Question 271/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(4x^2-4kx+k^2=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(4x^2-4kx+k^2=0\)?
#quadratic-equations
#perfect-square
#nature-of-roots
#hard
A दो समान वास्तविक मूल / Two equal real roots
B दो भिन्न वास्तविक मूल / Two distinct real roots
C कोई वास्तविक मूल नहीं / No real roots
D निर्धारित नहीं हो सकता / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. दो समान वास्तविक मूल / Two equal real roots
Step 1
Concept
It can be written as ((2x-k)2 =0). Therefore both roots are equal and real.
Step 2
Why this answer is correct
The correct answer is A. दो समान वास्तविक मूल / Two equal real roots. It can be written as ((2x-k)2 =0). Therefore both roots are equal and real.
Step 3
Exam Tip
यह ((2x-k)2 =0) के रूप में लिखा जा सकता है। इसलिए दोनों मूल समान वास्तविक होते हैं।
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Question 272/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि ((x+a)2 =x-2 +10x+25) है, तो (a) का मान क्या होगा?
If ((x+a)2 =x-2 +10x+25), what is the value of (a)?
#quadratic-equations
#identity
#coefficient-comparison
#hard
A (5)
B (-5)
C (10)
D (25)
Explanation opens after your attempt
Step 1
Concept
((x+a)2 =x-2 +2ax+a-2 ). From (2a=10), we get (a=5).
Step 2
Why this answer is correct
The correct answer is A. (5). ((x+a)2 =x-2 +2ax+a-2 ). From (2a=10), we get (a=5).
Step 3
Exam Tip
((x+a)2 =x-2 +2ax+a-2 ) होता है। (2a=10) से (a=5) मिलता है।
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Question 273/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस समीकरण में मूल बराबर और धनात्मक होंगे?
Which equation will have equal and positive roots?
#quadratic-equations
#equal-positive-roots
#perfect-square
#hard
A \(x^2-12x+36=0\)
B \(x^2+12x+36=0\)
C \(x^2-36=0\)
D \(x^2+36=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+36=0\)
Step 1
Concept
(x-2 -12x+36=(x-6)2 ), so both roots are (6). Both roots are equal and positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+36=0\). (x-2 -12x+36=(x-6)2 ), so both roots are (6). Both roots are equal and positive.
Step 3
Exam Tip
(x-2 -12x+36=(x-6)2 ), इसलिए दोनों मूल (6) हैं। दोनों मूल बराबर और धनात्मक हैं।
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Question 274/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2-5x+6=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या है?
If \(\alpha,\beta\) are roots of \(x^2-5x+6=0\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#quadratic-equations
#reciprocal-roots
#hard
A \( \frac{5}{6} \)
B \( \frac{6}{5} \)
C (5)
D (6)
Explanation opens after your attempt
Correct Answer
A. \( \frac{5}{6} \)
Step 1
Concept
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the value is \(\frac{5}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{5}{6} \). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the value is \(\frac{5}{6}\).
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ मान \(\frac{5}{6}\) है।
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Question 275/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
दो संख्याओं का योग (11) और गुणनफल (30) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (11) and product (30). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#hard
A \(x^2-11x+30=0\)
B \(x^2+11x+30=0\)
C \(x^2-30x+11=0\)
D \(x^2+30x-11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+30=0\)
Step 1
Concept
If the sum of roots is (11) and product is (30), the equation is \(x^2-11x+30=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). If the sum of roots is (11) and product is (30), the equation is \(x^2-11x+30=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (11) और गुणनफल (30) है, तो समीकरण \(x^2-11x+30=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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Question 276/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
एक आयत की लंबाई (x+5) और चौड़ाई (x-2) है। क्षेत्रफल (84) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+5) and breadth (x-2). If its area is (84), which equation is correct?
