द्विघात व्यंजक \(3x^2-5x+7\) में \(x^2\) का गुणांक क्या है?
What is the coefficient of \(x^2\) in the quadratic expression \(3x^2-5x+7\)?
#quadratic expressions
#coefficient
#algebraic identities
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A (3) / coefficient of \(x^2\)
B (-5) / coefficient of (x)
C (7) / constant term
D (5) / sign ignored
Explanation opens after your attempt
Correct Answer
A. (3) / coefficient of \(x^2\)
Step 1
Concept
The number attached to \(x^2\) is (3). In exams check the sign while finding coefficients.
Step 2
Why this answer is correct
The correct answer is A. (3) / coefficient of \(x^2\). The number attached to \(x^2\) is (3). In exams check the sign while finding coefficients.
Step 3
Exam Tip
\(x^2\) के साथ लगी संख्या (3) है। परीक्षा में गुणांक निकालते समय चिन्ह भी देखें।
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व्यंजक \(2y^2+9y-11\) में स्थिर पद कौन-सा है?
Which is the constant term in the expression \(2y^2+9y-11\)?
#quadratic expressions
#constant term
#standard form
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A (2) / quadratic coefficient
B (9) / linear coefficient
C (-11) / constant term
D (11y) / wrong term
Explanation opens after your attempt
Correct Answer
C. (-11) / constant term
Step 1
Concept
The term without a variable is the constant term. In exams write the constant term with its sign.
Step 2
Why this answer is correct
The correct answer is C. (-11) / constant term. The term without a variable is the constant term. In exams write the constant term with its sign.
Step 3
Exam Tip
जिस पद में चर नहीं होता वही स्थिर पद होता है। परीक्षा में चिन्ह सहित स्थिर पद लिखें।
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\(x^2+6x+9\) को पूर्ण वर्ग के रूप में लिखें।
Write \(x^2+6x+9\) as a perfect square.
#quadratic expressions
#perfect square
#identity
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A ((x+3)2 ) / correct perfect square
B ((x-3)2 ) / wrong sign
C ((x+6)2 ) / middle doubled wrongly
D ((x+9)2 ) / constant used as side
Explanation opens after your attempt
Correct Answer
A. ((x+3)2 ) / correct perfect square
Step 1
Concept
\(9=3^2\) and \(6x=2\cdot x\cdot3\). In exams take the square root of the last term and match the middle term.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)2 ) / correct perfect square. \(9=3^2\) and \(6x=2\cdot x\cdot3\). In exams take the square root of the last term and match the middle term.
Step 3
Exam Tip
\(9=3^2\) और \(6x=2\cdot x\cdot3\) है। परीक्षा में अंतिम पद का वर्गमूल लेकर मध्य पद मिलाएं।
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\(a^2-10a+25\) किस पूर्ण वर्ग के बराबर है?
Which perfect square is equal to \(a^2-10a+25\)?
#quadratic expressions
#perfect square
#negative middle
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A ((a+5)2 ) / sign error
B ((a-5)2 ) / correct square
C ((a-10)2 ) / wrong constant
D ((a-25)2 ) / wrong side
Explanation opens after your attempt
Correct Answer
B. ((a-5)2 ) / correct square
Step 1
Concept
The middle term \(-10a=-2\cdot a\cdot5\), so the sign is negative. In exams the sign of the middle term is decisive.
Step 2
Why this answer is correct
The correct answer is B. ((a-5)2 ) / correct square. The middle term \(-10a=-2\cdot a\cdot5\), so the sign is negative. In exams the sign of the middle term is decisive.
Step 3
Exam Tip
मध्य पद \(-10a=-2\cdot a\cdot5\) है इसलिए चिन्ह ऋणात्मक होगा। परीक्षा में मध्य पद का चिन्ह निर्णायक होता है।
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\(x^2+8x+15\) के दो रैखिक गुणनखंड कौन-से हैं?
What are the two linear factors of \(x^2+8x+15\)?
#quadratic expressions
#factorisation
#split middle
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A ((x+3)(x+5)) / correct factors
B ((x+1)(x+15)) / middle term wrong
C ((x-3)(x-5)) / sign wrong
D ((x+4)(x+4)) / constant wrong
Explanation opens after your attempt
Correct Answer
A. ((x+3)(x+5)) / correct factors
Step 1
Concept
(3+5=8) and \(3\cdot5=15\). In exams match the sum with the middle coefficient and the product with the constant term.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)(x+5)) / correct factors. (3+5=8) and \(3\cdot5=15\). In exams match the sum with the middle coefficient and the product with the constant term.
Step 3
Exam Tip
(3+5=8) और \(3\cdot5=15\) है। परीक्षा में योग मध्य गुणांक और गुणनफल स्थिर पद से मिलाएं।
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\(x^2-7x+12\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2-7x+12\)?
#quadratic expressions
#negative factors
#factorisation
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A ((x+3)(x+4)) / sign error
B ((x-3)(x-4)) / correct factors
C ((x-2)(x-6)) / middle term wrong
D ((x-1)(x-12)) / middle term wrong
Explanation opens after your attempt
Correct Answer
B. ((x-3)(x-4)) / correct factors
Step 1
Concept
(-3+(-4)=-7) and ((-3)(-4)=12). In exams use two negative signs when the constant is positive and the middle term is negative.
Step 2
Why this answer is correct
The correct answer is B. ((x-3)(x-4)) / correct factors. (-3+(-4)=-7) and ((-3)(-4)=12). In exams use two negative signs when the constant is positive and the middle term is negative.
Step 3
Exam Tip
(-3+(-4)=-7) और ((-3)(-4)=12) है। परीक्षा में धनात्मक स्थिर पद और ऋणात्मक मध्य पद पर दोनों चिन्ह ऋणात्मक लें।
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\(x^2+x-20\) को गुणनखंडित करें।
Factorise \(x^2+x-20\).
#quadratic expressions
#signed factors
#factorisation
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A ((x+5)(x-4)) / correct factors
B ((x-5)(x+4)) / middle sign wrong
C ((x+2)(x-10)) / middle term wrong
D ((x+1)(x-20)) / middle term wrong
Explanation opens after your attempt
Correct Answer
A. ((x+5)(x-4)) / correct factors
Step 1
Concept
(5+(-4)=1) and (5\cdot(-4)=-20). In exams look for opposite signs when the constant is negative.
