गुणनखंड विधि से \(x^2-15x+56=0\) के मूल क्या होंगे?
Using factorisation method, what will be the roots of \(x^2-15x+56=0\)?
#quadratic
#factorisation
#roots
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A (x=7,8)
B (x=-7,-8)
C (x=4,14)
D (x=1,56)
Explanation opens after your attempt
Correct Answer
A. (x=7,8)
Step 1
Concept
(x-2 -15x+56=(x-7)(x-8)), so the roots are (7) and (8). In exams, check both sum and product.
Step 2
Why this answer is correct
The correct answer is A. (x=7,8). (x-2 -15x+56=(x-7)(x-8)), so the roots are (7) and (8). In exams, check both sum and product.
Step 3
Exam Tip
(x-2 -15x+56=(x-7)(x-8)), इसलिए मूल (7) और (8) हैं। परीक्षा में योग और गुणनफल दोनों मिलाकर जांचें।
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\(x^2+13x+42=0\) का सही गुणनखंड रूप कौनसा है?
What is the correct factorised form of \(x^2+13x+42=0\)?
#quadratic
#factorisation
#standard-form
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A ((x+6)(x+7)=0)
B ((x-6)(x-7)=0)
C ((x+3)(x+14)=0)
D ((x+2)(x+21)=0)
Explanation opens after your attempt
Correct Answer
A. ((x+6)(x+7)=0)
Step 1
Concept
(6+7=13) and \(6\times7=42\), so the correct factors are ((x+6)(x+7)). In exams, match the signs carefully.
Step 2
Why this answer is correct
The correct answer is A. ((x+6)(x+7)=0). (6+7=13) and \(6\times7=42\), so the correct factors are ((x+6)(x+7)). In exams, match the signs carefully.
Step 3
Exam Tip
(6+7=13) और \(6\times7=42\), इसलिए सही गुणनखंड ((x+6)(x+7)) हैं। परीक्षा में संकेतों को ध्यान से मिलाएं।
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\(x^2-49=0\) को हल करने की सबसे सरल विधि कौनसी है?
Which is the simplest method to solve \(x^2-49=0\)?
#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 सारणीयन विधि / Tabulation method
Explanation opens after your attempt
Correct Answer
A. वर्गों के अंतर की विधि / Difference of squares method
Step 1
Concept
\(x^2-49=x^2-7^2\), so the difference of squares method is fastest. In exams, recognizing \(a^2-b^2\) is useful.
Step 2
Why this answer is correct
The correct answer is A. वर्गों के अंतर की विधि / Difference of squares method. \(x^2-49=x^2-7^2\), so the difference of squares method is fastest. In exams, recognizing \(a^2-b^2\) is useful.
Step 3
Exam Tip
\(x^2-49=x^2-7^2\), इसलिए वर्गों के अंतर की विधि सबसे तेज है। परीक्षा में \(a^2-b^2\) पहचानना उपयोगी है।
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वर्गमूल विधि से \(x^2=144\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=144\)?
#quadratic
#square-root-method
#solutions
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A \(x=\pm12\)
B (x=12)
C (x=-12)
D \(x=\pm72\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm12\)
Step 1
Concept
\(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm12\). \(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{144}=\pm12\) होता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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शून्य गुणनफल नियम से ((x-9)(x+2)=0) के मूल क्या हैं?
Using zero product rule, what are the roots of ((x-9)(x+2)=0)?
#quadratic
#zero-product-rule
#roots
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A (x=9,-2)
B (x=-9,2)
C (x=9,2)
D (x=-9,-2)
Explanation opens after your attempt
Correct Answer
A. (x=9,-2)
Step 1
Concept
((x-9)=0) or ((x+2)=0), so (x=9) or (x=-2). In exams, set each factor equal to zero separately.
Step 2
Why this answer is correct
The correct answer is A. (x=9,-2). ((x-9)=0) or ((x+2)=0), so (x=9) or (x=-2). In exams, set each factor equal to zero separately.
Step 3
Exam Tip
((x-9)=0) या ((x+2)=0), इसलिए (x=9) या (x=-2) है। परीक्षा में हर गुणनखंड को अलग-अलग शून्य रखें।
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पूर्ण वर्ग विधि में \(x^2+20x+13=0\) के लिए कौनसी संख्या जोड़ी और घटाई जाएगी?
In completing square method for \(x^2+20x+13=0\), which number will be added and subtracted?
