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
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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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\(x^2+11x+30=0\) को गुणनखंड रूप में कैसे लिखा जाएगा?
How will \(x^2+11x+30=0\) be written in factorised form?
#quadratic
#factorisation
#signs
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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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\(x^2-36=0\) को हल करने की सबसे तेज विधि कौनसी है?
Which is the fastest method to solve \(x^2-36=0\)?
#quadratic
#difference-of-squares
#method
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A वर्गों के अंतर की विधि / Difference of squares method
B पूर्ण वर्ग विधि / Completing square method
C द्विघात सूत्र विधि / Quadratic formula method
D ग्राफ विधि / Graph method
Explanation opens after your attempt
Correct Answer
A. वर्गों के अंतर की विधि / Difference of squares method
Step 1
Concept
\(x^2-36=x^2-6^2\), so it is solved quickly by difference of squares. In exams, recognizing \(a^2-b^2\) saves time.
Step 2
Why this answer is correct
The correct answer is A. वर्गों के अंतर की विधि / Difference of squares method. \(x^2-36=x^2-6^2\), so it is solved quickly by difference of squares. In exams, recognizing \(a^2-b^2\) saves time.
Step 3
Exam Tip
\(x^2-36=x^2-6^2\), इसलिए वर्गों के अंतर से तुरंत हल होता है। परीक्षा में \(a^2-b^2\) पहचानना समय बचाता 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
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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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शून्य गुणनफल नियम से ((x-7)(x+1)=0) के मूल क्या हैं?
Using zero product rule, what are the roots of ((x-7)(x+1)=0)?
#quadratic
#zero-product-rule
#roots
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A (x=7,-1)
B (x=-7,1)
C (x=7,1)
D (x=-7,-1)
Explanation opens after your attempt
Correct Answer
A. (x=7,-1)
Step 1
Concept
((x-7)=0) or ((x+1)=0), so (x=7) or (x=-1). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=7,-1). ((x-7)=0) or ((x+1)=0), so (x=7) or (x=-1). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
((x-7)=0) या ((x+1)=0), इसलिए (x=7) या (x=-1) है। परीक्षा में हर गुणनखंड को अलग से शून्य रखें।
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पूर्ण वर्ग विधि में \(x^2+12x+7=0\) के लिए कौनसी संख्या जोड़ी और घटाई जाएगी?
In completing square method for \(x^2+12x+7=0\), which number will be added and subtracted?
#quadratic
#completing-square
#number-choice
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A (36)
B (12)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Half of (12) is (6), and \(6^2=36\). In exams, use (\left\(\frac{b}{2}\right\)2 ).
Step 2
Why this answer is correct
The correct answer is A. (36). Half of (12) is (6), and \(6^2=36\). In exams, use (\left\(\frac{b}{2}\right\)2 ).
Step 3
Exam Tip
(12) का आधा (6) है और \(6^2=36\) होता है। परीक्षा में (\left\(\frac{b}{2}\right\)2 ) का प्रयोग करें।
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द्विघात सूत्र लगाने के लिए \(4x^2-3x-1=0\) में (a), (b), (c) क्या हैं?
For applying the quadratic formula to \(4x^2-3x-1=0\), what are (a), (b), and (c)?
#quadratic
#coefficients
#formula
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A (a=4,b=-3,c=-1)
B (a=4,b=3,c=-1)
C (a=-3,b=4,c=-1)
D (a=4,b=-1,c=-3)
Explanation opens after your attempt
Correct Answer
A. (a=4,b=-3,c=-1)
Step 1
Concept
From standard form \(ax^2+bx+c=0\), (a=4), (b=-3), and (c=-1). In exams, write the signs of (b) and (c) carefully.
Step 2
Why this answer is correct
The correct answer is A. (a=4,b=-3,c=-1). From standard form \(ax^2+bx+c=0\), (a=4), (b=-3), and (c=-1). In exams, write the signs of (b) and (c) carefully.
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) से (a=4), (b=-3), (c=-1) हैं। परीक्षा में (b) और (c) के चिन्ह जरूर लिखें।
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\(6x^2-24=0\) को पहले सरल करने पर क्या मिलेगा?
What will be obtained first after simplifying \(6x^2-24=0\)?
