Question 301/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक समकोण त्रिभुज का आधार (x), ऊँचाई (x+2) और क्षेत्रफल (24) है। सही द्विघात समीकरण कौन-सा है?
A right triangle has base (x), height (x+2), and area (24). Which quadratic equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#medium
A \(x^2+2x-48=0\)
B \(x^2+2x-24=0\)
C \(2x^2+2x-24=0\)
D \(x^2-2x-48=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x-48=0\)
Step 1
Concept
The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x-48=0\). The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}x(x+2)=24) होगा। इसलिए (x(x+2)=48) और \(x^2+2x-48=0\) मिलता है।
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Question 302/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस समीकरण में मूलों का योग (0) होगा?
Which equation will have sum of roots (0)?
#quadratic-equations
#sum-of-roots
#pure-quadratic
#medium
A \(x^2-36=0\)
B \(x^2+6x+9=0\)
C \(x^2-6x+9=0\)
D \(x^2+x-36=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-36=0\)
Step 1
Concept
In \(x^2-36=0\), (b=0), so the sum of roots is \(-\frac{b}{a}=0\). If the (x) term is absent, the sum can be (0).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-36=0\). In \(x^2-36=0\), (b=0), so the sum of roots is \(-\frac{b}{a}=0\). If the (x) term is absent, the sum can be (0).
Step 3
Exam Tip
\(x^2-36=0\) में (b=0), इसलिए मूलों का योग \(-\frac{b}{a}=0\) है। (x) पद अनुपस्थित हो तो योग (0) हो सकता है।
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Question 303/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(7x^2-5x+2=0\) में \(b^2+ac\) का मान क्या है?
What is the value of \(b^2+ac\) in \(7x^2-5x+2=0\)?
#quadratic-equations
#coefficients
#expression-value
#medium
A (39)
B (11)
C (29)
D (-11)
Explanation opens after your attempt
Step 1
Concept
Here (a=7), (b=-5), (c=2), so \(b^2+ac=25+14=39\). In \(b^2\), the negative sign becomes positive after squaring.
Step 2
Why this answer is correct
The correct answer is A. (39). Here (a=7), (b=-5), (c=2), so \(b^2+ac=25+14=39\). In \(b^2\), the negative sign becomes positive after squaring.
Step 3
Exam Tip
यहाँ (a=7), (b=-5), (c=2) हैं, इसलिए \(b^2+ac=25+14=39\) है। \(b^2\) में ऋण चिन्ह वर्ग के कारण धनात्मक हो जाता है।
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Question 304/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=4) समीकरण \(x^2+sx-32=0\) का मूल है, तो (s) का मान क्या होगा?
If (x=4) is a root of \(x^2+sx-32=0\), what is the value of (s)?
#quadratic-equations
#parameter
#root-substitution
#medium
A (4)
B (-4)
C (12)
D (-12)
Explanation opens after your attempt
Step 1
Concept
Putting (x=4) gives (16+4s-32=0), so (s=4). When a root is given, substitute directly.
Step 2
Why this answer is correct
The correct answer is A. (4). Putting (x=4) gives (16+4s-32=0), so (s=4). When a root is given, substitute directly.
Step 3
Exam Tip
(x=4) रखने पर (16+4s-32=0) मिलता है, इसलिए (s=4) है। मूल दिया हो तो सीधा प्रतिस्थापन करें।
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Question 305/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((3x+4)2 =7x+16) का मानक रूप कौन-सा है?
What is the standard form of ((3x+4)2 =7x+16)?
#quadratic-equations
#standard-form
#identity
#medium
A \(9x^2+17x=0\)
B \(9x^2+31x=0\)
C \(9x^2+17x+32=0\)
D \(9x^2+24x-16=0\)
Explanation opens after your attempt
Correct Answer
A. \(9x^2+17x=0\)
Step 1
Concept
((3x+4)2 =9x-2 +24x+16). Subtracting (7x+16) gives \(9x^2+17x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(9x^2+17x=0\). ((3x+4)2 =9x-2 +24x+16). Subtracting (7x+16) gives \(9x^2+17x=0\).