Step 2
Why this answer is correct
The correct answer is A. ((x+5)(x-4)) / correct factors. (5+(-4)=1) and (5\cdot(-4)=-20). In exams look for opposite signs when the constant is negative.
Step 3
Exam Tip
(5+(-4)=1) और (5\cdot(-4)=-20) है। परीक्षा में ऋणात्मक स्थिर पद के लिए विपरीत चिन्ह खोजें।
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\(3x^2+10x+8\) का सही गुणनखंड रूप चुनें।
Choose the correct factorised form of \(3x^2+10x+8\).
#quadratic expressions
#factorisation
#cross terms
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A ((3x+4)(x+2)) / correct factors
B ((3x+2)(x+4)) / middle term wrong
C ((x+4)(x+2)) / leading coefficient wrong
D ((3x-4)(x-2)) / sign wrong
Explanation opens after your attempt
Correct Answer
A. ((3x+4)(x+2)) / correct factors
Step 1
Concept
In ((3x+4)(x+2)), the cross terms are (6x) and (4x). In exams their sum should be the middle term.
Step 2
Why this answer is correct
The correct answer is A. ((3x+4)(x+2)) / correct factors. In ((3x+4)(x+2)), the cross terms are (6x) and (4x). In exams their sum should be the middle term.
Step 3
Exam Tip
((3x+4)(x+2)) में क्रॉस पद (6x) और (4x) हैं। परीक्षा में क्रॉस पदों का योग मध्य पद होना चाहिए।
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\(4x^2-9\) को द्विघात पहचान से लिखें।
Write \(4x^2-9\) using a quadratic identity.
#quadratic expressions
#difference of squares
#identity
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A ((2x-3)2 ) / square of difference
B ((2x-3)(2x+3)) / difference of squares
C ((4x-9)(4x+9)) / wrong roots
D ((2x+3)2 ) / square of sum
Explanation opens after your attempt
Correct Answer
B. ((2x-3)(2x+3)) / difference of squares
Step 1
Concept
(4x-2 =(2x)2 ) and \(9=3^2\). In exams write difference of squares as ((a-b)(a+b)).
Step 2
Why this answer is correct
The correct answer is B. ((2x-3)(2x+3)) / difference of squares. (4x-2 =(2x)2 ) and \(9=3^2\). In exams write difference of squares as ((a-b)(a+b)).
Step 3
Exam Tip
(4x-2 =(2x)2 ) और \(9=3^2\) है। परीक्षा में वर्गों के अंतर को ((a-b)(a+b)) लिखें।
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\(9y^2-24y+16\) किसके बराबर है?
What is \(9y^2-24y+16\) equal to?
#quadratic expressions
#perfect square
#identity
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A ((3y+4)2 ) / sign error
B ((3y-4)2 ) / correct perfect square
C ((9y-4)2 ) / wrong root
D ((3y-8)(3y-2)) / middle term wrong
Explanation opens after your attempt
Correct Answer
B. ((3y-4)2 ) / correct perfect square
Step 1
Concept
(9y-2 =(3y)2 ), \(16=4^2\), and the middle term is (-24y). In exams match both square terms and the middle term.
Step 2
Why this answer is correct
The correct answer is B. ((3y-4)2 ) / correct perfect square. (9y-2 =(3y)2 ), \(16=4^2\), and the middle term is (-24y). In exams match both square terms and the middle term.
Step 3
Exam Tip
(9y-2 =(3y)2 ), \(16=4^2\), और मध्य पद (-24y) है। परीक्षा में दोनों वर्ग पद और मध्य पद मिलाएं।
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(x=2) होने पर \(x^2-5x+6\) का मान क्या होगा?
What is the value of \(x^2-5x+6\) when (x=2)?
#quadratic expressions
#evaluation
#substitution
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A (0) / correct value
B (2) / substitution error
C (4) / square only
D (6) / constant only
Explanation opens after your attempt
Correct Answer
A. (0) / correct value
Step 1
Concept
Substitution gives \(2^2-5\cdot2+6=0\). In exams do powers first, then multiplication, then addition or subtraction.
Step 2
Why this answer is correct
The correct answer is A. (0) / correct value. Substitution gives \(2^2-5\cdot2+6=0\). In exams do powers first, then multiplication, then addition or subtraction.
Step 3
Exam Tip
मान रखने पर \(2^2-5\cdot2+6=0\) मिलता है। परीक्षा में पहले घात फिर गुणा और फिर जोड़-घटाव करें।
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(t=-1) पर \(2t^2+3t-5\) का मान क्या है?
What is the value of \(2t^2+3t-5\) at (t=-1)?
#quadratic expressions
#evaluation
#negative value
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A (-6) / correct value
B (0) / sign error
C (6) / negative ignored
D (-10) / square error
Explanation opens after your attempt
Correct Answer
A. (-6) / correct value
Step 1
Concept
(2(-1)2 +3(-1)-5=2-3-5=-6). In exams the square of a negative number is positive.
Step 2
Why this answer is correct
The correct answer is A. (-6) / correct value. (2(-1)2 +3(-1)-5=2-3-5=-6). In exams the square of a negative number is positive.
Step 3
Exam Tip
(2(-1)2 +3(-1)-5=2-3-5=-6) है। परीक्षा में ऋणात्मक संख्या का वर्ग धनात्मक होता है।
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\(x^2+kx+16\) पूर्ण वर्ग हो और (k) धनात्मक हो तो (k) का मान क्या है?
If \(x^2+kx+16\) is a perfect square and (k) is positive, what is the value of (k)?
#quadratic expressions
#perfect square
#unknown coefficient
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A (4) / square root only
B (8) / correct middle coefficient
C (16) / constant term
D (32) / doubled twice
Explanation opens after your attempt
Correct Answer
B. (8) / correct middle coefficient
Step 1
Concept
\(16=4^2\), and in a perfect square the middle term is \(2\cdot x\cdot4=8x\). In exams double the root of the constant term.
Step 2
Why this answer is correct
The correct answer is B. (8) / correct middle coefficient. \(16=4^2\), and in a perfect square the middle term is \(2\cdot x\cdot4=8x\). In exams double the root of the constant term.
