Question 1/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(25x^2-10x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(25x^2-10x+1=0\)?
#roots
#repeated_root
#fraction_root
A \(\frac{1}{5}\)
B -\(\frac{1}{5}\)
C (5)
D (-5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5}\)
Step 1
Concept
(25x-2 -10x+1=(5x-1)2 ). Therefore the repeated root is \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). (25x-2 -10x+1=(5x-1)2 ). Therefore the repeated root is \(\frac{1}{5}\).
Step 3
Exam Tip
(25x-2 -10x+1=(5x-1)2 ) है। इसलिए दोहराया मूल \(\frac{1}{5}\) है।
Login to save your score, XP, coins and progress.
Question 2/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x=\frac{5}{2}\) समीकरण \(2x^2-9x+c=0\) का मूल है तो (c) का मान क्या है?
If \(x=\frac{5}{2}\) is a root of \(2x^2-9x+c=0\), what is the value of (c)?
#roots
#parameter
#fraction_root
A (10)
B (-10)
C \(\frac{5}{2}\)
D -\(\frac{5}{2}\)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{5}{2}\) gives \(\frac{25}{2}-\frac{45}{2}+c=0\), so (c=10). Calculate each term carefully with a fractional root.
Step 2
Why this answer is correct
The correct answer is A. (10). Putting \(x=\frac{5}{2}\) gives \(\frac{25}{2}-\frac{45}{2}+c=0\), so (c=10). Calculate each term carefully with a fractional root.
Step 3
Exam Tip
\(x=\frac{5}{2}\) रखने पर \(\frac{25}{2}-\frac{45}{2}+c=0\) इसलिए (c=10) है। भिन्न मूल में हर पद सावधानी से निकालें।
Login to save your score, XP, coins and progress.
Question 3/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(4x^2-17x+4=0\) के मूल कौन से हैं?
What are the roots of \(4x^2-17x+4=0\)?
#roots
#factorisation
#fraction_root
A (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\)
B (-4) और \(-\frac{1}{4}\) / (-4) and \(-\frac{1}{4}\)
C (2) और (4) / (2) and (4)
D \(\frac{4}{3}\) और (1) / \(\frac{4}{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\)
Step 1
Concept
(4x-2 -17x+4=(4x-1)(x-4)). Therefore the roots are \(\frac{1}{4}\) and (4).
Step 2
Why this answer is correct
The correct answer is A. (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\). (4x-2 -17x+4=(4x-1)(x-4)). Therefore the roots are \(\frac{1}{4}\) and (4).
Step 3
Exam Tip
(4x-2 -17x+4=(4x-1)(x-4)) है। इसलिए मूल \(\frac{1}{4}\) और (4) हैं।
Login to save your score, XP, coins and progress.
Question 4/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(16x^2-8x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(16x^2-8x+1=0\)?
#roots
#repeated_root
#fraction_root
A \(\frac{1}{4}\)
B -\(\frac{1}{4}\)
C (4)
D (-4)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{4}\)
Step 1
Concept
(16x-2 -8x+1=(4x-1)2 ). Therefore the repeated root is \(\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{4}\). (16x-2 -8x+1=(4x-1)2 ). Therefore the repeated root is \(\frac{1}{4}\).
Step 3
Exam Tip
(16x-2 -8x+1=(4x-1)2 ) है। इसलिए दोहराया मूल \(\frac{1}{4}\) है।
Login to save your score, XP, coins and progress.
Question 5/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x=\frac{4}{3}\) समीकरण \(3x^2-7x+c=0\) का मूल है तो (c) का मान क्या है?
If \(x=\frac{4}{3}\) is a root of \(3x^2-7x+c=0\), what is the value of (c)?
#roots
#parameter
#fraction_root
A (4)
B (-4)
C \(\frac{4}{3}\)
D -\(\frac{4}{3}\)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{4}{3}\) gives \(\frac{16}{3}-\frac{28}{3}+c=0\), so (c=4). For fractional roots calculate each term separately.
Step 2
Why this answer is correct
The correct answer is A. (4). Putting \(x=\frac{4}{3}\) gives \(\frac{16}{3}-\frac{28}{3}+c=0\), so (c=4). For fractional roots calculate each term separately.