#quadratic
#simplification
#square-root-method
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A \(x^2-4=0\)
B \(x^2+4=0\)
C \(6x^2=0\)
D \(x^2-24=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4=0\)
Step 1
Concept
Dividing both sides by (6) gives \(x^2-4=0\). In exams, simplify the equation first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4=0\). Dividing both sides by (6) gives \(x^2-4=0\). In exams, simplify the equation first.
Step 3
Exam Tip
दोनों पक्षों को (6) से भाग देने पर \(x^2-4=0\) मिलता है। परीक्षा में पहले समीकरण सरल 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
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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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\(x^2+14x+49=0\) को किस रूप में लिखा जा सकता है?
In which form can \(x^2+14x+49=0\) be written?
#quadratic
#perfect-square
#factorisation
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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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\(x^2+14x+49=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+14x+49=0\)?
#quadratic
#repeated-root
#perfect-square
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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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सामान्य गुणनखंड निकालकर \(5x^2+15x=0\) को कैसे लिखा जाएगा?
By taking common factor, how will \(5x^2+15x=0\) be written?
#quadratic
#common-factor
#factorisation
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A (5x(x+3)=0)
B (5(x+3)=0)
C (x(5x-15)=0)
D (5x(x-3)=0)
Explanation opens after your attempt
Correct Answer
A. (5x(x+3)=0)
Step 1
Concept
(5x) is the common factor, so (5x(x+3)=0). In exams, take out the greatest common factor.
Step 2
Why this answer is correct
The correct answer is A. (5x(x+3)=0). (5x) is the common factor, so (5x(x+3)=0). In exams, take out the greatest common factor.
Step 3
Exam Tip
(5x) सामान्य गुणनखंड है, इसलिए (5x(x+3)=0) मिलता है। परीक्षा में सबसे बड़ा सामान्य गुणनखंड निकालें।
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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
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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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\(x^2-11x+28=0\) के मूल क्या होंगे?
What will be the roots of \(x^2-11x+28=0\)?
#quadratic
#factorisation
#roots
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A (x=4,7)
B (x=-4,-7)
C (x=2,14)
D (x=1,28)
Explanation opens after your attempt
Correct Answer
A. (x=4,7)
Step 1
Concept
(x-2 -11x+28=(x-4)(x-7)), so the roots are (4) and (7). In exams, check (4+7=11) and \(4\times7=28\).
Step 2
Why this answer is correct
The correct answer is A. (x=4,7). (x-2 -11x+28=(x-4)(x-7)), so the roots are (4) and (7). In exams, check (4+7=11) and \(4\times7=28\).
Step 3
Exam Tip
(x-2 -11x+28=(x-4)(x-7)), इसलिए मूल (4) और (7) हैं। परीक्षा में (4+7=11) और \(4\times7=28\) जांचें।
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\(x^2+2x-24=0\) में मध्य पद बनाने के लिए कौनसी संख्या जोड़ी सही है?
Which number pair is correct for making the middle term in \(x^2+2x-24=0\)?
#quadratic
#middle-term-splitting
#pair
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A (6) और (-4) / (6) and (-4)
B (8) और (-3) / (8) and (-3)
C (-6) और (4) / (-6) and (4)
D (12) और (-2) / (12) and (-2)
Explanation opens after your attempt
Correct Answer
A. (6) और (-4) / (6) and (-4)
Step 1
Concept
(6+(-4)=2) and \(6\times(-4)=-24\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 2
Why this answer is correct
The correct answer is A. (6) और (-4) / (6) and (-4). (6+(-4)=2) and \(6\times(-4)=-24\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 3
Exam Tip
(6+(-4)=2) और \(6\times(-4)=-24\), इसलिए यह जोड़ी सही है। परीक्षा में योग (b) और गुणनफल (c) से मिलाएं।
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मध्य पद विभाजन में \(3x^2+10x+3=0\) के लिए (ac) का मान क्या है?
In middle term splitting for \(3x^2+10x+3=0\), what is the value of (ac)?
#quadratic
#ac-method
#middle-term
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A (9)
B (10)
C (13)
D (3)
Explanation opens after your attempt
Step 1
Concept
Here (a=3) and (c=3), so (ac=9). In exams, finding (ac) first helps.
Step 2
Why this answer is correct
The correct answer is A. (9). Here (a=3) and (c=3), so (ac=9). In exams, finding (ac) first helps.