Step 3
Exam Tip
\(16=4^2\) और पूर्ण वर्ग में मध्य पद \(2\cdot x\cdot4=8x\) होगा। परीक्षा में स्थिर पद की जड़ को दोगुना करें।
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\(x^2-14x+c\) पूर्ण वर्ग हो तो (c) क्या होगा?
If \(x^2-14x+c\) is a perfect square, what is (c)?
#quadratic expressions
#complete square
#unknown constant
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A (14) / middle coefficient
B (7) / half coefficient
C (49) / square of half coefficient
D (196) / square of middle coefficient
Explanation opens after your attempt
Correct Answer
C. (49) / square of half coefficient
Step 1
Concept
\(-14x=-2\cdot x\cdot7\), so \(c=7^2=49\). In exams take half of the middle coefficient and square it.
Step 2
Why this answer is correct
The correct answer is C. (49) / square of half coefficient. \(-14x=-2\cdot x\cdot7\), so \(c=7^2=49\). In exams take half of the middle coefficient and square it.
Step 3
Exam Tip
\(-14x=-2\cdot x\cdot7\), इसलिए \(c=7^2=49\) है। परीक्षा में मध्य गुणांक का आधा लेकर उसका वर्ग करें।
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\(x^2+px+21\) के गुणनखंड ((x+3)(x+7)) हैं। (p) का मान क्या है?
The factors of \(x^2+px+21\) are ((x+3)(x+7)). What is the value of (p)?
#quadratic expressions
#unknown coefficient
#factorisation
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A (3) / one factor only
B (7) / one factor only
C (10) / sum of constants
D (21) / product of constants
Explanation opens after your attempt
Correct Answer
C. (10) / sum of constants
Step 1
Concept
((x+3)(x+7)=x-2 +10x+21), so (p=10). In exams the middle term comes from the sum of the constants.
Step 2
Why this answer is correct
The correct answer is C. (10) / sum of constants. ((x+3)(x+7)=x-2 +10x+21), so (p=10). In exams the middle term comes from the sum of the constants.
Step 3
Exam Tip
((x+3)(x+7)=x-2 +10x+21), इसलिए (p=10) है। परीक्षा में मध्य पद स्थिर संख्याओं के योग से बनता है।
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\(x^2-11x+30\) के शून्य बनाने वाले गुणनखंड कौन-से हैं?
Which factors make \(x^2-11x+30\) zero?
#quadratic expressions
#factorisation
#zeros
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A ((x-5)(x-6)) / correct factors
B ((x+5)(x+6)) / sign error
C ((x-3)(x-10)) / middle term wrong
D ((x-2)(x-15)) / middle term wrong
Explanation opens after your attempt
Correct Answer
A. ((x-5)(x-6)) / correct factors
Step 1
Concept
(-5+(-6)=-11) and ((-5)(-6)=30). In exams verify by expanding the factors.
Step 2
Why this answer is correct
The correct answer is A. ((x-5)(x-6)) / correct factors. (-5+(-6)=-11) and ((-5)(-6)=30). In exams verify by expanding the factors.
Step 3
Exam Tip
(-5+(-6)=-11) और ((-5)(-6)=30) है। परीक्षा में गुणनखंड फैलाकर सत्यापन करें।
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कौन-सा द्विघात व्यंजक ((x-4)(x+2)) से प्राप्त होता है?
Which quadratic expression is obtained from ((x-4)(x+2))?
#quadratic expressions
#expansion
#factor product
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A \(x^2-2x-8\) / correct expansion
B \(x^2+6x+8\) / sign error
C \(x^2-6x-8\) / middle error
D \(x^2+2x-8\) / one term missed
Explanation opens after your attempt
Correct Answer
A. \(x^2-2x-8\) / correct expansion
Step 1
Concept
The cross terms (2x) and (-4x) combine to (-2x). In exams add outer and inner terms.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-2x-8\) / correct expansion. The cross terms (2x) and (-4x) combine to (-2x). In exams add outer and inner terms.
Step 3
Exam Tip
क्रॉस पद (2x) और (-4x) मिलकर (-2x) देते हैं। परीक्षा में बाहरी और अंदरूनी पद जोड़ें।
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((2x-3)(x+5)) का विस्तार क्या है?
What is the expansion of ((2x-3)(x+5))?
#quadratic expressions
#expansion
#cross terms
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A \(2x^2+7x-15\) / correct expansion
B \(2x^2+10x-15\) / one cross term missed
C \(2x^2-13x-15\) / sign error
D \(2x^2+7x+15\) / constant sign wrong
Explanation opens after your attempt
Correct Answer
A. \(2x^2+7x-15\) / correct expansion
Step 1
Concept
\(2x\cdot5=10x\) and \(-3\cdot x=-3x\), so the middle term is (7x). In exams write all four products.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+7x-15\) / correct expansion. \(2x\cdot5=10x\) and \(-3\cdot x=-3x\), so the middle term is (7x). In exams write all four products.
Step 3
Exam Tip
\(2x\cdot5=10x\) और \(-3\cdot x=-3x\), इसलिए मध्य पद (7x) है। परीक्षा में चारों गुणनफल लिखें।
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\(5x^2-20x+20\) में समान गुणनखंड निकालने के बाद अंदर कौन-सा द्विघात व्यंजक बचेगा?
After taking the common factor from \(5x^2-20x+20\), which quadratic expression remains inside?
#quadratic expressions
#common factor
#perfect square
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A \(x^2-4x+4\) / correct inside expression
B \(5x^2-4x+4\) / common factor incomplete
C \(x^2+4x+4\) / sign error
D \(x^2-20x+20\) / not simplified
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x+4\) / correct inside expression
Step 1
Concept
Taking common (5) gives (5\(x^2-4x+4\)). In exams first take the common factor and then apply identities.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+4\) / correct inside expression. Taking common (5) gives (5\(x^2-4x+4\)). In exams first take the common factor and then apply identities.
Step 3
Exam Tip
समान गुणनखंड (5) निकालने पर (5\(x^2-4x+4\)) मिलता है। परीक्षा में पहले समान गुणनखंड निकालें फिर पहचान लगाएं।
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\(3x^2-27\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(3x^2-27\)?