Step 3
Exam Tip
((3x+4)2 =9x-2 +24x+16) है। (7x+16) घटाने पर \(9x^2+17x=0\) मिलता है।
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Question 306/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2-18x+81=0\) को किस पूर्ण वर्ग रूप में लिखा जा सकता है?
In which perfect square form can \(x^2-18x+81=0\) be written?
#quadratic-equations
#perfect-square
#identity
#medium
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 ). To identify a perfect square, match the middle term with (2ab).
Step 2
Why this answer is correct
The correct answer is A. ((x-9)2 =0). (x-2 -18x+81=(x-9)2 ). To identify a perfect square, match the middle term with (2ab).
Step 3
Exam Tip
(x-2 -18x+81=(x-9)2 ) होता है। पूर्ण वर्ग पहचानने के लिए मध्य पद (2ab) से मिलाएं।
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Question 307/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक वर्ग का क्षेत्रफल उसकी भुजा के (10) गुने से (21) अधिक है। यदि भुजा (x) है, तो समीकरण कौन-सा है?
The area of a square is (21) more than (10) times its side. If the side is (x), which equation is correct?
#quadratic-equations
#word-problem
#square-area
#medium
A \(x^2-10x-21=0\)
B \(x^2+10x+21=0\)
C \(x^2-21x-10=0\)
D \(10x^2-x+21=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x-21=0\)
Step 1
Concept
The area is \(x^2\), and it is given that \(x^2=10x+21\). Therefore \(x^2-10x-21=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x-21=0\). The area is \(x^2\), and it is given that \(x^2=10x+21\). Therefore \(x^2-10x-21=0\) is correct.
Step 3
Exam Tip
क्षेत्रफल \(x^2\) है और दिया है \(x^2=10x+21\)। इसलिए \(x^2-10x-21=0\) सही है।
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Question 308/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+vx+28=0\) के मूल (-4) और (-7) हैं, तो (v) का मान क्या है?
If the roots of \(x^2+vx+28=0\) are (-4) and (-7), what is the value of (v)?
#quadratic-equations
#parameter
#roots
#medium
A (11)
B (-11)
C (28)
D (-28)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-11), and in \(x^2+vx+28=0\), the sum is (-v). Therefore (v=11).
Step 2
Why this answer is correct
The correct answer is A. (11). The sum of roots is (-11), and in \(x^2+vx+28=0\), the sum is (-v). Therefore (v=11).
Step 3
Exam Tip
मूलों का योग (-11) है और \(x^2+vx+28=0\) में योग (-v) होता है। इसलिए (v=11) है।
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Question 309/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2-17x+72=0\) के गुणनखंड पहचानने के लिए कौन-सी संख्या जोड़ी उपयोगी है?
Which pair of numbers is useful to identify the factors of \(x^2-17x+72=0\)?
#quadratic-equations
#factorisation
#number-pair
#medium
A (8) और (9) / (8) and (9)
B (6) और (12) / (6) and (12)
C (4) और (18) / (4) and (18)
D (3) और (24) / (3) and (24)
Explanation opens after your attempt
Correct Answer
A. (8) और (9) / (8) and (9)
Step 1
Concept
\(8\cdot9=72\) and (8+9=17). Because the middle term is negative, the factors are ((x-8)(x-9)).
Step 2
Why this answer is correct
The correct answer is A. (8) और (9) / (8) and (9). \(8\cdot9=72\) and (8+9=17). Because the middle term is negative, the factors are ((x-8)(x-9)).
Step 3
Exam Tip
\(8\cdot9=72\) और (8+9=17) होता है। ऋण मध्य पद के कारण गुणनखंड ((x-8)(x-9)) बनते हैं।
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Question 310/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(8x^2-32x+24=0\) को (8) से भाग दें, तो सरल समीकरण कौन-सा होगा?
If \(8x^2-32x+24=0\) is divided by (8), which simplified equation is obtained?