Step 3
Exam Tip
यहां (a=3) और (c=3), इसलिए (ac=9) है। परीक्षा में पहले (ac) निकालना मदद करता है।
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\(3x^2+10x+3=0\) में मध्य पद का सही विभाजन कौनसा है?
Which is the correct splitting of the middle term in \(3x^2+10x+3=0\)?
#quadratic
#middle-term-splitting
#steps
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A \(3x^2+9x+x+3=0\)
B \(3x^2+5x+5x+3=0\)
C \(3x^2+7x+3x+3=0\)
D \(3x^2+12x-2x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2+9x+x+3=0\)
Step 1
Concept
(9+1=10) and \(9\times1=9=ac\), so (10x) is split as (9x+x). In exams, keep both sum and product correct.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2+9x+x+3=0\). (9+1=10) and \(9\times1=9=ac\), so (10x) is split as (9x+x). In exams, keep both sum and product correct.
Step 3
Exam Tip
(9+1=10) और \(9\times1=9=ac\), इसलिए (10x) को (9x+x) में तोड़ते हैं। परीक्षा में योग और गुणनफल दोनों सही रखें।
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\(x^2-81=0\) के हल क्या हैं?
What are the solutions of \(x^2-81=0\)?
#quadratic
#difference-of-squares
#roots
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A \(x=\pm9\)
B \(x=\pm81\)
C (x=9)
D (x=-9)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm9\)
Step 1
Concept
(x-2 -81=(x-9)(x+9)), so \(x=\pm9\). In exams, recognize \(81=9^2\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm9\). (x-2 -81=(x-9)(x+9)), so \(x=\pm9\). In exams, recognize \(81=9^2\).
Step 3
Exam Tip
(x-2 -81=(x-9)(x+9)), इसलिए \(x=\pm9\) है। परीक्षा में \(81=9^2\) पहचानें।
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\(7x^2-14x=0\) को गुणनखंड करके क्या मिलेगा?
What will be obtained by factoring \(7x^2-14x=0\)?
#quadratic
#common-factor
#factorisation
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A (7x(x-2)=0)
B (7(x-2)=0)
C (x(7x+14)=0)
D (7x(x+2)=0)
Explanation opens after your attempt
Correct Answer
A. (7x(x-2)=0)
Step 1
Concept
Taking common factor (7x) gives (7x(x-2)=0). In exams, you can check by expanding after factoring.
Step 2
Why this answer is correct
The correct answer is A. (7x(x-2)=0). Taking common factor (7x) gives (7x(x-2)=0). In exams, you can check by expanding after factoring.
Step 3
Exam Tip
सामान्य गुणनखंड (7x) निकालने पर (7x(x-2)=0) मिलता है। परीक्षा में गुणनखंड निकालने के बाद विस्तार करके जांच सकते हैं।
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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
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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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\(x^2-2x+1=0\) का विविक्तकर (D) क्या होगा?
What will be the discriminant (D) of \(x^2-2x+1=0\)?
#quadratic
#discriminant
#calculation
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A (0)
B (1)
C (2)
D (4)
Explanation opens after your attempt
Step 1
Concept
Here (D=(-2)2 -4(1)(1)=0). In exams, square (b) with its sign.
Step 2
Why this answer is correct
The correct answer is A. (0). Here (D=(-2)2 -4(1)(1)=0). In exams, square (b) with its sign.
Step 3
Exam Tip
यहां (D=(-2)2 -4(1)(1)=0) है। परीक्षा में (b) का चिन्ह लगाकर वर्ग करें।
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यदि किसी द्विघात समीकरण का (D=25) है, तो वास्तविक मूलों के बारे में सही कथन क्या है?
If a quadratic equation has (D=25), what is the correct statement about real roots?
#quadratic
#discriminant
#nature-of-roots
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A दो अलग वास्तविक मूल मिलेंगे / Two distinct real roots will be obtained
B दो समान वास्तविक मूल मिलेंगे / Two equal real roots will be obtained
C कोई वास्तविक मूल नहीं मिलेगा / No real root will be obtained
D अनंत वास्तविक मूल मिलेंगे / Infinitely many real roots will be obtained
Explanation opens after your attempt
Correct Answer
A. दो अलग वास्तविक मूल मिलेंगे / Two distinct real roots will be obtained
Step 1
Concept
(D=25>0), so two distinct real roots are obtained. In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. दो अलग वास्तविक मूल मिलेंगे / Two distinct real roots will be obtained. (D=25>0), so two distinct real roots are obtained. In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
(D=25>0), इसलिए दो अलग वास्तविक मूल मिलते हैं। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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यदि किसी द्विघात समीकरण में (D=-4) हो, तो कौनसा निष्कर्ष सही है?