Step 3
Exam Tip
\(x=\frac{4}{3}\) रखने पर \(\frac{16}{3}-\frac{28}{3}+c=0\) इसलिए (c=4) है। भिन्न मूल में हर पद अलग निकालें।
Login to save your score, XP, coins and progress.
Question 6/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(3x^2-13x+4=0\) के मूल कौन से हैं?
What are the roots of \(3x^2-13x+4=0\)?
#roots
#factorisation
#fraction_root
A (4) और \(\frac{1}{3}\) / (4) and \(\frac{1}{3}\)
B (-4) और \(-\frac{1}{3}\) / (-4) and \(-\frac{1}{3}\)
C (3) और (4) / (3) and (4)
D \(\frac{4}{3}\) और (1) / \(\frac{4}{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (4) और \(\frac{1}{3}\) / (4) and \(\frac{1}{3}\)
Step 1
Concept
(3x-2 -13x+4=(3x-1)(x-4)). Therefore the roots are \(\frac{1}{3}\) and (4).
Step 2
Why this answer is correct
The correct answer is A. (4) और \(\frac{1}{3}\) / (4) and \(\frac{1}{3}\). (3x-2 -13x+4=(3x-1)(x-4)). Therefore the roots are \(\frac{1}{3}\) and (4).
Step 3
Exam Tip
(3x-2 -13x+4=(3x-1)(x-4)) है। इसलिए मूल \(\frac{1}{3}\) और (4) हैं।
Login to save your score, XP, coins and progress.
Question 7/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(9x^2-6x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(9x^2-6x+1=0\)?
#roots
#repeated_root
#fraction_root
A \(\frac{1}{3}\)
B -\(\frac{1}{3}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{3}\)
Step 1
Concept
(9x-2 -6x+1=(3x-1)2 ). Therefore the repeated root is \(\frac{1}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{3}\). (9x-2 -6x+1=(3x-1)2 ). Therefore the repeated root is \(\frac{1}{3}\).
Step 3
Exam Tip
(9x-2 -6x+1=(3x-1)2 ) है। इसलिए दोहराया मूल \(\frac{1}{3}\) है।
Login to save your score, XP, coins and progress.
Question 8/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x=\frac{3}{2}\) समीकरण \(2x^2-5x+c=0\) का मूल है तो (c) का मान क्या है?
If \(x=\frac{3}{2}\) is a root of \(2x^2-5x+c=0\), what is the value of (c)?
#roots
#parameter
#fraction_root
A (3)
B (-3)
C \(\frac{3}{2}\)
D \(-\frac{3}{2}\)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{3}{2}\) gives \(\frac{9}{2}-\frac{15}{2}+c=0\), so (c=3). For fractional roots calculate each term separately.
Step 2
Why this answer is correct
The correct answer is A. (3). Putting \(x=\frac{3}{2}\) gives \(\frac{9}{2}-\frac{15}{2}+c=0\), so (c=3). For fractional roots calculate each term separately.
Step 3
Exam Tip
\(x=\frac{3}{2}\) रखने पर \(\frac{9}{2}-\frac{15}{2}+c=0\) इसलिए (c=3) है। भिन्न मूल में हर पद अलग निकालें।
Login to save your score, XP, coins and progress.
Question 9/12
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(2x^2-7x+3=0\) के मूल कौन से हैं?
What are the roots of \(2x^2-7x+3=0\)?
#roots
#factorisation
#fraction_root
A (3) और \(\frac{1}{2}\) / (3) and \(\frac{1}{2}\)
B (-3) और \(-\frac{1}{2}\) / (-3) and \(-\frac{1}{2}\)
C (2) और (3) / (2) and (3)
D \(\frac{3}{2}\) और (1) / \(\frac{3}{2}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (3) और \(\frac{1}{2}\) / (3) and \(\frac{1}{2}\)
Step 1
Concept
(2x-2 -7x+3=(2x-1)(x-3)). Therefore the roots are \(\frac{1}{2}\) and (3).