#quadratic-equations
#simplification
#common-factor
#medium
A \(x^2-4x+3=0\)
B \(x^2+4x+3=0\)
C \(8x^2-4x+3=0\)
D \(x^2-32x+24=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x+3=0\)
Step 1
Concept
Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+3=0\). Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (8) से भाग देने पर \(x^2-4x+3=0\) मिलता है। समान अशून्य गुणनखंड से भाग देने पर मूल नहीं बदलते।
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Question 311/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस विकल्प में स्थिर पद अनुपस्थित है लेकिन समीकरण द्विघात है?
In which option is the constant term absent but the equation is quadratic?
#quadratic-equations
#missing-constant-term
#identify
#medium
A \(2x^2+7x=0\)
B (2x+7=0)
C \(2x^3+7x=0\)
D \(\frac{2}{x^2}+7x=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+7x=0\)
Step 1
Concept
In \(2x^2+7x=0\), the \(x^2\) term is present and the constant term is absent. An equation can be quadratic even without a constant term.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+7x=0\). In \(2x^2+7x=0\), the \(x^2\) term is present and the constant term is absent. An equation can be quadratic even without a constant term.
Step 3
Exam Tip
\(2x^2+7x=0\) में \(x^2\) पद है और स्थिर पद अनुपस्थित है। स्थिर पद न होने पर भी समीकरण द्विघात हो सकता है।
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Question 312/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2+kx+81=0\) में यदि मूल समान और ऋणात्मक हैं, तो (k) का संभव मान कौन-सा है?
In \(x^2+kx+81=0\), if the roots are equal and negative, which possible value of (k) is correct?
#quadratic-equations
#equal-negative-roots
#parameter
#medium
A (18)
B (-18)
C (9)
D (-9)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(k^2=324\), and the equal root is \(-\frac{k}{2}\). For a negative root, (k=18) is correct.
Step 2
Why this answer is correct
The correct answer is A. (18). For equal roots, \(k^2=324\), and the equal root is \(-\frac{k}{2}\). For a negative root, (k=18) is correct.
Step 3
Exam Tip
समान मूलों के लिए \(k^2=324\) और समान मूल \(-\frac{k}{2}\) होगा। ऋणात्मक मूल के लिए (k=18) सही है।
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Question 313/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि द्विघात समीकरण \(x^2-18x+k=0\) के मूल समान हैं, तो (k) क्या होगा?
If the roots of \(x^2-18x+k=0\) are equal, what is (k)?
#quadratic-equations
#equal-roots
#discriminant
#medium
A (18)
B (81)
C (162)
D (324)
Explanation opens after your attempt
Step 1
Concept
For equal roots, (D=0), so (324-4k=0) and (k=81). For equal roots, set the discriminant to zero.
Step 2
Why this answer is correct
The correct answer is B. (81). For equal roots, (D=0), so (324-4k=0) and (k=81). For equal roots, set the discriminant to zero.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए (324-4k=0) और (k=81)। समान मूल की शर्त में विवेचक शून्य रखें।
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Question 314/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2-15x+56=0\), तो (x=7) रखने पर बायां पक्ष क्या बनेगा?
If \(x^2-15x+56=0\), what will the left side become when (x=7)?
#quadratic-equations
#substitution
#root-verification
#medium
A (0)
B (7)
C (-7)
D (14)
Explanation opens after your attempt
Step 1
Concept
\(7^2-15\cdot7+56=49-105+56=0\). If the left side becomes (0), the given value is a root.
Step 2
Why this answer is correct
The correct answer is A. (0). \(7^2-15\cdot7+56=49-105+56=0\). If the left side becomes (0), the given value is a root.
Step 3
Exam Tip
\(7^2-15\cdot7+56=49-105+56=0\) है। बायां पक्ष (0) हो तो दिया मान मूल है।
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Question 315/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((x-4)(x+9)=5x) का मानक रूप क्या है?
What is the standard form of ((x-4)(x+9)=5x)?
#quadratic-equations
#standard-form
#expansion
#medium
A \(x^2-36=0\)
B \(x^2+5x-36=0\)
C \(x^2-5x-36=0\)
D \(x^2+14x-36=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-36=0\)
Step 1
Concept
The left side gives \(x^2+5x-36\). Subtracting (5x) gives \(x^2-36=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-36=0\). The left side gives \(x^2+5x-36\). Subtracting (5x) gives \(x^2-36=0\).