#quadratic expressions
#complete factorisation
#difference of squares
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A (3(x-3)(x+3)) / correct complete form
B (3\(x^2-27\)) / wrong common factor
C (3(x-9)(x+9)) / wrong square root
D (3(x-3)2 ) / wrong identity
Explanation opens after your attempt
Correct Answer
A. (3(x-3)(x+3)) / correct complete form
Step 1
Concept
First take out (3), and \(x^2-9\) is a difference of squares. In exams factorise the inside expression too.
Step 2
Why this answer is correct
The correct answer is A. (3(x-3)(x+3)) / correct complete form. First take out (3), and \(x^2-9\) is a difference of squares. In exams factorise the inside expression too.
Step 3
Exam Tip
पहले (3) निकालें और \(x^2-9\) वर्गों का अंतर है। परीक्षा में अंदर की अभिव्यक्ति को भी गुणनखंडित करें।
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\(4x^2+12x+k\) पूर्ण वर्ग हो तो (k) क्या होगा?
If \(4x^2+12x+k\) is a perfect square, what is (k)?
#quadratic expressions
#unknown constant
#perfect square
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A (3) / half of (6)
B (9) / correct constant
C (12) / middle coefficient
D (36) / square of (6)
Explanation opens after your attempt
Correct Answer
B. (9) / correct constant
Step 1
Concept
(4x-2 =(2x)2 ) and \(12x=2\cdot2x\cdot3\), so (k=9). In exams identify the second part using (2ab).
Step 2
Why this answer is correct
The correct answer is B. (9) / correct constant. (4x-2 =(2x)2 ) and \(12x=2\cdot2x\cdot3\), so (k=9). In exams identify the second part using (2ab).
Step 3
Exam Tip
(4x-2 =(2x)2 ) और \(12x=2\cdot2x\cdot3\), इसलिए (k=9) है। परीक्षा में (2ab) से दूसरा भाग पहचानें।
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\(9x^2+kx+25\) पूर्ण वर्ग हो और (k) ऋणात्मक हो तो (k) क्या है?
If \(9x^2+kx+25\) is a perfect square and (k) is negative, what is (k)?
#quadratic expressions
#perfect square
#unknown coefficient
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A (30) / positive middle
B (-30) / correct middle
C (15) / half middle
D (-15) / half value
Explanation opens after your attempt
Correct Answer
B. (-30) / correct middle
Step 1
Concept
(9x-2 =(3x)2 ), \(25=5^2\), so the middle term is \(-2\cdot3x\cdot5=-30x\). In exams follow the given sign.
Step 2
Why this answer is correct
The correct answer is B. (-30) / correct middle. (9x-2 =(3x)2 ), \(25=5^2\), so the middle term is \(-2\cdot3x\cdot5=-30x\). In exams follow the given sign.
Step 3
Exam Tip
(9x-2 =(3x)2 ), \(25=5^2\), इसलिए मध्य पद \(-2\cdot3x\cdot5=-30x\) है। परीक्षा में दिए गए चिन्ह का पालन करें।
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\(x^2+2xy-15y^2\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2+2xy-15y^2\)?
#quadratic expressions
#variables
#factorisation
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A ((x+5y)(x-3y)) / correct factors
B ((x-5y)(x+3y)) / middle sign wrong
C ((x+15y)(x-y)) / middle term wrong
D ((x+2y)(x-15y)) / middle term wrong
Explanation opens after your attempt
Correct Answer
A. ((x+5y)(x-3y)) / correct factors
Step 1
Concept
(5y+(-3y)=2y) and (5y\cdot(-3y)=-15y-2 ). In exams add the factors along with variables.
Step 2
Why this answer is correct
The correct answer is A. ((x+5y)(x-3y)) / correct factors. (5y+(-3y)=2y) and (5y\cdot(-3y)=-15y-2 ). In exams add the factors along with variables.
Step 3
Exam Tip
(5y+(-3y)=2y) और (5y\cdot(-3y)=-15y-2 ) है। परीक्षा में चर सहित गुणनखंडों का योग करें।
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\(2a^2+ab-6b^2\) के गुणनखंड कौन-से हैं?
What are the factors of \(2a^2+ab-6b^2\)?
#quadratic expressions
#two variables
#factorisation
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A ((2a-3b)(a+2b)) / correct factors
B ((2a+3b)(a-2b)) / middle sign wrong
C ((a-3b)(2a+2b)) / leading term wrong
D ((2a-b)(a+6b)) / middle term wrong
Explanation opens after your attempt
Correct Answer
A. ((2a-3b)(a+2b)) / correct factors
Step 1
Concept
((2a-3b)(a+2b)=2a-2 +ab-6b-2 ). In exams add cross terms to check the middle term.
Step 2
Why this answer is correct
The correct answer is A. ((2a-3b)(a+2b)) / correct factors. ((2a-3b)(a+2b)=2a-2 +ab-6b-2 ). In exams add cross terms to check the middle term.
Step 3
Exam Tip
((2a-3b)(a+2b)=2a-2 +ab-6b-2 ) है। परीक्षा में क्रॉस पदों को जोड़कर मध्य पद जांचें।
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\(6p^2+5pq-q^2\) का सही गुणनखंड रूप चुनें।
Choose the correct factorised form of \(6p^2+5pq-q^2\).
#quadratic expressions
#cross terms
#factorisation
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A ((3p-q)(2p+q)) / correct factors
B ((3p+q)(2p-q)) / middle sign wrong
C ((6p-q)(p+q)) / middle term wrong
D ((2p-q)(3p-q)) / constant sign wrong
Explanation opens after your attempt
Correct Answer
A. ((3p-q)(2p+q)) / correct factors
Step 1
Concept
Actually ((3p-q)(2p+q)) gives \(6p^2+pq-q^2\). Check the cross terms carefully in exams.
Step 2
Why this answer is correct
The correct answer is A. ((3p-q)(2p+q)) / correct factors. Actually ((3p-q)(2p+q)) gives \(6p^2+pq-q^2\). Check the cross terms carefully in exams.
Step 3
Exam Tip
क्रॉस पद (3pq) और (-2pq) नहीं, सही रूप में \(3p\cdot q=3pq\) और \(-q\cdot2p=-2pq\) से (pq) मिलता है?
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\(6p^2+5pq-q^2\) का सही गुणनखंड रूप कौन-सा है?