#quadratic
#completing-square
#number-choice
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A (100)
B (20)
C (10)
D (13)
Explanation opens after your attempt
Step 1
Concept
Half of (20) is (10), and \(10^2=100\). In exams, use (\left\(\frac{b}{2}\right\)2 ).
Step 2
Why this answer is correct
The correct answer is A. (100). Half of (20) is (10), and \(10^2=100\). In exams, use (\left\(\frac{b}{2}\right\)2 ).
Step 3
Exam Tip
(20) का आधा (10) है और \(10^2=100\) होता है। परीक्षा में (\left\(\frac{b}{2}\right\)2 ) का प्रयोग करें।
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द्विघात सूत्र लगाने के लिए \(5x^2+2x-7=0\) में (a), (b), (c) क्या हैं?
For applying the quadratic formula to \(5x^2+2x-7=0\), what are (a), (b), and (c)?
#quadratic
#coefficients
#formula
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A (a=5,b=2,c=-7)
B (a=2,b=5,c=-7)
C (a=5,b=-2,c=7)
D (a=-7,b=2,c=5)
Explanation opens after your attempt
Correct Answer
A. (a=5,b=2,c=-7)
Step 1
Concept
From standard form \(ax^2+bx+c=0\), (a=5), (b=2), and (c=-7). In exams, always check the sign of (c).
Step 2
Why this answer is correct
The correct answer is A. (a=5,b=2,c=-7). From standard form \(ax^2+bx+c=0\), (a=5), (b=2), and (c=-7). In exams, always check the sign of (c).
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) से (a=5), (b=2), (c=-7) हैं। परीक्षा में (c) का संकेत जरूर देखें।
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\(10x^2-90=0\) को पहले सरल करने पर क्या मिलेगा?
What will be obtained first after simplifying \(10x^2-90=0\)?
#quadratic
#simplification
#square-root-method
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A \(x^2-9=0\)
B \(x^2+9=0\)
C \(10x^2=0\)
D \(x^2-90=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-9=0\)
Step 1
Concept
Dividing both sides by (10) gives \(x^2-9=0\). In exams, simplifying the equation first saves time.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-9=0\). Dividing both sides by (10) gives \(x^2-9=0\). In exams, simplifying the equation first saves time.
Step 3
Exam Tip
दोनों पक्षों को (10) से भाग देने पर \(x^2-9=0\) मिलता है। परीक्षा में पहले समीकरण को सरल करना समय बचाता है।
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\(x^2-6x-27=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2-6x-27=0\)?
#quadratic
#factorisation
#mixed-signs
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A ((x-9)(x+3))
B ((x+9)(x-3))
C ((x-27)(x+1))
D ((x+9)(x+3))
Explanation opens after your attempt
Correct Answer
A. ((x-9)(x+3))
Step 1
Concept
Since (-9+3=-6) and \(-9\times3=-27\), ((x-9)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 2
Why this answer is correct
The correct answer is A. ((x-9)(x+3)). Since (-9+3=-6) and \(-9\times3=-27\), ((x-9)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 3
Exam Tip
(-9+3=-6) और \(-9\times3=-27\), इसलिए ((x-9)(x+3)) सही है। परीक्षा में मिश्रित चिन्ह वाले गुणनखंड ध्यान से चुनें।
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\(x^2+18x+81=0\) को किस रूप में लिखा जा सकता है?
In which form can \(x^2+18x+81=0\) be written?
#quadratic
#perfect-square
#factorisation
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A ((x+9)2 =0)
B ((x-9)2 =0)
C ((x+18)2 =0)
D ((x-18)2 =0)
Explanation opens after your attempt
Correct Answer
A. ((x+9)2 =0)
Step 1
Concept
(x-2 +18x+81=(x+9)2 ), so it is a perfect square. In exams, match it using \(2\cdot9x=18x\).
Step 2
Why this answer is correct
The correct answer is A. ((x+9)2 =0). (x-2 +18x+81=(x+9)2 ), so it is a perfect square. In exams, match it using \(2\cdot9x=18x\).
Step 3
Exam Tip
(x-2 +18x+81=(x+9)2 ), इसलिए यह पूर्ण वर्ग है। परीक्षा में \(2\cdot9x=18x\) से मिलान करें।
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\(x^2+18x+81=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+18x+81=0\)?