Step 2
Why this answer is correct
The correct answer is A. (3) और \(\frac{1}{2}\) / (3) and \(\frac{1}{2}\). (2x-2 -7x+3=(2x-1)(x-3)). Therefore the roots are \(\frac{1}{2}\) and (3).
Step 3
Exam Tip
(2x-2 -7x+3=(2x-1)(x-3)) है। इसलिए मूल \(\frac{1}{2}\) और (3) हैं।
Login to save your score, XP, coins and progress.
Question 10/12
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(4x^2-12x+9=0\) के मूल कौन से हैं?
What are the roots of \(4x^2-12x+9=0\)?
#roots
#factorisation
#fraction_root
A \( \frac{3}{2} \) और \( \frac{3}{2} \) / \( \frac{3}{2} \) and \( \frac{3}{2} \)
B \(-\frac{3}{2}\) और \(-\frac{3}{2}\) / \(-\frac{3}{2}\) and \(-\frac{3}{2}\)
C (2) और (3) / (2) and (3)
D \(\frac{2}{3}\) और \(\frac{2}{3}\) / \(\frac{2}{3}\) and \(\frac{2}{3}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{3}{2} \) और \( \frac{3}{2} \) / \( \frac{3}{2} \) and \( \frac{3}{2} \)
Step 1
Concept
(4x-2 -12x+9=(2x-3)2 ). Therefore the repeated root is \(\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{3}{2} \) और \( \frac{3}{2} \) / \( \frac{3}{2} \) and \( \frac{3}{2} \). (4x-2 -12x+9=(2x-3)2 ). Therefore the repeated root is \(\frac{3}{2}\).
Step 3
Exam Tip
(4x-2 -12x+9=(2x-3)2 ) है। इसलिए दोहराया मूल \(\frac{3}{2}\) है।
Login to save your score, XP, coins and progress.
Question 11/12
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x=\frac{1}{2}\) समीकरण \(4x^2+px-3=0\) का मूल है तो (p) का मान क्या है?
If \(x=\frac{1}{2}\) is a root of \(4x^2+px-3=0\), what is the value of (p)?
#roots
#parameter
#fraction_root
A (2)
B (4)
C (-4)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{1}{2}\) gives \(1+\frac{p}{2}-3=0\), so (p=4). When substituting a fractional root, solve each term carefully.
Step 2
Why this answer is correct
The correct answer is B. (4). Putting \(x=\frac{1}{2}\) gives \(1+\frac{p}{2}-3=0\), so (p=4). When substituting a fractional root, solve each term carefully.
Step 3
Exam Tip
\(x=\frac{1}{2}\) रखने पर \(1+\frac{p}{2}-3=0\) इसलिए (p=4)। भिन्न मूल रखते समय हर पद सावधानी से हल करें।
Login to save your score, XP, coins and progress.
Question 12/12
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(3x^2-10x+3=0\) के मूल कौन से हैं?
What are the roots of \(3x^2-10x+3=0\)?
#roots
#factorisation
#fraction_root
A (3) और \(\frac{1}{3}\) / (3) and \(\frac{1}{3}\)
B (1) और (3) / (1) and (3)
C (-3) और \(-\frac{1}{3}\) / (-3) and \(-\frac{1}{3}\)
D \(\frac{3}{2}\) और (2) / \(\frac{3}{2}\) and (2)
Explanation opens after your attempt
Correct Answer
A. (3) और \(\frac{1}{3}\) / (3) and \(\frac{1}{3}\)
Step 1
Concept
(3x-2 -10x+3=(3x-1)(x-3)). Therefore the roots are \(\frac{1}{3}\) and (3).
Step 2
Why this answer is correct
The correct answer is A. (3) और \(\frac{1}{3}\) / (3) and \(\frac{1}{3}\). (3x-2 -10x+3=(3x-1)(x-3)). Therefore the roots are \(\frac{1}{3}\) and (3).
Step 3
Exam Tip
(3x-2 -10x+3=(3x-1)(x-3)) है। इसलिए मूल \(\frac{1}{3}\) और (3) हैं।
Login to save your score, XP, coins and progress.