Which is the correct factorised form of \(6p^2+5pq-q^2\)?
#quadratic expressions
#two variables
#factorisation
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A ((6p-q)(p+q)) / correct factors
B ((3p-q)(2p+q)) / middle term wrong
C ((6p+q)(p-q)) / middle sign wrong
D ((2p-q)(3p+q)) / middle term wrong
Explanation opens after your attempt
Correct Answer
A. ((6p-q)(p+q)) / correct factors
Step 1
Concept
((6p-q)(p+q)=6p-2 +5pq-q-2 ). In exams always expand the option and check cross terms.
Step 2
Why this answer is correct
The correct answer is A. ((6p-q)(p+q)) / correct factors. ((6p-q)(p+q)=6p-2 +5pq-q-2 ). In exams always expand the option and check cross terms.
Step 3
Exam Tip
((6p-q)(p+q)=6p-2 +5pq-q-2 ) है। परीक्षा में विकल्प को फैलाकर क्रॉस पद जरूर जांचें।
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\(x^2+4x+4-y^2\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(x^2+4x+4-y^2\)?
#quadratic expressions
#grouping
#difference of squares
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A ((x+2-y)(x+2+y)) / correct form
B ((x-2-y)(x-2+y)) / sign error
C ((x+2)2 -y-2 ) / partial form
D ((x+y+2)2 ) / wrong square
Explanation opens after your attempt
Correct Answer
A. ((x+2-y)(x+2+y)) / correct form
Step 1
Concept
First (x-2 +4x+4=(x+2)2 ), then apply difference of squares. In exams do not stop at the partial form.
Step 2
Why this answer is correct
The correct answer is A. ((x+2-y)(x+2+y)) / correct form. First (x-2 +4x+4=(x+2)2 ), then apply difference of squares. In exams do not stop at the partial form.
Step 3
Exam Tip
पहले (x-2 +4x+4=(x+2)2 ), फिर वर्गों का अंतर लगाएं। परीक्षा में आंशिक रूप पर न रुकें।
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\(x^2-6x+9-z^2\) को गुणनखंडित करें।
Factorise \(x^2-6x+9-z^2\).
#quadratic expressions
#perfect square
#difference of squares
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A ((x-3-z)(x-3+z)) / correct form
B ((x+3-z)(x+3+z)) / sign error
C ((x-3)2 -z-2 ) / partial form
D ((x-3-z)2 ) / wrong square
Explanation opens after your attempt
Correct Answer
A. ((x-3-z)(x-3+z)) / correct form
Step 1
Concept
First (x-2 -6x+9=(x-3)2 ), then use the \(a^2-b^2\) identity. In exams identify the perfect-square group.
Step 2
Why this answer is correct
The correct answer is A. ((x-3-z)(x-3+z)) / correct form. First (x-2 -6x+9=(x-3)2 ), then use the \(a^2-b^2\) identity. In exams identify the perfect-square group.
Step 3
Exam Tip
पहले (x-2 -6x+9=(x-3)2 ), फिर (\(a^2-b^2\)) पहचान लगती है। परीक्षा में पूर्ण वर्ग समूह पहचानें।
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\(x^2+ax+24\) का एक गुणनखंड (x+6) है। दूसरा गुणनखंड (x+4) हो तो (a) क्या है?
One factor of \(x^2+ax+24\) is (x+6). If the other factor is (x+4), what is (a)?
#quadratic expressions
#unknown coefficient
#factors
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A (10) / correct coefficient
B (24) / constant term
C (6) / one factor constant
D (4) / second factor constant
Explanation opens after your attempt
Correct Answer
A. (10) / correct coefficient
Step 1
Concept
((x+6)(x+4)=x-2 +10x+24), so (a=10). In exams add the constant parts of the factors.
Step 2
Why this answer is correct
The correct answer is A. (10) / correct coefficient. ((x+6)(x+4)=x-2 +10x+24), so (a=10). In exams add the constant parts of the factors.
Step 3
Exam Tip
((x+6)(x+4)=x-2 +10x+24), इसलिए (a=10) है। परीक्षा में गुणनखंडों के स्थिर भाग जोड़ें।
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\(x^2-bx+18\) के गुणनखंड ((x-3)(x-6)) हैं। (b) क्या है?
The factors of \(x^2-bx+18\) are ((x-3)(x-6)). What is (b)?
#quadratic expressions
#unknown coefficient
#negative middle
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A (3) / one constant
B (6) / one constant
C (9) / correct value
D (18) / product
Explanation opens after your attempt
Correct Answer
C. (9) / correct value
Step 1
Concept
((x-3)(x-6)=x-2 -9x+18), so (b=9). In exams (b) remains positive because the expression has (-bx).
Step 2
Why this answer is correct
The correct answer is C. (9) / correct value. ((x-3)(x-6)=x-2 -9x+18), so (b=9). In exams (b) remains positive because the expression has (-bx).
Step 3
Exam Tip
((x-3)(x-6)=x-2 -9x+18), इसलिए (b=9) है। परीक्षा में ऋण चिन्ह के कारण (b) धनात्मक रहता है।
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\(2x^2+mx+10\) का गुणनखंड रूप ((2x+5)(x+2)) है। (m) क्या होगा?
The factorised form of \(2x^2+mx+10\) is ((2x+5)(x+2)). What is (m)?
#quadratic expressions
#cross terms
#unknown coefficient
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A (4) / one cross term
B (5) / one cross term
C (9) / correct middle coefficient
D (10) / constant term
Explanation opens after your attempt
Correct Answer
C. (9) / correct middle coefficient
Step 1
Concept
The cross terms are (4x) and (5x), whose sum is (9x). In exams add both cross terms.
Step 2
Why this answer is correct
The correct answer is C. (9) / correct middle coefficient. The cross terms are (4x) and (5x), whose sum is (9x). In exams add both cross terms.
Step 3
Exam Tip
क्रॉस पद (4x) और (5x) हैं, जिनका योग (9x) है। परीक्षा में दोनों क्रॉस पद जोड़ें।
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कौन-सा (k), \(x^2+kx+81\) को पूर्ण वर्ग बनाएगा और (k) ऋणात्मक है?
Which (k) makes \(x^2+kx+81\) a perfect square and (k) is negative?