#quadratic
#repeated-root
#perfect-square
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A (x=-9)
B (x=9)
C (x=-18)
D (x=18)
Explanation opens after your attempt
Step 1
Concept
((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-9). ((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 3
Exam Tip
((x+9)2 =0), इसलिए दोहराया हुआ मूल (-9) है। परीक्षा में पूर्ण वर्ग समीकरण में दोनों मूल समान होते हैं।
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सामान्य गुणनखंड निकालकर \(4x^2+28x=0\) को कैसे लिखा जाएगा?
By taking common factor, how will \(4x^2+28x=0\) be written?
#quadratic
#common-factor
#factorisation
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A (4x(x+7)=0)
B (4(x+7)=0)
C (x(4x-28)=0)
D (4x(x-7)=0)
Explanation opens after your attempt
Correct Answer
A. (4x(x+7)=0)
Step 1
Concept
(4x) is the common factor, so (4x(x+7)=0). In exams, take out the greatest common factor.
Step 2
Why this answer is correct
The correct answer is A. (4x(x+7)=0). (4x) is the common factor, so (4x(x+7)=0). In exams, take out the greatest common factor.
Step 3
Exam Tip
(4x) सामान्य गुणनखंड है, इसलिए (4x(x+7)=0) मिलता है। परीक्षा में सबसे बड़ा सामान्य गुणनखंड निकालें।
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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
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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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\(x^2-17x+72=0\) के मूल क्या होंगे?
What will be the roots of \(x^2-17x+72=0\)?
#quadratic
#factorisation
#roots
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A (x=8,9)
B (x=-8,-9)
C (x=6,12)
D (x=4,18)
Explanation opens after your attempt
Correct Answer
A. (x=8,9)
Step 1
Concept
(x-2 -17x+72=(x-8)(x-9)), so the roots are (8) and (9). In exams, check (8+9=17) and \(8\times9=72\).
Step 2
Why this answer is correct
The correct answer is A. (x=8,9). (x-2 -17x+72=(x-8)(x-9)), so the roots are (8) and (9). In exams, check (8+9=17) and \(8\times9=72\).
Step 3
Exam Tip
(x-2 -17x+72=(x-8)(x-9)), इसलिए मूल (8) और (9) हैं। परीक्षा में (8+9=17) और \(8\times9=72\) जांचें।
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\(x^2+3x-40=0\) में मध्य पद बनाने के लिए कौनसी संख्या जोड़ी सही है?
Which number pair is correct for making the middle term in \(x^2+3x-40=0\)?
#quadratic
#middle-term-splitting
#pair
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A (8) और (-5) / (8) and (-5)
B (10) और (-4) / (10) and (-4)
C (-8) और (5) / (-8) and (5)
D (20) और (-2) / (20) and (-2)
Explanation opens after your attempt
Correct Answer
A. (8) और (-5) / (8) and (-5)
Step 1
Concept
(8+(-5)=3) and \(8\times(-5)=-40\), 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. (8) और (-5) / (8) and (-5). (8+(-5)=3) and \(8\times(-5)=-40\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 3
Exam Tip
(8+(-5)=3) और \(8\times(-5)=-40\), इसलिए यह जोड़ी सही है। परीक्षा में योग (b) और गुणनफल (c) से मिलाएं।
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मध्य पद विभाजन में \(4x^2+13x+3=0\) के लिए (ac) का मान क्या है?
In middle term splitting for \(4x^2+13x+3=0\), what is the value of (ac)?
#quadratic
#ac-method
#middle-term
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A (12)
B (13)
C (7)
D (3)
Explanation opens after your attempt
Step 1
Concept
Here (a=4) and (c=3), so (ac=12). In exams, finding (ac) first helps.
Step 2
Why this answer is correct
The correct answer is A. (12). Here (a=4) and (c=3), so (ac=12). In exams, finding (ac) first helps.
Step 3
Exam Tip
यहां (a=4) और (c=3), इसलिए (ac=12) है। परीक्षा में पहले (ac) निकालना मदद करता है।
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\(4x^2+13x+3=0\) में मध्य पद का सही विभाजन कौनसा है?
Which is the correct splitting of the middle term in \(4x^2+13x+3=0\)?
#quadratic
#middle-term-splitting
#steps
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A \(4x^2+12x+x+3=0\)
B \(4x^2+9x+4x+3=0\)
C \(4x^2+10x+3x+3=0\)
D \(4x^2+15x-2x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(4x^2+12x+x+3=0\)
Step 1
Concept
(12+1=13) and \(12\times1=12=ac\), so (13x) is split as (12x+x). In exams, keep both sum and product correct.