Step 3
Exam Tip
बाईं ओर \(x^2+5x-36\) मिलता है। (5x) घटाने पर \(x^2-36=0\) बनता है।
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Question 316/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((x+4)2 -3(x+4)-10=0) को सरल करने पर कौन-सा द्विघात समीकरण मिलेगा?
Which quadratic equation is obtained by simplifying ((x+4)2 -3(x+4)-10=0)?
#quadratic-equations
#simplification
#identity
#medium
A \(x^2+5x-6=0\)
B \(x^2+11x-6=0\)
C \(x^2+5x+6=0\)
D \(x^2+8x-10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x-6=0\)
Step 1
Concept
((x+4)2 =x-2 +8x+16) and (-3(x+4)=-3x-12). Simplifying gives \(x^2+5x-6=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x-6=0\). ((x+4)2 =x-2 +8x+16) and (-3(x+4)=-3x-12). Simplifying gives \(x^2+5x-6=0\).
Step 3
Exam Tip
((x+4)2 =x-2 +8x+16) और (-3(x+4)=-3x-12) है। सरल करने पर \(x^2+5x-6=0\) मिलता है।
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Question 317/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+ux+64=0\) के मूल (8) और (8) हैं, तो (u) का मान क्या होगा?
If the roots of \(x^2+ux+64=0\) are (8) and (8), what is the value of (u)?
#quadratic-equations
#equal-roots
#parameter
#medium
A (16)
B (-16)
C (8)
D (-8)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (16), and in \(x^2+ux+64=0\), the sum is (-u). Therefore (u=-16).
Step 2
Why this answer is correct
The correct answer is B. (-16). The sum of roots is (16), and in \(x^2+ux+64=0\), the sum is (-u). Therefore (u=-16).
Step 3
Exam Tip
मूलों का योग (16) है और \(x^2+ux+64=0\) में योग (-u) होता है। इसलिए (u=-16) है।
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Question 318/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2-3x-40=0\) के मूलों का सही युग्म कौन-सा है?
Which pair is the correct roots of \(x^2-3x-40=0\)?
#quadratic-equations
#roots
#factorisation
#medium
A (8,-5)
B (5,-8)
C (-8,-5)
D (8,5)
Explanation opens after your attempt
Step 1
Concept
From ((x-8)(x+5)=0), the roots are (8) and (-5). Check both product and sum.
Step 2
Why this answer is correct
The correct answer is A. (8,-5). From ((x-8)(x+5)=0), the roots are (8) and (-5). Check both product and sum.
Step 3
Exam Tip
((x-8)(x+5)=0) से मूल (8) और (-5) मिलते हैं। गुणनफल और योग दोनों मिलाकर जांचें।
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Question 319/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(qx^2-8x+12=0\) में मूलों का गुणनफल (4) है, तो (q) का मान क्या है?
If the product of roots of \(qx^2-8x+12=0\) is (4), what is the value of (q)?
#quadratic-equations
#parameter
#product-of-roots
#medium
A (3)
B (4)
C (6)
D (-3)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{12}{q}=4\), so (q=3). Use \(\frac{c}{a}\) for the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (3). The product of roots is \(\frac{12}{q}=4\), so (q=3). Use \(\frac{c}{a}\) for the product of roots.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{12}{q}=4\), इसलिए (q=3) है। गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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Question 320/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(4x^2+mx+20=0\) में मूलों का योग (-6) है, तो (m) क्या है?
If the sum of roots of \(4x^2+mx+20=0\) is (-6), what is (m)?
#quadratic-equations
#parameter
#sum-of-roots
#medium
A (24)
B (-24)
C (6)
D (-6)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is \(-\frac{m}{4}=-6\), so (m=24). Do not forget the negative sign in the sum formula.
Step 2
Why this answer is correct
The correct answer is A. (24). The sum of roots is \(-\frac{m}{4}=-6\), so (m=24). Do not forget the negative sign in the sum formula.