#quadratic expressions
#perfect square
#unknown coefficient
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A (18) / positive middle
B (-18) / correct middle
C (9) / root only
D (-9) / half value
Explanation opens after your attempt
Correct Answer
B. (-18) / correct middle
Step 1
Concept
\(81=9^2\), so the negative middle term is \(-2\cdot x\cdot9=-18x\). In exams do not forget the given sign.
Step 2
Why this answer is correct
The correct answer is B. (-18) / correct middle. \(81=9^2\), so the negative middle term is \(-2\cdot x\cdot9=-18x\). In exams do not forget the given sign.
Step 3
Exam Tip
\(81=9^2\), इसलिए ऋणात्मक मध्य पद \(-2\cdot x\cdot9=-18x\) होगा। परीक्षा में दिए गए चिन्ह को न भूलें।
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\(x^2-4x+5\) को पूर्ण वर्ग जोड़कर किस रूप में लिखा जा सकता है?
How can \(x^2-4x+5\) be written by completing the square?
#quadratic expressions
#completing square
#standard form
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A ((x-2)2 +1) / correct form
B ((x+2)2 +1) / sign error
C ((x-2)2 -1) / constant error
D ((x-4)2 +5) / wrong half coefficient
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 +1) / correct form
Step 1
Concept
(x-2 -4x+4=(x-2)2 ), so write (5) as (4+1). In exams take half of the middle coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x-2)2 +1) / correct form. (x-2 -4x+4=(x-2)2 ), so write (5) as (4+1). In exams take half of the middle coefficient.
Step 3
Exam Tip
(x-2 -4x+4=(x-2)2 ), इसलिए (5) को (4+1) लिखते हैं। परीक्षा में मध्य गुणांक का आधा लें।
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\(x^2+6x+11\) को पूर्ण वर्ग रूप में लिखें।
Write \(x^2+6x+11\) in completed square form.
#quadratic expressions
#completing square
#identity
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A ((x+3)2 +2) / correct form
B ((x-3)2 +2) / sign error
C ((x+6)2 -25) / wrong half coefficient
D ((x+3)2 -2) / constant error
Explanation opens after your attempt
Correct Answer
A. ((x+3)2 +2) / correct form
Step 1
Concept
(x-2 +6x+9=(x+3)2 ), so (11=9+2). In exams form the needed square and keep the remainder separate.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)2 +2) / correct form. (x-2 +6x+9=(x+3)2 ), so (11=9+2). In exams form the needed square and keep the remainder separate.
Step 3
Exam Tip
(x-2 +6x+9=(x+3)2 ), इसलिए (11=9+2) है। परीक्षा में आवश्यक वर्ग जोड़कर शेष अलग रखें।
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\(x^2-8x+13\) का पूर्ण वर्ग रूप कौन-सा है?
Which is the completed square form of \(x^2-8x+13\)?
#quadratic expressions
#completing square
#hard
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A ((x-4)2 -3) / correct form
B ((x+4)2 -3) / sign error
C ((x-8)2 +13) / wrong half coefficient
D ((x-4)2 +3) / constant error
Explanation opens after your attempt
Correct Answer
A. ((x-4)2 -3) / correct form
Step 1
Concept
(x-2 -8x+16=(x-4)2 ) and (13=16-3). In exams balance the square that was added.
Step 2
Why this answer is correct
The correct answer is A. ((x-4)2 -3) / correct form. (x-2 -8x+16=(x-4)2 ) and (13=16-3). In exams balance the square that was added.
Step 3
Exam Tip
(x-2 -8x+16=(x-4)2 ) और (13=16-3) है। परीक्षा में जो वर्ग जोड़ा गया है उसे संतुलित करें।
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यदि (x+y=7) और (xy=10), तो \(x^2+y^2\) का मान क्या है?
If (x+y=7) and (xy=10), what is the value of \(x^2+y^2\)?
#quadratic expressions
#identity
#value based
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A (29) / correct value
B (49) / square of sum only
C (39) / one product subtracted
D (69) / product added twice
Explanation opens after your attempt
Correct Answer
A. (29) / correct value
Step 1
Concept
(x-2 +y-2 =(x+y)2 -2xy=49-20=29). In exams use the square of sum identity.
Step 2
Why this answer is correct
The correct answer is A. (29) / correct value. (x-2 +y-2 =(x+y)2 -2xy=49-20=29). In exams use the square of sum identity.
Step 3
Exam Tip
(x-2 +y-2 =(x+y)2 -2xy=49-20=29) है। परीक्षा में योग के वर्ग की पहचान उपयोग करें।
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यदि (a-b=5) और (ab=6), तो \(a^2+b^2\) क्या होगा?
If (a-b=5) and (ab=6), what is \(a^2+b^2\)?
#quadratic expressions
#identity
#value based
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A (13) / product added once
B (25) / square of difference only
C (37) / correct value
D (31) / wrong sign
Explanation opens after your attempt
Correct Answer
C. (37) / correct value
Step 1
Concept
(a-2 +b-2 =(a-b)2 +2ab=25+12=37). In exams add (2ab) when using the square of difference identity.
Step 2
Why this answer is correct
The correct answer is C. (37) / correct value. (a-2 +b-2 =(a-b)2 +2ab=25+12=37). In exams add (2ab) when using the square of difference identity.
Step 3
Exam Tip
(a-2 +b-2 =(a-b)2 +2ab=25+12=37) है। परीक्षा में अंतर के वर्ग की पहचान में (+2ab) जोड़ें।
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यदि (p+q=9) और \(p^2+q^2=41\), तो (pq) क्या है?
If (p+q=9) and \(p^2+q^2=41\), what is (pq)?
#quadratic expressions
#identity
#pq value
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A (20) / correct value
B (40) / double product
C (81) / square of sum
D (22) / subtraction error
Explanation opens after your attempt
Correct Answer
A. (20) / correct value
Step 1
Concept
((p+q)2 =p-2 +q-2 +2pq), so (81=41+2pq) and (pq=20). In exams find (2pq) first.
Step 2
Why this answer is correct
The correct answer is A. (20) / correct value. ((p+q)2 =p-2 +q-2 +2pq), so (81=41+2pq) and (pq=20). In exams find (2pq) first.