Step 2
Why this answer is correct
The correct answer is A. \(4x^2+12x+x+3=0\). (12+1=13) and \(12\times1=12=ac\), so (13x) is split as (12x+x). In exams, keep both sum and product correct.
Step 3
Exam Tip
(12+1=13) और \(12\times1=12=ac\), इसलिए (13x) को (12x+x) में तोड़ते हैं। परीक्षा में योग और गुणनफल दोनों सही रखें।
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\(x^2-121=0\) के हल क्या हैं?
What are the solutions of \(x^2-121=0\)?
#quadratic
#difference-of-squares
#roots
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A \(x=\pm11\)
B \(x=\pm121\)
C (x=11)
D (x=-11)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm11\)
Step 1
Concept
(x-2 -121=(x-11)(x+11)), so \(x=\pm11\). In exams, recognize \(121=11^2\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm11\). (x-2 -121=(x-11)(x+11)), so \(x=\pm11\). In exams, recognize \(121=11^2\).
Step 3
Exam Tip
(x-2 -121=(x-11)(x+11)), इसलिए \(x=\pm11\) है। परीक्षा में \(121=11^2\) पहचानें।
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\(9x^2-27x=0\) को गुणनखंड करके क्या मिलेगा?
What will be obtained by factoring \(9x^2-27x=0\)?
#quadratic
#common-factor
#factorisation
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A (9x(x-3)=0)
B (9(x-3)=0)
C (x(9x+27)=0)
D (9x(x+3)=0)
Explanation opens after your attempt
Correct Answer
A. (9x(x-3)=0)
Step 1
Concept
Taking common factor (9x) gives (9x(x-3)=0). In exams, check by expanding after factoring.
Step 2
Why this answer is correct
The correct answer is A. (9x(x-3)=0). Taking common factor (9x) gives (9x(x-3)=0). In exams, check by expanding after factoring.
Step 3
Exam Tip
सामान्य गुणनखंड (9x) निकालने पर (9x(x-3)=0) मिलता है। परीक्षा में गुणनखंड निकालने के बाद विस्तार करके जांचें।
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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
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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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\(x^2+2x+1=0\) 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=22 -4(1)(1)=0). In exams, (D=0) gives equal roots.
Step 2
Why this answer is correct
The correct answer is A. (0). Here (D=22 -4(1)(1)=0). In exams, (D=0) gives equal roots.
Step 3
Exam Tip
यहां (D=22 -4(1)(1)=0) है। परीक्षा में (D=0) से समान मूल मिलते हैं।
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यदि किसी द्विघात समीकरण का (D=36) है, तो वास्तविक मूलों के बारे में सही कथन क्या है?
If a quadratic equation has (D=36), 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=36>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=36>0), so two distinct real roots are obtained. In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
(D=36>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=-9) हो, तो कौनसा निष्कर्ष सही है?
If a quadratic equation has (D=-9), 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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\(x^2+7x+10=0\) को हल करने के लिए सबसे आसान विधि कौनसी है?
Which is the easiest method to solve \(x^2+7x+10=0\)?
#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+5)(x+2)=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+5)(x+2)=0). In exams, choose factorisation when coefficients are small.
Step 3
Exam Tip
यह ((x+5)(x+2)=0) में आसानी से टूटता है। परीक्षा में छोटे गुणांक देखकर गुणनखंड विधि चुनें।
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\(x^2+7x+10=0\) के मूल क्या हैं?
What are the roots of \(x^2+7x+10=0\)?
#quadratic
#roots
#factorisation
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A (x=-5,-2)
B (x=5,2)
C (x=-7,-10)
D (x=0,-10)
Explanation opens after your attempt
Correct Answer
A. (x=-5,-2)
Step 1
Concept
((x+5)(x+2)=0), so (x=-5) 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=-5,-2). ((x+5)(x+2)=0), so (x=-5) and (x=-2). In exams, from ((x+a)=0), write (x=-a).
Step 3
Exam Tip
((x+5)(x+2)=0), इसलिए (x=-5) और (x=-2) हैं। परीक्षा में ((x+a)=0) से (x=-a) लिखें।
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पूर्ण वर्ग बनाते समय \(x^2-22x\) में कौनसा पद जोड़ना होगा?
While completing the square, which term must be added to \(x^2-22x\)?
