \(\frac{3}{2}\) and \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{2}\) और \(\frac{1}{5}\) / \(\frac{3}{2}\) and \(\frac{1}{5}\). (10x-2 -17x+3=(2x-3)(5x-1)). Therefore the roots are \(\frac{3}{2}\) and \(\frac{1}{5}\).
Step 3
Exam Tip
(10x-2 -17x+3=(2x-3)(5x-1)) है। इसलिए मूल \(\frac{3}{2}\) और \(\frac{1}{5}\) हैं।
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समीकरण \(x^2-8x-9=0\) के मूलों का अंतर कितना है?
What is the difference between the roots of \(x^2-8x-9=0\)?
#roots
#difference_of_roots
#factorisation
A (10)
B (8)
C (9)
D (1)
Explanation opens after your attempt
Step 1
Concept
(x-2 -8x-9=(x-9)(x+1)). The roots are (9) and (-1), so the difference is (10).
Step 2
Why this answer is correct
The correct answer is A. (10). (x-2 -8x-9=(x-9)(x+1)). The roots are (9) and (-1), so the difference is (10).
Step 3
Exam Tip
(x-2 -8x-9=(x-9)(x+1)) है। मूल (9) और (-1) हैं इसलिए अंतर (10) है।
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समीकरण \(x^2-7x-30=0\) का धनात्मक मूल कौन सा है?
What is the positive root of \(x^2-7x-30=0\)?
#roots
#positive_root
#factorisation
A (10)
B (-3)
C (3)
D (-10)
Explanation opens after your attempt
Step 1
Concept
(x-2 -7x-30=(x-10)(x+3)). Therefore the roots are (10) and (-3), and the positive root is (10).
Step 2
Why this answer is correct
The correct answer is A. (10). (x-2 -7x-30=(x-10)(x+3)). Therefore the roots are (10) and (-3), and the positive root is (10).
Step 3
Exam Tip
(x-2 -7x-30=(x-10)(x+3)) है। इसलिए मूल (10) और (-3) हैं तथा धनात्मक मूल (10) है।
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समीकरण \(4x^2+3x-10=0\) के मूल कौन से हैं?
What are the roots of \(4x^2+3x-10=0\)?
#roots
#factorisation
#medium
A \(\frac{5}{4}\) और (-2) / \(\frac{5}{4}\) and (-2)
B -\(\frac{5}{4}\) और (2) / \(-\frac{5}{4}\) and (2)
C (5) और (-2) / (5) and (-2)
D (2) और (-5) / (2) and (-5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{4}\) और (-2) / \(\frac{5}{4}\) and (-2)
Step 1
Concept
(4x-2 +3x-10=(4x-5)(x+2)). Therefore the roots are \(\frac{5}{4}\) and (-2).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{4}\) और (-2) / \(\frac{5}{4}\) and (-2). (4x-2 +3x-10=(4x-5)(x+2)). Therefore the roots are \(\frac{5}{4}\) and (-2).
Step 3
Exam Tip
(4x-2 +3x-10=(4x-5)(x+2)) है। इसलिए मूल \(\frac{5}{4}\) और (-2) हैं।
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समीकरण \(x^2+2x-35=0\) के मूल कौन से हैं?
What are the roots of \(x^2+2x-35=0\)?
#roots
#factorisation
#mixed_signs
A (5) और (-7) / (5) and (-7)
B (7) और (-5) / (7) and (-5)
C (5) और (7) / (5) and (7)
D (-5) और (-7) / (-5) and (-7)
Explanation opens after your attempt
Correct Answer
A. (5) और (-7) / (5) and (-7)
Step 1
Concept
(x-2 +2x-35=(x+7)(x-5)). Therefore the roots are (-7) and (5).
Step 2
Why this answer is correct
The correct answer is A. (5) और (-7) / (5) and (-7). (x-2 +2x-35=(x+7)(x-5)). Therefore the roots are (-7) and (5).
Step 3
Exam Tip
(x-2 +2x-35=(x+7)(x-5)) है। इसलिए मूल (-7) और (5) हैं।
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समीकरण \(5x^2+16x+3=0\) के मूल कौन से हैं?
What are the roots of \(5x^2+16x+3=0\)?