Step 3
Exam Tip
((p+q)2 =p-2 +q-2 +2pq), इसलिए (81=41+2pq) और (pq=20) है। परीक्षा में (2pq) पहले निकालें।
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यदि (m-n=4) और \(m^2+n^2=34\), तो (mn) क्या है?
If (m-n=4) and \(m^2+n^2=34\), what is (mn)?
#quadratic expressions
#identity
#product value
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A (9) / correct value
B (18) / double product
C (25) / square difference added
D (7) / subtraction error
Explanation opens after your attempt
Correct Answer
A. (9) / correct value
Step 1
Concept
((m-n)2 =m-2 +n-2 -2mn), so (16=34-2mn) and (mn=9). In exams pay attention to the sign.
Step 2
Why this answer is correct
The correct answer is A. (9) / correct value. ((m-n)2 =m-2 +n-2 -2mn), so (16=34-2mn) and (mn=9). In exams pay attention to the sign.
Step 3
Exam Tip
((m-n)2 =m-2 +n-2 -2mn), इसलिए (16=34-2mn) और (mn=9) है। परीक्षा में चिन्ह पर ध्यान दें।
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\(2x^2-3x-2\) का गुणनखंड रूप क्या है?
What is the factorised form of \(2x^2-3x-2\)?
#quadratic expressions
#factorisation
#signed factors
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A ((2x+1)(x-2)) / correct factors
B ((2x-1)(x+2)) / middle sign wrong
C ((x-1)(2x+2)) / constant wrong
D ((2x-2)(x+1)) / leading factor not correct
Explanation opens after your attempt
Correct Answer
A. ((2x+1)(x-2)) / correct factors
Step 1
Concept
((2x+1)(x-2)=2x-2 -3x-2). In exams check opposite signs when the constant term is negative.
Step 2
Why this answer is correct
The correct answer is A. ((2x+1)(x-2)) / correct factors. ((2x+1)(x-2)=2x-2 -3x-2). In exams check opposite signs when the constant term is negative.
Step 3
Exam Tip
((2x+1)(x-2)=2x-2 -3x-2) है। परीक्षा में ऋणात्मक स्थिर पद पर विपरीत चिन्हों को जांचें।
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\(5x^2+13x+6\) के गुणनखंड कौन-से हैं?
What are the factors of \(5x^2+13x+6\)?
#quadratic expressions
#factorisation
#cross terms
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A ((5x+3)(x+2)) / correct factors
B ((5x+2)(x+3)) / middle term wrong
C ((x+3)(x+2)) / leading coefficient wrong
D ((5x-3)(x-2)) / sign wrong
Explanation opens after your attempt
Correct Answer
A. ((5x+3)(x+2)) / correct factors
Step 1
Concept
The cross terms (10x) and (3x) combine to (13x). In exams check options using both leading coefficient and constant term.
Step 2
Why this answer is correct
The correct answer is A. ((5x+3)(x+2)) / correct factors. The cross terms (10x) and (3x) combine to (13x). In exams check options using both leading coefficient and constant term.
Step 3
Exam Tip
क्रॉस पद (10x) और (3x) मिलकर (13x) देते हैं। परीक्षा में अग्र गुणांक और स्थिर पद दोनों से विकल्प जांचें।
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\(6x^2-x-2\) का सही गुणनखंड रूप चुनें।
Choose the correct factorised form of \(6x^2-x-2\).
#quadratic expressions
#split middle
#factorisation
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A ((3x-2)(2x+1)) / correct factors
B ((3x+2)(2x-1)) / middle sign wrong
C ((6x-2)(x+1)) / middle term wrong
D ((2x-2)(3x+1)) / constant wrong
Explanation opens after your attempt
Correct Answer
A. ((3x-2)(2x+1)) / correct factors
Step 1
Concept
((3x-2)(2x+1)=6x-2 -x-2). In exams add cross terms (3x) and (-4x).
Step 2
Why this answer is correct
The correct answer is A. ((3x-2)(2x+1)) / correct factors. ((3x-2)(2x+1)=6x-2 -x-2). In exams add cross terms (3x) and (-4x).
Step 3
Exam Tip
((3x-2)(2x+1)=6x-2 -x-2) है। परीक्षा में क्रॉस पद (3x) और (-4x) जोड़ें।
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\(12x^2+7x-12\) का गुणनखंड रूप क्या है?
What is the factorised form of \(12x^2+7x-12\)?
#quadratic expressions
#factorisation
#hard
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A ((3x+4)(4x-3)) / correct factors
B ((3x-4)(4x+3)) / middle sign wrong
C ((12x-4)(x+3)) / middle term wrong
D ((6x+3)(2x-4)) / not correct
Explanation opens after your attempt
Correct Answer
A. ((3x+4)(4x-3)) / correct factors
Step 1
Concept
((3x+4)(4x-3)=12x-2 +7x-12). In exams check that (16x-9x=7x).
Step 2
Why this answer is correct
The correct answer is A. ((3x+4)(4x-3)) / correct factors. ((3x+4)(4x-3)=12x-2 +7x-12). In exams check that (16x-9x=7x).
Step 3
Exam Tip
((3x+4)(4x-3)=12x-2 +7x-12) है। परीक्षा में (16x-9x=7x) जांचें।
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\(7-4x+2x^2\) को मानक द्विघात रूप में लिखें।
Write \(7-4x+2x^2\) in standard quadratic form.
#quadratic expressions
#standard form
#ordering
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A \(2x^2-4x+7\) / correct standard form
B \(7x^2-4x+2\) / coefficients swapped
C \(-4x+2x^2+7\) / not standard order
D \(2x^2+4x+7\) / sign error
Explanation opens after your attempt
Correct Answer
A. \(2x^2-4x+7\) / correct standard form
Step 1
Concept
In standard form, the \(x^2\) term comes first, then the (x) term, then the constant term. In exams arrange terms in descending powers.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-4x+7\) / correct standard form. In standard form, the \(x^2\) term comes first, then the (x) term, then the constant term. In exams arrange terms in descending powers.
Step 3
Exam Tip
मानक रूप में \(x^2\) वाला पद पहले, फिर (x) वाला पद और फिर स्थिर पद आता है। परीक्षा में पदों को घात के घटते क्रम में रखें।
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\(x^2+10x+24\) को ((x+a)(x+b)) रूप में लिखने पर (a+b) क्या होगा?