#quadratic
#completing-square
#term
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A (121)
B (22)
C (11)
D (44)
Explanation opens after your attempt
Step 1
Concept
Half of (-22) is (-11), and ((-11)2 =121). In exams, the square of half the coefficient is always positive.
Step 2
Why this answer is correct
The correct answer is A. (121). Half of (-22) is (-11), and ((-11)2 =121). In exams, the square of half the coefficient is always positive.
Step 3
Exam Tip
(-22) का आधा (-11) है और ((-11)2 =121) होता है। परीक्षा में आधे गुणांक का वर्ग हमेशा धनात्मक होता है।
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\(x^2=169\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=169\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
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A \(x=\pm13\)
B (x=13)
C (x=-13)
D \(x=\pm169\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm13\)
Step 1
Concept
\(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm13\). \(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{169}=\pm13\) होता है। परीक्षा में केवल (13) लिखना अधूरा उत्तर है।
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\(16x^2-25=0\) का सही गुणनखंड रूप कौनसा है?
What is the correct factorised form of \(16x^2-25=0\)?
#quadratic
#difference-of-squares
#factorisation
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A ((4x-5)(4x+5)=0)
B ((16x-5)(x+5)=0)
C ((4x-5)2 =0)
D ((x-5)(16x+5)=0)
Explanation opens after your attempt
Correct Answer
A. ((4x-5)(4x+5)=0)
Step 1
Concept
(16x-2 -25=(4x)2 -52 ), so ((4x-5)(4x+5)) is obtained. In exams, identify both squares first.
Step 2
Why this answer is correct
The correct answer is A. ((4x-5)(4x+5)=0). (16x-2 -25=(4x)2 -52 ), so ((4x-5)(4x+5)) is obtained. In exams, identify both squares first.
Step 3
Exam Tip
(16x-2 -25=(4x)2 -52 ), इसलिए ((4x-5)(4x+5)) मिलता है। परीक्षा में दोनों वर्गों को पहले पहचानें।
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\(16x^2-25=0\) के मूल क्या हैं?
What are the roots of \(16x^2-25=0\)?
#quadratic
#fraction-roots
#difference-of-squares
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A \(x=\frac{5}{4},-\frac{5}{4}\)
B \(x=\frac{4}{5},-\frac{4}{5}\)
C (x=5,-5)
D (x=4,-4)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{4},-\frac{5}{4}\)
Step 1
Concept
((4x-5)(4x+5)=0), so \(x=\frac{5}{4}\) or \(x=-\frac{5}{4}\). In exams, solve linear factors carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{4},-\frac{5}{4}\). ((4x-5)(4x+5)=0), so \(x=\frac{5}{4}\) or \(x=-\frac{5}{4}\). In exams, solve linear factors carefully.
Step 3
Exam Tip
((4x-5)(4x+5)=0), इसलिए \(x=\frac{5}{4}\) या \(x=-\frac{5}{4}\) है। परीक्षा में रैखिक गुणनखंड सावधानी से हल करें।
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\(x^2-144=0\) में कौनसी पहचान सीधे लागू होती है?
Which identity directly applies to \(x^2-144=0\)?
#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
\(144=12^2\), 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)). \(144=12^2\), so it is a difference of squares form. In exams, choosing the correct identity is the fastest method.
Step 3
Exam Tip
\(144=12^2\), इसलिए यह वर्गों के अंतर का रूप है। परीक्षा में सही पहचान चुनना सबसे तेज तरीका है।
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\(x^2+9x+20=0\) के मूल कौनसे हैं?
Which are the roots of \(x^2+9x+20=0\)?
#quadratic
#roots
#signs
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A (x=-4,-5)
B (x=4,5)
C (x=-2,-10)
D (x=2,10)
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, a positive middle term can give negative roots.
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, a positive middle term can give negative roots.
Step 3
Exam Tip
(x-2 +9x+20=(x+4)(x+5)), इसलिए मूल (-4) और (-5) हैं। परीक्षा में धनात्मक मध्य पद से ऋणात्मक मूल मिल सकते हैं।
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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
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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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यदि \(x^2-11x=0\) को (x) से भाग देकर केवल (x-11=0) लिखा जाए, तो कौनसा मूल छूटेगा?
If \(x^2-11x=0\) is divided by (x) and only (x-11=0) is written, which root is missed?