If a quadratic equation has (D=-4), which conclusion is correct?
#quadratic
#discriminant
#no-real-roots
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A वास्तविक मूल नहीं होंगे / There will be no real roots
B दो समान वास्तविक मूल होंगे / There will be two equal real roots
C दो अलग वास्तविक मूल होंगे / There will be two distinct real roots
D एक मूल (0) होगा / One root will be (0)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं होंगे / There will be no real roots
Step 1
Concept
When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं होंगे / There will be no real roots. When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.
Step 3
Exam Tip
(D<0) होने पर वास्तविक वर्गमूल नहीं मिलता। परीक्षा में ऋणात्मक विविक्तकर का अर्थ याद रखें।
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\(x^2+4x+3=0\) को हल करने के लिए सबसे आसान विधि कौनसी है?
Which is the easiest method to solve \(x^2+4x+3=0\)?
#quadratic
#method-selection
#factorisation
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A गुणनखंड विधि / Factorisation method
B लंबा भाग विधि / Long division method
C सारणी विधि / Table method
D केवल ग्राफ विधि / Only graph method
Explanation opens after your attempt
Correct Answer
A. गुणनखंड विधि / Factorisation method
Step 1
Concept
It factors easily as ((x+1)(x+3)=0). In exams, choose factorisation when coefficients are small.
Step 2
Why this answer is correct
The correct answer is A. गुणनखंड विधि / Factorisation method. It factors easily as ((x+1)(x+3)=0). In exams, choose factorisation when coefficients are small.
Step 3
Exam Tip
यह ((x+1)(x+3)=0) में आसानी से टूटता है। परीक्षा में छोटे गुणांक देखकर गुणनखंड विधि चुनें।
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\(x^2+4x+3=0\) के मूल क्या हैं?
What are the roots of \(x^2+4x+3=0\)?
#quadratic
#roots
#factorisation
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A (x=-1,-3)
B (x=1,3)
C (x=-4,-3)
D (x=0,-3)
Explanation opens after your attempt
Correct Answer
A. (x=-1,-3)
Step 1
Concept
((x+1)(x+3)=0), so (x=-1) and (x=-3). In exams, from ((x+a)=0), write (x=-a).
Step 2
Why this answer is correct
The correct answer is A. (x=-1,-3). ((x+1)(x+3)=0), so (x=-1) and (x=-3). In exams, from ((x+a)=0), write (x=-a).
Step 3
Exam Tip
((x+1)(x+3)=0), इसलिए (x=-1) और (x=-3) हैं। परीक्षा में ((x+a)=0) से (x=-a) लिखें।
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पूर्ण वर्ग बनाते समय \(x^2-18x\) में कौनसा पद जोड़ना होगा?
While completing the square, which term must be added to \(x^2-18x\)?
#quadratic
#completing-square
#term
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A (81)
B (18)
C (9)
D (36)
Explanation opens after your attempt
Step 1
Concept
Half of (-18) is (-9), and ((-9)2 =81). In exams, the square of half the coefficient is always positive.
Step 2
Why this answer is correct
The correct answer is A. (81). Half of (-18) is (-9), and ((-9)2 =81). In exams, the square of half the coefficient is always positive.
Step 3
Exam Tip
(-18) का आधा (-9) है और ((-9)2 =81) होता है। परीक्षा में आधे गुणांक का वर्ग हमेशा धनात्मक होता है।
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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
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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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\(9x^2-16=0\) का सही गुणनखंड रूप कौनसा है?
What is the correct factorised form of \(9x^2-16=0\)?
#quadratic
#difference-of-squares
#factorisation
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A ((3x-4)(3x+4)=0)
B ((9x-4)(x+4)=0)
C ((3x-4)2 =0)
D ((x-4)(9x+4)=0)
Explanation opens after your attempt
Correct Answer
A. ((3x-4)(3x+4)=0)
Step 1
Concept
(9x-2 -16=(3x)2 -42 ), so ((3x-4)(3x+4)) is obtained. In exams, identify both squares first.