#roots
#factorisation
#negative_fraction
A (-3) और \(-\frac{1}{5}\) / (-3) and \(-\frac{1}{5}\)
B (3) और \(\frac{1}{5}\) / (3) and \(\frac{1}{5}\)
C (-5) और (-3) / (-5) and (-3)
D (5) और (3) / (5) and (3)
Explanation opens after your attempt
Correct Answer
A. (-3) और \(-\frac{1}{5}\) / (-3) and \(-\frac{1}{5}\)
Step 1
Concept
(5x-2 +16x+3=(5x+1)(x+3)). Therefore the roots are \(-\frac{1}{5}\) and (-3).
Step 2
Why this answer is correct
The correct answer is A. (-3) और \(-\frac{1}{5}\) / (-3) and \(-\frac{1}{5}\). (5x-2 +16x+3=(5x+1)(x+3)). Therefore the roots are \(-\frac{1}{5}\) and (-3).
Step 3
Exam Tip
(5x-2 +16x+3=(5x+1)(x+3)) है। इसलिए मूल \(-\frac{1}{5}\) और (-3) हैं।
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समीकरण \(4x^2-17x+4=0\) के मूल कौन से हैं?
What are the roots of \(4x^2-17x+4=0\)?
#roots
#factorisation
#fraction_root
A (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\)
B (-4) और \(-\frac{1}{4}\) / (-4) and \(-\frac{1}{4}\)
C (2) और (4) / (2) and (4)
D \(\frac{4}{3}\) और (1) / \(\frac{4}{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\)
Step 1
Concept
(4x-2 -17x+4=(4x-1)(x-4)). Therefore the roots are \(\frac{1}{4}\) and (4).
Step 2
Why this answer is correct
The correct answer is A. (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\). (4x-2 -17x+4=(4x-1)(x-4)). Therefore the roots are \(\frac{1}{4}\) and (4).
Step 3
Exam Tip
(4x-2 -17x+4=(4x-1)(x-4)) है। इसलिए मूल \(\frac{1}{4}\) और (4) हैं।
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समीकरण \(x^2-14x+45=0\) के मूल कौन से हैं?
What are the roots of \(x^2-14x+45=0\)?
#roots
#factorisation
#numerical
A (5) और (9) / (5) and (9)
B (3) और (15) / (3) and (15)
C (-5) और (-9) / (-5) and (-9)
D (0) और (14) / (0) and (14)
Explanation opens after your attempt
Correct Answer
A. (5) और (9) / (5) and (9)
Step 1
Concept
(x-2 -14x+45=(x-5)(x-9)). Therefore the roots are (5) and (9).
Step 2
Why this answer is correct
The correct answer is A. (5) और (9) / (5) and (9). (x-2 -14x+45=(x-5)(x-9)). Therefore the roots are (5) and (9).
Step 3
Exam Tip
(x-2 -14x+45=(x-5)(x-9)) है। इसलिए मूल (5) और (9) हैं।
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समीकरण \(8x^2-10x+3=0\) के मूल कौन से हैं?
What are the roots of \(8x^2-10x+3=0\)?
#roots
#factorisation
#fraction_roots
A \(\frac{1}{2}\) और \(\frac{3}{4}\) / \(\frac{1}{2}\) and \(\frac{3}{4}\)
B -\(\frac{1}{2}\) और \(-\frac{3}{4}\) / \(-\frac{1}{2}\) and \(-\frac{3}{4}\)
C (2) और (3) / (2) and (3)
D \(\frac{4}{3}\) और \(\frac{1}{2}\) / \(\frac{4}{3}\) and \(\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\) और \(\frac{3}{4}\) / \(\frac{1}{2}\) and \(\frac{3}{4}\)
Step 1
Concept
(8x-2 -10x+3=(2x-1)(4x-3)). Therefore the roots are \(\frac{1}{2}\) and \(\frac{3}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\) और \(\frac{3}{4}\) / \(\frac{1}{2}\) and \(\frac{3}{4}\). (8x-2 -10x+3=(2x-1)(4x-3)). Therefore the roots are \(\frac{1}{2}\) and \(\frac{3}{4}\).
Step 3
Exam Tip
(8x-2 -10x+3=(2x-1)(4x-3)) है। इसलिए मूल \(\frac{1}{2}\) और \(\frac{3}{4}\) हैं।
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