#quadratic
#division-by-variable
#common-mistake
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A (x=0)
B (x=11)
C (x=-11)
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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\(x^2-19x+90=0\) के गुणनखंड कौनसे होंगे?
What will be the factors of \(x^2-19x+90=0\)?
#quadratic
#factorisation
#negative-middle-term
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A ((x-9)(x-10))
B ((x+9)(x+10))
C ((x-6)(x-15))
D ((x+6)(x+15))
Explanation opens after your attempt
Correct Answer
A. ((x-9)(x-10))
Step 1
Concept
(9+10=19) and \(9\times10=90\), so ((x-9)(x-10)) 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-9)(x-10)). (9+10=19) and \(9\times10=90\), so ((x-9)(x-10)) is correct. In exams, take both negative signs for a negative middle term.
Step 3
Exam Tip
(9+10=19) और \(9\times10=90\), इसलिए ((x-9)(x-10)) सही है। परीक्षा में ऋणात्मक मध्य पद के लिए दोनों ऋणात्मक चिन्ह लें।
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\(x^2-19x+90=0\) के मूल क्या हैं?
What are the roots of \(x^2-19x+90=0\)?
#quadratic
#roots
#factorisation
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A (x=9,10)
B (x=-9,-10)
C (x=6,15)
D (x=-6,-15)
Explanation opens after your attempt
Correct Answer
A. (x=9,10)
Step 1
Concept
((x-9)(x-10)=0), so (x=9) and (x=10). In exams, roots are obtained by taking opposite signs of factors.
Step 2
Why this answer is correct
The correct answer is A. (x=9,10). ((x-9)(x-10)=0), so (x=9) and (x=10). In exams, roots are obtained by taking opposite signs of factors.
Step 3
Exam Tip
((x-9)(x-10)=0), इसलिए (x=9) और (x=10) हैं। परीक्षा में गुणनखंड के विपरीत चिन्ह से मूल मिलते हैं।
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\(3x^2-11x+6=0\) में मध्य पद का सही विभाजन क्या है?
What is the correct splitting of the middle term in \(3x^2-11x+6=0\)?
#quadratic
#middle-term-splitting
#signs
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A \(3x^2-9x-2x+6=0\)
B \(3x^2-5x-6x+6=0\)
C \(3x^2+9x-20x+6=0\)
D \(3x^2-x-10x+6=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-9x-2x+6=0\)
Step 1
Concept
((-9)+(-2)=-11) and ((-9)(-2)=18=ac), so (-9x-2x) is correct. In exams, split the middle term by checking (ac).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-9x-2x+6=0\). ((-9)+(-2)=-11) and ((-9)(-2)=18=ac), so (-9x-2x) is correct. In exams, split the middle term by checking (ac).
Step 3
Exam Tip
((-9)+(-2)=-11) और ((-9)(-2)=18=ac), इसलिए (-9x-2x) सही है। परीक्षा में (ac) देखकर मध्य पद तोड़ें।
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\(3x^2-11x+6=0\) के गुणनखंड कौनसे हैं?
What are the factors of \(3x^2-11x+6=0\)?
#quadratic
#factorisation
#middle-term
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A ((3x-2)(x-3))
B ((3x+2)(x+3))
C ((3x-3)(x-2))
D ((x-6)(x-3))
Explanation opens after your attempt
Correct Answer
A. ((3x-2)(x-3))
Step 1
Concept
(3x-2 -11x+6=(3x-2)(x-3)). In exams, expand back to check (-11x).
Step 2
Why this answer is correct
The correct answer is A. ((3x-2)(x-3)). (3x-2 -11x+6=(3x-2)(x-3)). In exams, expand back to check (-11x).
Step 3
Exam Tip
(3x-2 -11x+6=(3x-2)(x-3)) है। परीक्षा में विस्तार करके (-11x) वापस जांचें।
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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
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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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((x-4)2 =25) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x-4)2 =25) gives which values?
#quadratic
#square-root-method
#roots
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A (x=9,-1)
B (x=5,-5)
C (x=4,25)
D (x=1,9)
Explanation opens after your attempt
Correct Answer
A. (x=9,-1)
Step 1
Concept
\(x-4=\pm5\), so (x=9) or (x=-1). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=9,-1). \(x-4=\pm5\), so (x=9) or (x=-1). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x-4=\pm5\), इसलिए (x=9) या (x=-1) है। परीक्षा में दोनों \(\pm\) स्थितियां लिखें।
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पूर्ण वर्ग विधि में \(x^2+24x+17=0\) के लिए किस संख्या का प्रयोग होगा?