Step 2
Why this answer is correct
The correct answer is A. ((3x-4)(3x+4)=0). (9x-2 -16=(3x)2 -42 ), so ((3x-4)(3x+4)) is obtained. In exams, identify both squares first.
Step 3
Exam Tip
(9x-2 -16=(3x)2 -42 ), इसलिए ((3x-4)(3x+4)) मिलता है। परीक्षा में दोनों वर्गों को पहले पहचानें।
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\(9x^2-16=0\) के मूल क्या हैं?
What are the roots of \(9x^2-16=0\)?
#quadratic
#fraction-roots
#difference-of-squares
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A \(x=\frac{4}{3},-\frac{4}{3}\)
B \(x=\frac{3}{4},-\frac{3}{4}\)
C (x=4,-4)
D (x=3,-3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{4}{3},-\frac{4}{3}\)
Step 1
Concept
((3x-4)(3x+4)=0), so \(x=\frac{4}{3}\) or \(x=-\frac{4}{3}\). In exams, solve linear factors carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{4}{3},-\frac{4}{3}\). ((3x-4)(3x+4)=0), so \(x=\frac{4}{3}\) or \(x=-\frac{4}{3}\). In exams, solve linear factors carefully.
Step 3
Exam Tip
((3x-4)(3x+4)=0), इसलिए \(x=\frac{4}{3}\) या \(x=-\frac{4}{3}\) है। परीक्षा में रैखिक गुणनखंड सावधानी से हल करें।
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\(x^2-100=0\) में कौनसी पहचान सीधे लागू होती है?
Which identity directly applies to \(x^2-100=0\)?
#quadratic
#identity
#difference-of-squares
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A (a-2 -b-2 =(a-b)(a+b))
B ((a+b)2 =a-2 +2ab+b-2 )
C ((a-b)2 =a-2 -2ab+b-2 )
D (a-2 +b-2 =(a+b)2 )
Explanation opens after your attempt
Correct Answer
A. (a-2 -b-2 =(a-b)(a+b))
Step 1
Concept
\(100=10^2\), so it is a difference of squares form. In exams, choosing the correct identity is the fastest method.
Step 2
Why this answer is correct
The correct answer is A. (a-2 -b-2 =(a-b)(a+b)). \(100=10^2\), so it is a difference of squares form. In exams, choosing the correct identity is the fastest method.
Step 3
Exam Tip
\(100=10^2\), इसलिए यह वर्गों के अंतर का रूप है। परीक्षा में सही पहचान चुनना सबसे तेज तरीका है।
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\(x^2+8x+12=0\) के मूल कौनसे हैं?
Which are the roots of \(x^2+8x+12=0\)?
#quadratic
#roots
#signs
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A (x=-2,-6)
B (x=2,6)
C (x=-3,-4)
D (x=3,4)
Explanation opens after your attempt
Correct Answer
A. (x=-2,-6)
Step 1
Concept
(x-2 +8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-2,-6). (x-2 +8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.
Step 3
Exam Tip
(x-2 +8x+12=(x+2)(x+6)), इसलिए मूल (-2) और (-6) हैं। परीक्षा में धनात्मक मध्य पद से ऋणात्मक मूल मिल सकते हैं।
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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
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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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यदि \(x^2-5x=0\) को (x) से भाग देकर केवल (x-5=0) लिखा जाए, तो कौनसा मूल छूटेगा?
If \(x^2-5x=0\) is divided by (x) and only (x-5=0) is written, which root is missed?
#quadratic
#division-by-variable
#common-mistake
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A (x=0)
B (x=5)
C (x=-5)
D (x=1)
Explanation opens after your attempt
Step 1
Concept
Dividing by (x) misses the root (x=0). In exams, writing the factor form first is safer.
Step 2
Why this answer is correct
The correct answer is A. (x=0). Dividing by (x) misses the root (x=0). In exams, writing the factor form first is safer.
Step 3
Exam Tip
(x) से भाग देने पर (x=0) वाला मूल छूट जाता है। परीक्षा में पहले गुणनखंड रूप लिखना सुरक्षित है।
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\(x^2-13x+40=0\) के गुणनखंड कौनसे होंगे?
What will be the factors of \(x^2-13x+40=0\)?