Step 3
Exam Tip
मूलों का योग \(-\frac{m}{4}=-6\), इसलिए (m=24) है। योग के सूत्र में ऋण चिन्ह न भूलें।
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Question 321/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(6x^2+5x-14=0\) में मूलों का गुणनफल क्या है?
What is the product of roots of \(6x^2+5x-14=0\)?
#quadratic-equations
#product-of-roots
#medium
A \(-\frac{7}{3}\)
B \(\frac{7}{3}\)
C \(-\frac{5}{6}\)
D \(\frac{5}{6}\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{7}{3}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{-14}{6}=-\frac{7}{3}\). Write the answer in simplified form.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{7}{3}\). The product of roots is \(\frac{c}{a}=\frac{-14}{6}=-\frac{7}{3}\). Write the answer in simplified form.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{-14}{6}=-\frac{7}{3}\) है। उत्तर को सरल रूप में लिखें।
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Question 322/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(5x^2-9x+4=0\) में मूलों का योग क्या है?
What is the sum of roots of \(5x^2-9x+4=0\)?
#quadratic-equations
#sum-of-roots
#medium
A \(\frac{9}{5}\)
B \(-\frac{9}{5}\)
C \(\frac{4}{5}\)
D (4)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{5}\)
Step 1
Concept
The sum of roots is \(-\frac{b}{a}=-\frac{-9}{5}=\frac{9}{5}\). Pay attention to the sign of (b) in the formula.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{5}\). The sum of roots is \(-\frac{b}{a}=-\frac{-9}{5}=\frac{9}{5}\). Pay attention to the sign of (b) in the formula.
Step 3
Exam Tip
मूलों का योग \(-\frac{b}{a}=-\frac{-9}{5}=\frac{9}{5}\) है। सूत्र में (b) का चिन्ह ध्यान से रखें।
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Question 323/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस द्विघात समीकरण में मूलों का योग (-8) और गुणनफल (15) है?
Which quadratic equation has sum of roots (-8) and product (15)?
#quadratic-equations
#sum-product
#forming-equation
#medium
A \(x^2+8x+15=0\)
B \(x^2-8x+15=0\)
C \(x^2+15x+8=0\)
D \(x^2-15x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x+15=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-8) gives (x^2+8x+15=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+8x+15=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-8) gives (x^2+8x+15=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-8) रखने पर (x^2+8x+15=0) मिलता है\)।
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Question 324/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि मूल (-4) और (-9) हैं, तो मूलों का गुणनफल क्या है?
If the roots are (-4) and (-9), what is the product of the roots?
#quadratic-equations
#roots
#product
#medium
A (36)
B (-36)
C (13)
D (-13)
Explanation opens after your attempt
Step 1
Concept
The product of roots is ((-4)(-9)=36). The product of two negative numbers is positive.
Step 2
Why this answer is correct
The correct answer is A. (36). The product of roots is ((-4)(-9)=36). The product of two negative numbers is positive.
Step 3
Exam Tip
मूलों का गुणनफल ((-4)(-9)=36) है। दो ऋणात्मक संख्याओं का गुणनफल धनात्मक होता है।
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Question 325/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=-6) और (x=1) मूल हैं, तो मूलों का योग क्या है?
If (x=-6) and (x=1) are roots, what is the sum of the roots?
#quadratic-equations
#roots
#sum
#medium
A (-5)
B (5)
C (-7)
D (7)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-6+1=-5). For numbers with opposite signs, the sign of the larger magnitude is used.
Step 2
Why this answer is correct
The correct answer is A. (-5). The sum of roots is (-6+1=-5). For numbers with opposite signs, the sign of the larger magnitude is used.
Step 3
Exam Tip
मूलों का योग (-6+1=-5) है। विपरीत चिन्ह वाली संख्याओं में बड़े परिमाण का चिन्ह आता है।
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Question 326/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2+6\sqrt{5}x+45=0\) में (b) का मान क्या है?
What is the value of (b) in \(x^2+6\sqrt{5}x+45=0\)?