#quadratic-equations
#word-problem
#area
#hard
A \(x^2+3x-94=0\)
B \(x^2+3x-74=0\)
C \(x^2-3x-94=0\)
D \(x^2+7x-94=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-94=0\)
Step 1
Concept
The area is ((x+5)(x-2)=84). Expanding gives \(x^2+3x-10=84\) and \(x^2+3x-94=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-94=0\). The area is ((x+5)(x-2)=84). Expanding gives \(x^2+3x-10=84\) and \(x^2+3x-94=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+5)(x-2)=84) होगा। विस्तार से \(x^2+3x-10=84\) और \(x^2+3x-94=0\) मिलता है।
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Question 277/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस विकल्प में मूलों का योग और गुणनफल दोनों धनात्मक होंगे?
In which option will both the sum and product of roots be positive?
#quadratic-equations
#sum-product
#signs
#hard
A \(x^2-5x+6=0\)
B \(x^2+5x+6=0\)
C \(x^2-5x-6=0\)
D \(x^2+5x-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-5x+6=0\)
Step 1
Concept
In the first option, the sum is \(-\frac{b}{a}=5\) and the product is \(\frac{c}{a}=6\). Both are positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-5x+6=0\). In the first option, the sum is \(-\frac{b}{a}=5\) and the product is \(\frac{c}{a}=6\). Both are positive.
Step 3
Exam Tip
पहले विकल्प में योग \(-\frac{b}{a}=5\) और गुणनफल \(\frac{c}{a}=6\) है। दोनों धनात्मक हैं।
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Question 278/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि समीकरण \(x^2+bx+25=0\) के दोनों मूल समान हैं और उनका मान (-5) है, तो (b) क्या है?
If both roots of \(x^2+bx+25=0\) are equal and their value is (-5), what is (b)?
#quadratic-equations
#equal-roots
#coefficient
#hard
A (10)
B (-10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
Both roots are (-5), so their sum is (-10). Here the sum of roots is (-b), so (b=10).
Step 2
Why this answer is correct
The correct answer is A. (10). Both roots are (-5), so their sum is (-10). Here the sum of roots is (-b), so (b=10).
Step 3
Exam Tip
दोनों मूल (-5) हैं, इसलिए योग (-10) है। यहाँ मूलों का योग (-b) है, अतः (b=10)।
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Question 279/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+6x+c=0\) पूर्ण वर्ग द्विघात समीकरण है, तो (c) का मान क्या होगा?
If \(x^2+6x+c=0\) is a perfect square quadratic equation, what is the value of (c)?
#quadratic-equations
#perfect-square
#parameter
#hard
A (9)
B (6)
C (12)
D (36)
Explanation opens after your attempt
Step 1
Concept
For a perfect square, (x-2 +6x+c=(x+3)2 ) is needed. Hence (c=9).
Step 2
Why this answer is correct
The correct answer is A. (9). For a perfect square, (x-2 +6x+c=(x+3)2 ) is needed. Hence (c=9).
Step 3
Exam Tip
पूर्ण वर्ग के लिए (x-2 +6x+c=(x+3)2 ) होना चाहिए। इसलिए (c=9) होगा।
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Question 280/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(3x^2-10x+8=0\) में मूलों का अंतर क्या है?
What is the difference between the roots of \(3x^2-10x+8=0\)?
#quadratic-equations
#roots
#difference
#hard
A \( \frac{2}{3} \)
B \( \frac{4}{3} \)
C (2)
D \( \frac{8}{3} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{2}{3} \)
Step 1
Concept
The roots are (2) and \(\frac{4}{3}\). Their difference is \(2-\frac{4}{3}=\frac{2}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{2}{3} \). The roots are (2) and \(\frac{4}{3}\). Their difference is \(2-\frac{4}{3}=\frac{2}{3}\).
Step 3
Exam Tip
मूल (2) और \(\frac{4}{3}\) हैं। उनका अंतर \(2-\frac{4}{3}=\frac{2}{3}\) है।
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Question 281/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x-2 -(m+2)x+2m=0) का एक मूल (2) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+2)x+2m=0) is (2), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (2m)
C (m+2)
D (m-2 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 3
Exam Tip
गुणनफल (2m) है और एक मूल (2) है। इसलिए दूसरा मूल \(\frac{2m}{2}=m\) है।
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Question 282/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2-4x+6=0\) के लिए कौन-सा कथन सही है?