#quadratic
#factorisation
#negative-middle-term
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A ((x-5)(x-8))
B ((x+5)(x+8))
C ((x-4)(x-10))
D ((x+4)(x+10))
Explanation opens after your attempt
Correct Answer
A. ((x-5)(x-8))
Step 1
Concept
(5+8=13) and \(5\times8=40\), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.
Step 2
Why this answer is correct
The correct answer is A. ((x-5)(x-8)). (5+8=13) and \(5\times8=40\), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.
Step 3
Exam Tip
(5+8=13) और \(5\times8=40\), इसलिए ((x-5)(x-8)) सही है। परीक्षा में ऋणात्मक मध्य पद के लिए दोनों ऋणात्मक चिन्ह लें।
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\(x^2-13x+40=0\) के मूल क्या हैं?
What are the roots of \(x^2-13x+40=0\)?
#quadratic
#roots
#factorisation
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A (x=5,8)
B (x=-5,-8)
C (x=4,10)
D (x=-4,-10)
Explanation opens after your attempt
Correct Answer
A. (x=5,8)
Step 1
Concept
((x-5)(x-8)=0), so (x=5) and (x=8). In exams, roots are obtained by taking opposite signs of factors.
Step 2
Why this answer is correct
The correct answer is A. (x=5,8). ((x-5)(x-8)=0), so (x=5) and (x=8). In exams, roots are obtained by taking opposite signs of factors.
Step 3
Exam Tip
((x-5)(x-8)=0), इसलिए (x=5) और (x=8) हैं। परीक्षा में गुणनखंड के विपरीत चिन्ह से मूल मिलते हैं।
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\(2x^2-7x+5=0\) में मध्य पद का सही विभाजन क्या है?
What is the correct splitting of the middle term in \(2x^2-7x+5=0\)?
#quadratic
#middle-term-splitting
#signs
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A \(2x^2-2x-5x+5=0\)
B \(2x^2+2x-9x+5=0\)
C \(2x^2-3x-4x+5=0\)
D \(2x^2-x-6x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-2x-5x+5=0\)
Step 1
Concept
((-2)+(-5)=-7) and ((-2)(-5)=10=ac), so (-2x-5x) is correct. In exams, checking (ac) is important while splitting the middle term.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-5x+5=0\). ((-2)+(-5)=-7) and ((-2)(-5)=10=ac), so (-2x-5x) is correct. In exams, checking (ac) is important while splitting the middle term.
Step 3
Exam Tip
((-2)+(-5)=-7) और ((-2)(-5)=10=ac), इसलिए (-2x-5x) सही है। परीक्षा में मध्य पद तोड़ते समय (ac) देखना जरूरी है।
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\(2x^2-7x+5=0\) के गुणनखंड कौनसे हैं?
What are the factors of \(2x^2-7x+5=0\)?
#quadratic
#factorisation
#middle-term
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A ((2x-5)(x-1))
B ((2x+5)(x+1))
C ((2x-1)(x-5))
D ((x-5)(x-2))
Explanation opens after your attempt
Correct Answer
A. ((2x-5)(x-1))
Step 1
Concept
(2x-2 -7x+5=(2x-5)(x-1)). In exams, expand back to check (-7x).
Step 2
Why this answer is correct
The correct answer is A. ((2x-5)(x-1)). (2x-2 -7x+5=(2x-5)(x-1)). In exams, expand back to check (-7x).
Step 3
Exam Tip
(2x-2 -7x+5=(2x-5)(x-1)) है। परीक्षा में विस्तार करके (-7x) वापस जांचें।
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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
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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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((x+3)2 =16) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x+3)2 =16) gives which values?
#quadratic
#square-root-method
#roots
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A (x=1,-7)
B (x=4,-4)
C (x=7,-1)
D (x=3,16)
Explanation opens after your attempt
Correct Answer
A. (x=1,-7)
Step 1
Concept
\(x+3=\pm4\), so (x=1) or (x=-7). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=1,-7). \(x+3=\pm4\), so (x=1) or (x=-7). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x+3=\pm4\), इसलिए (x=1) या (x=-7) है। परीक्षा में दोनों \(\pm\) स्थितियां लिखें।
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पूर्ण वर्ग विधि में \(x^2+16x+11=0\) के लिए किस संख्या का प्रयोग होगा?
In completing square method for \(x^2+16x+11=0\), which number will be used?