#quadratic-equations
#coefficients
#radicals
#medium
A \(6\sqrt{5}\)
B \(-6\sqrt{5}\)
C (45)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{5}\)
Step 1
Concept
The coefficient attached to (x) is \(6\sqrt{5}\). Write radical coefficients with their signs too.
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{5}\). The coefficient attached to (x) is \(6\sqrt{5}\). Write radical coefficients with their signs too.
Step 3
Exam Tip
(x) के साथ लगा गुणांक \(6\sqrt{5}\) है। करणी वाले गुणांक भी चिन्ह सहित लिखें।
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Question 327/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
कौन-सा विकल्प सामान्य द्विघात समीकरण नहीं है?
Which option is not a usual quadratic equation?
#quadratic-equations
#non-quadratic
#negative-power
#medium
A \(5x^2-1=0\)
B \(x^2+4x=0\)
C \(\frac{1}{x^2}+x+2=0\)
D \(3x^2-x+6=0\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{1}{x^2}+x+2=0\)
Step 1
Concept
\(\frac{1}{x^2}=x^{-2}\), which is not polynomial form. A usual quadratic equation does not have a negative power of the variable.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{x^2}+x+2=0\). \(\frac{1}{x^2}=x^{-2}\), which is not polynomial form. A usual quadratic equation does not have a negative power of the variable.
Step 3
Exam Tip
\(\frac{1}{x^2}=x^{-2}\) है, जो बहुपद रूप नहीं है। सामान्य द्विघात समीकरण में चर की ऋणात्मक घात नहीं होती।
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Question 328/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(\frac{3}{4}x^2-\frac{1}{2}x+2=0\) का पूर्णांक गुणांकों वाला रूप कौन-सा है?
What is the form with integer coefficients for \(\frac{3}{4}x^2-\frac{1}{2}x+2=0\)?
#quadratic-equations
#fractional-coefficients
#medium
A \(3x^2-2x+8=0\)
B \(3x^2-2x+2=0\)
C \(4x^2-2x+8=0\)
D \(3x^2+2x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-2x+8=0\)
Step 1
Concept
Multiply the whole equation by (4) to remove the denominators. This gives \(3x^2-2x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-2x+8=0\). Multiply the whole equation by (4) to remove the denominators. This gives \(3x^2-2x+8=0\).
Step 3
Exam Tip
हर हटाने के लिए पूरे समीकरण को (4) से गुणा करें। इससे \(3x^2-2x+8=0\) मिलता है।
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Question 329/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(\frac{x^2}{5}+2x-7=0\) को बिना भिन्न के किस रूप में लिखा जाएगा?
How will \(\frac{x^2}{5}+2x-7=0\) be written without fractions?
#quadratic-equations
#fractions
#standard-form
#medium
A \(x^2+10x-35=0\)
B \(x^2+2x-7=0\)
C \(5x^2+2x-7=0\)
D \(x^2+10x+35=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+10x-35=0\)
Step 1
Concept
Multiplying the whole equation by (5) gives \(x^2+10x-35=0\). To remove fractions, multiply the whole equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+10x-35=0\). Multiplying the whole equation by (5) gives \(x^2+10x-35=0\). To remove fractions, multiply the whole equation.
Step 3
Exam Tip
पूरे समीकरण को (5) से गुणा करने पर \(x^2+10x-35=0\) मिलता है। भिन्न हटाने के लिए पूरे समीकरण पर गुणा करें।
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Question 330/600
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(4x^2+13x+3=0\) में (ac) का मान क्या है?
What is the value of (ac) in \(4x^2+13x+3=0\)?
#quadratic-equations
#coefficients
#ac-product
#medium
A (12)
B (13)
C (16)
D (3)
Explanation opens after your attempt
Step 1
Concept
Here (a=4) and (c=3), so (ac=12). The value (ac) is useful in splitting the middle term.
Step 2
Why this answer is correct
The correct answer is A. (12). Here (a=4) and (c=3), so (ac=12). The value (ac) is useful in splitting the middle term.
Step 3
Exam Tip
यहाँ (a=4) और (c=3), इसलिए (ac=12) है। मध्य पद विभाजन में (ac) उपयोगी होता है।
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