Which statement is correct for \(x^2-4x+6=0\)?
#quadratic-equations
#nature-of-roots
#hard
A वास्तविक मूल नहीं हैं / It has no real roots
B दो समान वास्तविक मूल हैं / It has two equal real roots
C दो भिन्न वास्तविक मूल हैं / It has two distinct real roots
D एकमात्र वास्तविक मूल है / It has only one real root
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं हैं / It has no real roots
Step 1
Concept
Here (D=(-4)2 -4\cdot1\cdot6=-8<0). Therefore it has no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं हैं / It has no real roots. Here (D=(-4)2 -4\cdot1\cdot6=-8<0). Therefore it has no real roots.
Step 3
Exam Tip
यहाँ (D=(-4)2 -4\cdot1\cdot6=-8<0) है। इसलिए वास्तविक मूल नहीं होंगे।
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Question 283/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+px+q=0\) के मूल (p) और (q) ही हैं, तो कौन-सा संबंध सही है?
If the roots of \(x^2+px+q=0\) are (p) and (q) themselves, which relation is correct?
#quadratic-equations
#roots
#relations
#hard
A (p+q=-p) और (pq=q) / (p+q=-p) and (pq=q)
B (p+q=p) और (pq=q) / (p+q=p) and (pq=q)
C (p+q=-q) और (pq=p) / (p+q=-q) and (pq=p)
D (p-q=-p) और (pq=q) / (p-q=-p) and (pq=q)
Explanation opens after your attempt
Correct Answer
A. (p+q=-p) और (pq=q) / (p+q=-p) and (pq=q)
Step 1
Concept
The sum of roots is (-p) and the product is (q). Since the roots are (p,q), we get (p+q=-p) and (pq=q).
Step 2
Why this answer is correct
The correct answer is A. (p+q=-p) और (pq=q) / (p+q=-p) and (pq=q). The sum of roots is (-p) and the product is (q). Since the roots are (p,q), we get (p+q=-p) and (pq=q).
Step 3
Exam Tip
मूलों का योग (-p) और गुणनफल (q) होता है। दिए मूल (p,q) हैं, इसलिए (p+q=-p) और (pq=q)।
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Question 284/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण ((x+1)(x+2)+(x+3)(x+4)=50) का मानक रूप कौन-सा है?
What is the standard form of ((x+1)(x+2)+(x+3)(x+4)=50)?
#quadratic-equations
#expansion
#standard-form
#hard
A \(2x^2+10x-36=0\)
B \(2x^2+10x+14=0\)
C \(x^2+5x-36=0\)
D \(2x^2+5x-50=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+10x-36=0\)
Step 1
Concept
The left side is \(x^2+3x+2+x^2+7x+12=2x^2+10x+14\). Subtracting (50) gives \(2x^2+10x-36=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+10x-36=0\). The left side is \(x^2+3x+2+x^2+7x+12=2x^2+10x+14\). Subtracting (50) gives \(2x^2+10x-36=0\).
Step 3
Exam Tip
बाईं ओर \(x^2+3x+2+x^2+7x+12=2x^2+10x+14\) है। (50) घटाने पर \(2x^2+10x-36=0\) मिलता है।
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Question 285/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस विकल्प में दिया गया समीकरण द्विघात नहीं रहेगा?
In which option will the given equation not remain quadratic?
#quadratic-equations
#parameter
#not-quadratic
#hard
A ((t-2)x-2 +5x+1=0), (t=2)
B \(3x^2+tx+1=0\), (t=0)
C \(x^2+t=0\), (t=0)
D \(2x^2+1=0\), (t=2)
Explanation opens after your attempt
Correct Answer
A. ((t-2)x-2 +5x+1=0), (t=2)
Step 1
Concept
In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
Step 2
Why this answer is correct
The correct answer is A. ((t-2)x-2 +5x+1=0), (t=2). In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
Step 3
Exam Tip
पहले विकल्प में (t=2) रखने पर \(x^2\) का गुणांक (0) हो जाता है। तब समीकरण रैखिक बन जाता है।
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Question 286/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2-8x+15=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो (\(\alpha-1\)\(\beta-1\)) का मान क्या है?