#quadratic
#completing-square
#number-choice
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A (64)
B (16)
C (8)
D (11)
Explanation opens after your attempt
Step 1
Concept
Half of (16) is (8), and \(8^2=64\). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. (64). Half of (16) is (8), and \(8^2=64\). In exams, add the square of half the coefficient.
Step 3
Exam Tip
(16) का आधा (8) है और \(8^2=64\) होता है। परीक्षा में आधे गुणांक का वर्ग जोड़ना होता है।
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\(x^2+16x+64\) किसके बराबर है?
What is \(x^2+16x+64\) equal to?
#quadratic
#perfect-square
#completing-square
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A ((x+8)2 )
B ((x-8)2 )
C ((x+16)2 )
D ((x-16)2 )
Explanation opens after your attempt
Correct Answer
A. ((x+8)2 )
Step 1
Concept
(x-2 +16x+64=(x+8)2 ). In exams, identify it using \(2\cdot8x=16x\).
Step 2
Why this answer is correct
The correct answer is A. ((x+8)2 ). (x-2 +16x+64=(x+8)2 ). In exams, identify it using \(2\cdot8x=16x\).
Step 3
Exam Tip
(x-2 +16x+64=(x+8)2 ) है। परीक्षा में \(2\cdot8x=16x\) से पहचान करें।
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\(x^2+5x-14=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2+5x-14=0\)?
#quadratic
#factorisation
#mixed-signs
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A ((x+7)(x-2))
B ((x-7)(x+2))
C ((x+14)(x-1))
D ((x-14)(x+1))
Explanation opens after your attempt
Correct Answer
A. ((x+7)(x-2))
Step 1
Concept
(7+(-2)=5) and \(7\times(-2)=-14\), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.
Step 2
Why this answer is correct
The correct answer is A. ((x+7)(x-2)). (7+(-2)=5) and \(7\times(-2)=-14\), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.
Step 3
Exam Tip
(7+(-2)=5) और \(7\times(-2)=-14\), इसलिए ((x+7)(x-2)) सही है। परीक्षा में एक चिन्ह धनात्मक और एक ऋणात्मक रखें।
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\(x^2+5x-14=0\) के मूल क्या हैं?
What are the roots of \(x^2+5x-14=0\)?
#quadratic
#roots
#mixed-signs
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A (x=2,-7)
B (x=-2,7)
C (x=14,-1)
D (x=-14,1)
Explanation opens after your attempt
Correct Answer
A. (x=2,-7)
Step 1
Concept
((x+7)(x-2)=0), so (x=-7) and (x=2). In exams, the sign changes while finding roots from factors.
Step 2
Why this answer is correct
The correct answer is A. (x=2,-7). ((x+7)(x-2)=0), so (x=-7) and (x=2). In exams, the sign changes while finding roots from factors.
Step 3
Exam Tip
((x+7)(x-2)=0), इसलिए (x=-7) और (x=2) हैं। परीक्षा में गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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\(x^2+25=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about the real roots of \(x^2+25=0\)?
#quadratic
#no-real-roots
#square-root-method
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A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक मूल (5) और (-5) / Two real roots (5) and (-5)
C एक वास्तविक मूल (25) / One real root (25)
D दो समान वास्तविक मूल (0) / Two equal real roots (0)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
\(x^2=-25\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. \(x^2=-25\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 3
Exam Tip
\(x^2=-25\) वास्तविक संख्याओं में संभव नहीं है। परीक्षा में \(x^2\) का मान वास्तविक (x) के लिए ऋणात्मक नहीं होता।
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\(8x^2=72\) को हल करने का सही सरल रूप कौनसा है?
What is the correct simplified form to solve \(8x^2=72\)?
#quadratic
#simplification
#square-root-method
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A \(x^2=9\)
B \(x^2=72\)
C \(x^2=64\)
D \(x^2=8\)
Explanation opens after your attempt
Correct Answer
A. \(x^2=9\)
Step 1
Concept
Dividing both sides by (8) gives \(x^2=9\). In exams, remove the coefficient first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2=9\). Dividing both sides by (8) gives \(x^2=9\). In exams, remove the coefficient first.
Step 3
Exam Tip
दोनों पक्षों को (8) से भाग देने पर \(x^2=9\) मिलता है। परीक्षा में पहले गुणांक हटाएं।
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\(8x^2=72\) के हल क्या हैं?