In completing square method for \(x^2+24x+17=0\), which number will be used?
#quadratic
#completing-square
#number-choice
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A (144)
B (24)
C (12)
D (17)
Explanation opens after your attempt
Step 1
Concept
Half of (24) is (12), and \(12^2=144\). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. (144). Half of (24) is (12), and \(12^2=144\). In exams, add the square of half the coefficient.
Step 3
Exam Tip
(24) का आधा (12) है और \(12^2=144\) होता है। परीक्षा में आधे गुणांक का वर्ग जोड़ना होता है।
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\(x^2+24x+144\) किसके बराबर है?
What is \(x^2+24x+144\) equal to?
#quadratic
#perfect-square
#completing-square
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A ((x+12)2 )
B ((x-12)2 )
C ((x+24)2 )
D ((x-24)2 )
Explanation opens after your attempt
Correct Answer
A. ((x+12)2 )
Step 1
Concept
(x-2 +24x+144=(x+12)2 ). In exams, identify it using \(2\cdot12x=24x\).
Step 2
Why this answer is correct
The correct answer is A. ((x+12)2 ). (x-2 +24x+144=(x+12)2 ). In exams, identify it using \(2\cdot12x=24x\).
Step 3
Exam Tip
(x-2 +24x+144=(x+12)2 ) है। परीक्षा में \(2\cdot12x=24x\) से पहचान करें।
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\(x^2+6x-55=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2+6x-55=0\)?
#quadratic
#factorisation
#mixed-signs
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A ((x+11)(x-5))
B ((x-11)(x+5))
C ((x+55)(x-1))
D ((x-55)(x+1))
Explanation opens after your attempt
Correct Answer
A. ((x+11)(x-5))
Step 1
Concept
(11+(-5)=6) and \(11\times(-5)=-55\), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.
Step 2
Why this answer is correct
The correct answer is A. ((x+11)(x-5)). (11+(-5)=6) and \(11\times(-5)=-55\), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.
Step 3
Exam Tip
(11+(-5)=6) और \(11\times(-5)=-55\), इसलिए ((x+11)(x-5)) सही है। परीक्षा में एक चिन्ह धनात्मक और एक ऋणात्मक रखें।
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\(x^2+6x-55=0\) के मूल क्या हैं?
What are the roots of \(x^2+6x-55=0\)?
#quadratic
#roots
#mixed-signs
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A (x=5,-11)
B (x=-5,11)
C (x=55,-1)
D (x=-55,1)
Explanation opens after your attempt
Correct Answer
A. (x=5,-11)
Step 1
Concept
((x+11)(x-5)=0), so (x=-11) and (x=5). In exams, the sign changes while finding roots from factors.
Step 2
Why this answer is correct
The correct answer is A. (x=5,-11). ((x+11)(x-5)=0), so (x=-11) and (x=5). In exams, the sign changes while finding roots from factors.
Step 3
Exam Tip
((x+11)(x-5)=0), इसलिए (x=-11) और (x=5) हैं। परीक्षा में गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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\(x^2+36=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about the real roots of \(x^2+36=0\)?
#quadratic
#no-real-roots
#square-root-method
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A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक मूल (6) और (-6) / Two real roots (6) and (-6)
C एक वास्तविक मूल (36) / One real root (36)
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
\(x^2=-36\) 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. \(x^2=-36\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 3
Exam Tip
\(x^2=-36\) वास्तविक संख्याओं में संभव नहीं है। परीक्षा में \(x^2\) का मान वास्तविक (x) के लिए ऋणात्मक नहीं होता।
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\(12x^2=108\) को हल करने का सही सरल रूप कौनसा है?
What is the correct simplified form to solve \(12x^2=108\)?
#quadratic
#simplification
#square-root-method
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A \(x^2=9\)
B \(x^2=108\)
C \(x^2=12\)
D \(x^2=96\)
Explanation opens after your attempt
Correct Answer
A. \(x^2=9\)
Step 1
Concept
Dividing both sides by (12) 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 (12) gives \(x^2=9\). In exams, remove the coefficient first.
Step 3
Exam Tip
दोनों पक्षों को (12) से भाग देने पर \(x^2=9\) मिलता है। परीक्षा में पहले गुणांक हटाएं।
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\(12x^2=108\) के हल क्या हैं?