If \(\alpha\) and \(\beta\) are roots of \(x^2-8x+15=0\), what is (\(\alpha-1\)\(\beta-1\))?
#quadratic-equations
#roots
#expression-value
#hard
A (8)
B (15)
C (6)
D (22)
Explanation opens after your attempt
Step 1
Concept
(\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1). Here (15-8+1=8).
Step 2
Why this answer is correct
The correct answer is A. (8). (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1). Here (15-8+1=8).
Step 3
Exam Tip
(\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1) होता है। यहाँ (15-8+1=8)।
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Question 287/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(2x^2+7x+3=0\) में मूलों के वर्गों का योग क्या है?
What is the sum of squares of the roots of \(2x^2+7x+3=0\)?
#quadratic-equations
#roots
#squares-sum
#hard
A \( \frac{37}{4} \)
B \( \frac{49}{4} \)
C \( \frac{13}{4} \)
D \( \frac{31}{4} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{37}{4} \)
Step 1
Concept
If the roots are \(\alpha,\beta\), then (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). Here (\left\(-\frac{7}{2}\right\)2 -2\cdot\frac{3}{2}=\frac{37}{4}).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{37}{4} \). If the roots are \(\alpha,\beta\), then (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). Here (\left\(-\frac{7}{2}\right\)2 -2\cdot\frac{3}{2}=\frac{37}{4}).
Step 3
Exam Tip
यदि मूल \(\alpha,\beta\) हैं, तो (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta)। यहाँ (\left\(-\frac{7}{2}\right\)2 -2\cdot\frac{3}{2}=\frac{37}{4})।
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Question 288/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+ax+12=0\) का एक मूल (3) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+12=0\) is (3), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
B दूसरा मूल (-4), (a=1) / other root (-4), (a=1)
C दूसरा मूल (4), (a=7) / other root (4), (a=7)
D दूसरा मूल (-3), (a=0) / other root (-3), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
Step 1
Concept
The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7). The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 3
Exam Tip
मूलों का गुणनफल (12) है, इसलिए दूसरा मूल (4) होगा। योग (7) है और (-a=7), इसलिए (a=-7)।
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Question 289/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस समीकरण का विवेचक (-15) है?
Which equation has discriminant (-15)?
#quadratic-equations
#discriminant
#identify
#hard
A \(x^2+x+4=0\)
B \(x^2+4x+1=0\)
C \(2x^2+x+2=0\)
D \(x^2-5x+10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+x+4=0\)
Step 1
Concept
For \(x^2+x+4=0\), \(D=1^2-4\cdot1\cdot4=-15\). Subtract the full (4ac) while finding the discriminant.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x+4=0\). For \(x^2+x+4=0\), \(D=1^2-4\cdot1\cdot4=-15\). Subtract the full (4ac) while finding the discriminant.
Step 3
Exam Tip
\(x^2+x+4=0\) के लिए \(D=1^2-4\cdot1\cdot4=-15\) है। विवेचक निकालते समय (4ac) पूरा घटाएं।
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Question 290/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2-6x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-6x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#hard
A (k<9)
B (k=9)
C (k>9)
D \(k\leq9\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (36-4k>0), so (k<9).
Step 2
Why this answer is correct
The correct answer is A. (k<9). For real and distinct roots, (D>0) is needed. Here (36-4k>0), so (k<9).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (36-4k>0), इसलिए (k<9)।
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Question 291/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (2) और (3) समीकरण \(x^2-sx+p=0\) के मूल हैं, तो (s+p) का मान क्या है?
If (2) and (3) are roots of \(x^2-sx+p=0\), what is the value of (s+p)?
#quadratic-equations
#roots
#parameters
#hard
A (11)
B (5)
C (6)
D (1)
Explanation opens after your attempt
Step 1
Concept
The sum of roots gives (s=5) and the product gives (p=6). Therefore (s+p=11).