What are the solutions of \(8x^2=72\)?
#quadratic
#square-root-method
#solutions
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A \(x=\pm3\)
B (x=3)
C (x=-3)
D \(x=\pm9\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm3\)
Step 1
Concept
From \(8x^2=72\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm3\). From \(8x^2=72\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 3
Exam Tip
\(8x^2=72\) से \(x^2=9\), इसलिए \(x=\pm3\) है। परीक्षा में अंतिम उत्तर में दोनों मान लिखें।
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द्विघात सूत्र में हर का सही रूप कौनसा होता है?
What is the correct denominator in the quadratic formula?
#quadratic
#formula
#denominator
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A (2a)
B (a)
C (2b)
D (4a)
Explanation opens after your attempt
Step 1
Concept
In \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), the denominator is (2a). In exams, forgetting (2a) is a common mistake.
Step 2
Why this answer is correct
The correct answer is A. (2a). In \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), the denominator is (2a). In exams, forgetting (2a) is a common mistake.
Step 3
Exam Tip
द्विघात सूत्र \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) में हर (2a) होता है। परीक्षा में (2a) भूलना सामान्य गलती है।
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यदि \(x^2-6x+k\) एक पूर्ण वर्ग हो और उसका रूप ((x-3)2 ) हो, तो (k) क्या होगा?
If \(x^2-6x+k\) is a perfect square and its form is ((x-3)2 ), what is (k)?
#quadratic
#perfect-square
#missing-term
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A (9)
B (3)
C (6)
D (12)
Explanation opens after your attempt
Step 1
Concept
((x-3)2 =x-2 -6x+9), so (k=9). In exams, remember perfect square expansion.
Step 2
Why this answer is correct
The correct answer is A. (9). ((x-3)2 =x-2 -6x+9), so (k=9). In exams, remember perfect square expansion.
Step 3
Exam Tip
((x-3)2 =x-2 -6x+9), इसलिए (k=9) है। परीक्षा में पूर्ण वर्ग विस्तार याद रखें।
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किस विधि में पहले \(b^2-4ac\) निकाला जाता है?
In which method is \(b^2-4ac\) found first?
#quadratic
#formula
#discriminant
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A द्विघात सूत्र विधि / Quadratic formula method
B गुणनखंड विधि / Factorisation method
C वर्गमूल विधि / Square root method
D सामान्य गुणनखंड विधि / Common factor method
Explanation opens after your attempt
Correct Answer
A. द्विघात सूत्र विधि / Quadratic formula method
Step 1
Concept
In the quadratic formula, \(b^2-4ac\) is used as the discriminant. In exams, identify (a), (b), and (c) before using the formula method.
Step 2
Why this answer is correct
The correct answer is A. द्विघात सूत्र विधि / Quadratic formula method. In the quadratic formula, \(b^2-4ac\) is used as the discriminant. In exams, identify (a), (b), and (c) before using the formula method.
Step 3
Exam Tip
द्विघात सूत्र में \(b^2-4ac\) विविक्तकर के रूप में प्रयोग होता है। परीक्षा में सूत्र विधि लगाने से पहले (a), (b), (c) पहचानें।
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\(4x^2+4x+1=0\) को किस पूर्ण वर्ग रूप में लिखा जा सकता है?
In which perfect square form can \(4x^2+4x+1=0\) be written?
#quadratic
#perfect-square
#factorisation
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A ((2x+1)2 =0)
B ((2x-1)2 =0)
C ((4x+1)2 =0)
D ((x+2)2 =0)
Explanation opens after your attempt
Correct Answer
A. ((2x+1)2 =0)
Step 1
Concept
(4x-2 +4x+1=(2x+1)2 ), so it is a perfect square equation. In exams, recognize ((a+b)2 ) to solve quickly.
Step 2
Why this answer is correct
The correct answer is A. ((2x+1)2 =0). (4x-2 +4x+1=(2x+1)2 ), so it is a perfect square equation. In exams, recognize ((a+b)2 ) to solve quickly.
Step 3
Exam Tip
(4x-2 +4x+1=(2x+1)2 ), इसलिए यह पूर्ण वर्ग समीकरण है। परीक्षा में ((a+b)2 ) पहचानकर जल्दी हल करें।
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