When \(x^2+10x+24\) is written as ((x+a)(x+b)), what is (a+b)?
#quadratic expressions
#factor form
#coefficients
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A (10) / correct sum
B (24) / product
C (14) / sum of factor pair and product mixed
D (34) / total of coefficients
Explanation opens after your attempt
Correct Answer
A. (10) / correct sum
Step 1
Concept
((x+a)(x+b)=x-2 +(a+b)x+ab), so (a+b=10). In exams treat the middle coefficient as (a+b).
Step 2
Why this answer is correct
The correct answer is A. (10) / correct sum. ((x+a)(x+b)=x-2 +(a+b)x+ab), so (a+b=10). In exams treat the middle coefficient as (a+b).
Step 3
Exam Tip
((x+a)(x+b)=x-2 +(a+b)x+ab), इसलिए (a+b=10) है। परीक्षा में मध्य गुणांक को (a+b) मानें।
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यदि (x-2 +sx+35=(x+5)(x+7)), तो (s) क्या है?
If (x-2 +sx+35=(x+5)(x+7)), what is (s)?
#quadratic expressions
#unknown coefficient
#factorisation
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A (5) / first constant
B (7) / second constant
C (12) / correct sum
D (35) / product
Explanation opens after your attempt
Correct Answer
C. (12) / correct sum
Step 1
Concept
((x+5)(x+7)=x-2 +12x+35), so (s=12). In exams the sum of constant parts is the middle coefficient.
Step 2
Why this answer is correct
The correct answer is C. (12) / correct sum. ((x+5)(x+7)=x-2 +12x+35), so (s=12). In exams the sum of constant parts is the middle coefficient.
Step 3
Exam Tip
((x+5)(x+7)=x-2 +12x+35), इसलिए (s=12) है। परीक्षा में स्थिर भागों का योग मध्य गुणांक होता है।
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\(x^2-16x+64\) के लिए सही कथन कौन-सा है?
Which statement is correct for \(x^2-16x+64\)?
#quadratic expressions
#perfect square
#identity
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A यह ((x-8)2 ) है / it is ((x-8)2 )
B यह ((x+8)2 ) है / it is ((x+8)2 )
C यह \(x^2-64\) है / it is \(x^2-64\)
D यह ((x-16)2 ) है / it is ((x-16)2 )
Explanation opens after your attempt
Correct Answer
A. यह ((x-8)2 ) है / it is ((x-8)2 )
Step 1
Concept
\(64=8^2\) and \(-16x=-2\cdot x\cdot8\). In exams check both the last term and the middle term in a perfect square.
Step 2
Why this answer is correct
The correct answer is A. यह ((x-8)2 ) है / it is ((x-8)2 ). \(64=8^2\) and \(-16x=-2\cdot x\cdot8\). In exams check both the last term and the middle term in a perfect square.
Step 3
Exam Tip
\(64=8^2\) और \(-16x=-2\cdot x\cdot8\) है। परीक्षा में पूर्ण वर्ग में अंतिम पद और मध्य पद दोनों जांचें।
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\(3x^2+14x+8\) का सही गुणनखंड रूप क्या है?
What is the correct factorised form of \(3x^2+14x+8\)?
#quadratic expressions
#factorisation
#cross terms
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A ((3x+2)(x+4)) / सही गुणनखंड
B ((3x+4)(x+2)) / मध्य पद गलत
C ((x+2)(x+4)) / अग्र गुणांक गलत
D ((3x-2)(x-4)) / चिन्ह गलत
Explanation opens after your attempt
Correct Answer
A. ((3x+2)(x+4)) / सही गुणनखंड
Step 1
Concept
((3x+2)(x+4)=3x-2 +14x+8). In exams check by adding the cross terms (12x) and (2x).
Step 2
Why this answer is correct
The correct answer is A. ((3x+2)(x+4)) / सही गुणनखंड. ((3x+2)(x+4)=3x-2 +14x+8). In exams check by adding the cross terms (12x) and (2x).
Step 3
Exam Tip
((3x+2)(x+4)=3x-2 +14x+8) है। परीक्षा में क्रॉस पदों (12x) और (2x) को जोड़कर जांचें।
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\(x^2-12x+20\) को पूर्ण वर्ग रूप में लिखें।
Write \(x^2-12x+20\) in completed square form.
#quadratic expressions
#completing square
#hard
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A ((x-6)2 -16) / सही रूप
B ((x+6)2 -16) / चिन्ह गलत
C ((x-12)2 +20) / आधा गुणांक नहीं लिया
D ((x-6)2 +16) / स्थिर पद गलत
Explanation opens after your attempt
Correct Answer
A. ((x-6)2 -16) / सही रूप
Step 1
Concept
(x-2 -12x+36=(x-6)2 ) and (20=36-16). In exams take half of the middle coefficient and form its square.
Step 2
Why this answer is correct
The correct answer is A. ((x-6)2 -16) / सही रूप. (x-2 -12x+36=(x-6)2 ) and (20=36-16). In exams take half of the middle coefficient and form its square.
Step 3
Exam Tip
(x-2 -12x+36=(x-6)2 ) और (20=36-16) है। परीक्षा में मध्य गुणांक का आधा लेकर उसका वर्ग बनाएं।
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यदि (u+v=11) और (uv=28), तो \(u^2+v^2\) का मान क्या होगा?
If (u+v=11) and (uv=28), what is the value of \(u^2+v^2\)?
#quadratic expressions
#identity
#value based
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A (65) / सही मान
B (121) / केवल योग का वर्ग
C (93) / एक बार गुणनफल घटाया
D (56) / केवल (2uv)
Explanation opens after your attempt
Correct Answer
A. (65) / सही मान
Step 1
Concept
(u-2 +v-2 =(u+v)2 -2uv=121-56=65). In exams subtract (2uv) from the square of the sum.
Step 2
Why this answer is correct
The correct answer is A. (65) / सही मान. (u-2 +v-2 =(u+v)2 -2uv=121-56=65). In exams subtract (2uv) from the square of the sum.
Step 3
Exam Tip
(u-2 +v-2 =(u+v)2 -2uv=121-56=65) है। परीक्षा में योग के वर्ग से (2uv) घटाएं।
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