Step 2
Why this answer is correct
The correct answer is A. (11). The sum of roots gives (s=5) and the product gives (p=6). Therefore (s+p=11).
Step 3
Exam Tip
मूलों का योग (s=5) और गुणनफल (p=6) है। इसलिए (s+p=11) है।
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Question 292/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(2x^2-5x+3=0\) के मूलों के व्युत्क्रमों का योग क्या है?
What is the sum of reciprocals of the roots of \(2x^2-5x+3=0\)?
#quadratic-equations
#roots
#reciprocal-sum
#hard
A \( \frac{5}{3} \)
B \( \frac{3}{5} \)
C \( \frac{2}{3} \)
D \( \frac{5}{2} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{5}{3} \)
Step 1
Concept
The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{5}{3} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\).
Step 3
Exam Tip
व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\) है।
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Question 293/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2+2kx+9=0\) के वास्तविक मूल होने की शर्त कौन-सी है?
What is the condition for \(x^2+2kx+9=0\) to have real roots?
#quadratic-equations
#real-roots
#parameter
#hard
A \(k\leq -3\) या \(k\geq 3\) / \(k\leq -3\) or \(k\geq 3\)
B (-3<k<3)
C (k=0)
D \(k\neq 3\)
Explanation opens after your attempt
Correct Answer
A. \(k\leq -3\) या \(k\geq 3\) / \(k\leq -3\) or \(k\geq 3\)
Step 1
Concept
For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\), so \(k\leq-3\) or \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq -3\) या \(k\geq 3\) / \(k\leq -3\) or \(k\geq 3\). For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\), so \(k\leq-3\) or \(k\geq3\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(4k^2-36\geq0\), इसलिए \(k\leq-3\) या \(k\geq3\)।
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Question 294/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2-2kx+16=0\) के मूल समान हैं, तो (k) के संभावित मान क्या हैं?
If the roots of \(x^2-2kx+16=0\) are equal, what are the possible values of (k)?
#quadratic-equations
#equal-roots
#discriminant
#hard
A \(k=\pm4\)
B \(k=\pm8\)
C (k=4)
D (k=-4)
Explanation opens after your attempt
Correct Answer
A. \(k=\pm4\)
Step 1
Concept
For equal roots, (D=0), so ((-2k)2 -64=0). This gives \(k^2=16\) and \(k=\pm4\).
Step 2
Why this answer is correct
The correct answer is A. \(k=\pm4\). For equal roots, (D=0), so ((-2k)2 -64=0). This gives \(k^2=16\) and \(k=\pm4\).
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((-2k)2 -64=0) मिलता है। इससे \(k^2=16\) और \(k=\pm4\) है।
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Question 295/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
मूलों का योग (7) और गुणनफल (10) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (7) and product (10)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-10x+7=0\)
D \(x^2+10x-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-7x+10=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-7x+10=0) सही है\)।
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Question 296/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x=-2) समीकरण \(3x^2+px+10=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-2) is a root of \(3x^2+px+10=0\), what is the value of (p)?
#quadratic-equations
#root-substitution
#parameter
#hard
A (11)
B (-11)
C (1)
D (-1)
Explanation opens after your attempt
Step 1
Concept
Putting (x=-2) gives (12-2p+10=0). Hence (p=11).
Step 2
Why this answer is correct
The correct answer is A. (11). Putting (x=-2) gives (12-2p+10=0). Hence (p=11).
Step 3
Exam Tip
(x=-2) रखने पर (12-2p+10=0) मिलता है। इससे (p=11) है।
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Question 297/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(\frac{x^2-3}{2}+\frac{x-1}{3}=4\) को पूर्णांक गुणांकों वाले मानक रूप में लिखिए।
Write \(\frac{x^2-3}{2}+\frac{x-1}{3}=4\) in standard form with integer coefficients.