What are the solutions of \(12x^2=108\)?
#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 \(12x^2=108\), \(x^2=9\), 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 \(12x^2=108\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 3
Exam Tip
\(12x^2=108\) से \(x^2=9\), इसलिए \(x=\pm3\) है। परीक्षा में अंतिम उत्तर में दोनों मान लिखें।
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द्विघात सूत्र में वर्गमूल के अंदर का सही भाग कौनसा होता है?
What is the correct part inside the square root in the quadratic formula?
#quadratic
#formula
#discriminant
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A \(b^2-4ac\)
B \(b^2+4ac\)
C \(a^2-4bc\)
D \(c^2-4ab\)
Explanation opens after your attempt
Correct Answer
A. \(b^2-4ac\)
Step 1
Concept
In the quadratic formula, the part inside the square root is \(b^2-4ac\). In exams, it is also called the discriminant (D).
Step 2
Why this answer is correct
The correct answer is A. \(b^2-4ac\). In the quadratic formula, the part inside the square root is \(b^2-4ac\). In exams, it is also called the discriminant (D).
Step 3
Exam Tip
द्विघात सूत्र में वर्गमूल के अंदर \(b^2-4ac\) होता है। परीक्षा में इसे विविक्तकर (D) भी कहते हैं।
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यदि \(x^2+10x+k\) एक पूर्ण वर्ग हो और उसका रूप ((x+5)2 ) हो, तो (k) क्या होगा?
If \(x^2+10x+k\) is a perfect square and its form is ((x+5)2 ), what is (k)?
#quadratic
#perfect-square
#missing-term
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A (25)
B (5)
C (10)
D (50)
Explanation opens after your attempt
Step 1
Concept
((x+5)2 =x-2 +10x+25), so (k=25). In exams, remember perfect square expansion.
Step 2
Why this answer is correct
The correct answer is A. (25). ((x+5)2 =x-2 +10x+25), so (k=25). In exams, remember perfect square expansion.
Step 3
Exam Tip
((x+5)2 =x-2 +10x+25), इसलिए (k=25) है। परीक्षा में पूर्ण वर्ग विस्तार याद रखें।
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किस विधि में समीकरण को पहले ((x+p)2 =q) जैसे रूप में बदला जाता है?
In which method is the equation first changed into a form like ((x+p)2 =q)?
#quadratic
#completing-square
#method
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A पूर्ण वर्ग विधि / Completing square method
B गुणनखंड विधि / Factorisation method
C वर्गों के अंतर की विधि / Difference of squares method
D शून्य गुणनफल नियम / Zero product rule
Explanation opens after your attempt
Correct Answer
A. पूर्ण वर्ग विधि / Completing square method
Step 1
Concept
In completing square method, the quadratic part is made into the form ((x+p)2 ). In exams, this method is useful when simple factors are not found quickly.
Step 2
Why this answer is correct
The correct answer is A. पूर्ण वर्ग विधि / Completing square method. In completing square method, the quadratic part is made into the form ((x+p)2 ). In exams, this method is useful when simple factors are not found quickly.
Step 3
Exam Tip
पूर्ण वर्ग विधि में द्विघात भाग को ((x+p)2 ) के रूप में बनाया जाता है। परीक्षा में यह विधि तब उपयोगी है जब सामान्य गुणनखंड जल्दी न मिलें।
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\(25x^2-1=0\) का सही गुणनखंड रूप कौनसा है?
What is the correct factorised form of \(25x^2-1=0\)?
#quadratic
#difference-of-squares
#factorisation
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A ((5x-1)(5x+1)=0)
B ((25x-1)(x+1)=0)
C ((5x-1)2 =0)
D ((x-1)(25x+1)=0)
Explanation opens after your attempt
Correct Answer
A. ((5x-1)(5x+1)=0)
Step 1
Concept
(25x-2 -1=(5x)2 -12 ), so ((5x-1)(5x+1)=0) is correct. In exams, quickly identify the difference of squares.
Step 2
Why this answer is correct
The correct answer is A. ((5x-1)(5x+1)=0). (25x-2 -1=(5x)2 -12 ), so ((5x-1)(5x+1)=0) is correct. In exams, quickly identify the difference of squares.
Step 3
Exam Tip
(25x-2 -1=(5x)2 -12 ), इसलिए ((5x-1)(5x+1)=0) सही है। परीक्षा में वर्गों के अंतर को जल्दी पहचानें।
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