#quadratic-equations
#fractions
#standard-form
#hard
A \(3x^2+2x-29=0\)
B \(3x^2+2x-35=0\)
C \(x^2+2x-29=0\)
D \(3x^2-2x-29=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2+2x-29=0\)
Step 1
Concept
Multiplying the whole equation by (6) gives \(3x^2-9+2x-2=24\). Thus the standard form is \(3x^2+2x-35=0\).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2+2x-29=0\). Multiplying the whole equation by (6) gives \(3x^2-9+2x-2=24\). Thus the standard form is \(3x^2+2x-35=0\).
Step 3
Exam Tip
पूरे समीकरण को (6) से गुणा करने पर \(3x^2-9+2x-2=24\) मिलता है। इसलिए \(3x^2+2x-35=0\) नहीं बल्कि सही रूप \(3x^2+2x-35=0\) होगा।
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Question 298/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण ((x-2)2 +(2x+1)2 =25) का मानक रूप कौन-सा है?
What is the standard form of ((x-2)2 +(2x+1)2 =25)?
#quadratic-equations
#identity
#simplification
#hard
A \(5x^2+1=25\)
B \(5x^2+1=0\)
C \(5x^2+20=0\)
D \(5x^2-20=0\)
Explanation opens after your attempt
Correct Answer
D. \(5x^2-20=0\)
Step 1
Concept
Expanding gives \(x^2-4x+4+4x^2+4x+1=25\). This gives \(5x^2-20=0\).
Step 2
Why this answer is correct
The correct answer is D. \(5x^2-20=0\). Expanding gives \(x^2-4x+4+4x^2+4x+1=25\). This gives \(5x^2-20=0\).
Step 3
Exam Tip
विस्तार करने पर \(x^2-4x+4+4x^2+4x+1=25\) मिलता है। इससे \(5x^2-20=0\) बनता है।
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Question 299/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (\(m^2-9\)x-2 +4x-5=0) द्विघात समीकरण है, तो (m) पर सही शर्त क्या है?
If (\(m^2-9\)x-2 +4x-5=0) is a quadratic equation, what is the correct condition on (m)?
#quadratic-equations
#parameter
#condition
#hard
A \(m\neq 3\)
B \(m\neq -3\)
C \(m\neq \pm3\)
D \(m=\pm3\)
Explanation opens after your attempt
Correct Answer
C. \(m\neq \pm3\)
Step 1
Concept
For the equation to be quadratic, \(m^2-9\neq0\) is needed. Hence both \(m\neq3\) and \(m\neq-3\) are necessary.
Step 2
Why this answer is correct
The correct answer is C. \(m\neq \pm3\). For the equation to be quadratic, \(m^2-9\neq0\) is needed. Hence both \(m\neq3\) and \(m\neq-3\) are necessary.
Step 3
Exam Tip
द्विघात होने के लिए \(m^2-9\neq0\) होना चाहिए। इसलिए \(m\neq3\) और \(m\neq-3\) दोनों जरूरी हैं।
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Question 300/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण ((5x-2)(2x+3)=7x) का मानक द्विघात रूप कौन-सा है?
What is the standard quadratic form of ((5x-2)(2x+3)=7x)?
#quadratic-equations
#standard-form
#expansion
#hard
A \(10x^2+4x-6=0\)
B \(10x^2+18x-6=0\)
C \(10x^2-4x-6=0\)
D \(10x^2+11x+6=0\)
Explanation opens after your attempt
Correct Answer
A. \(10x^2+4x-6=0\)
Step 1
Concept
Here ((5x-2)(2x+3)=10x-2 +11x-6) and subtracting (7x) gives \(10x^2+4x-6=0\). First expand and then bring all terms to one side.
Step 2
Why this answer is correct
The correct answer is A. \(10x^2+4x-6=0\). Here ((5x-2)(2x+3)=10x-2 +11x-6) and subtracting (7x) gives \(10x^2+4x-6=0\). First expand and then bring all terms to one side.
Step 3
Exam Tip
((5x-2)(2x+3)=10x-2 +11x-6) है और (7x) घटाने पर \(10x^2+4x-6=0\) मिलता है। पहले विस्तार करें फिर सभी पद एक ओर लाएं।
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