Question 1/6
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2=121\) के मूल कौन से हैं?
What are the roots of \(x^2=121\)?
Explanation opens after your attempt
Correct Answer
A. (11) और (-11)/(11) and (-11)
Step 1
Concept
From \(x^2=121\), we get \(x=\pm11\). Take both signs while finding square roots.
Step 2
Why this answer is correct
The correct answer is A. (11) और (-11) / (11) and (-11). From \(x^2=121\), we get \(x=\pm11\). Take both signs while finding square roots.
Step 3
Exam Tip
\(x^2=121\) से \(x=\pm11\) मिलता है। वर्गमूल लेते समय दोनों चिन्ह लें।
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Question 2/6
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(5x^2-45=0\) के मूल कौन से हैं?
What are the roots of \(5x^2-45=0\)?
Explanation opens after your attempt
Correct Answer
A. (3) और (-3)/(3) and (-3)
Step 1
Concept
From \(5x^2-45=0\), we get \(x^2=9\). Therefore \(x=\pm3\).
Step 2
Why this answer is correct
The correct answer is A. (3) और (-3) / (3) and (-3). From \(5x^2-45=0\), we get \(x^2=9\). Therefore \(x=\pm3\).
Step 3
Exam Tip
\(5x^2-45=0\) से \(x^2=9\) मिलता है। इसलिए \(x=\pm3\) है।
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Question 3/6
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2=49\) के मूल कौन से हैं?
What are the roots of \(x^2=49\)?
Explanation opens after your attempt
Correct Answer
A. (7) और (-7)/(7) and (-7)
Step 1
Concept
From \(x^2=49\), we get \(x=\pm7\). Take both signs while finding square roots.
Step 2
Why this answer is correct
The correct answer is A. (7) और (-7) / (7) and (-7). From \(x^2=49\), we get \(x=\pm7\). Take both signs while finding square roots.
Step 3
Exam Tip
\(x^2=49\) से \(x=\pm7\) मिलता है। वर्गमूल लेते समय दोनों चिन्ह लें।
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Question 4/6
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(3x^2-27=0\) के मूल कौन से हैं?
What are the roots of \(3x^2-27=0\)?
Explanation opens after your attempt
Correct Answer
A. (3) और (-3)/(3) and (-3)
Step 1
Concept
From \(3x^2-27=0\), we get \(x^2=9\). Therefore \(x=\pm3\).
Step 2
Why this answer is correct
The correct answer is A. (3) और (-3) / (3) and (-3). From \(3x^2-27=0\), we get \(x^2=9\). Therefore \(x=\pm3\).
Step 3
Exam Tip
\(3x^2-27=0\) से \(x^2=9\) मिलता है। इसलिए \(x=\pm3\) है।
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Question 5/6
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2=16\) के मूल कौन से हैं?
What are the roots of \(x^2=16\)?
Explanation opens after your attempt
Correct Answer
A. (4) और (-4)/(4) and (-4)
Step 1
Concept
From \(x^2=16\) we get \(x=\pm4\). In a square equation check both positive and negative roots.
Step 2
Why this answer is correct
The correct answer is A. (4) और (-4) / (4) and (-4). From \(x^2=16\) we get \(x=\pm4\). In a square equation check both positive and negative roots.
Step 3
Exam Tip
\(x^2=16\) से \(x=\pm4\) मिलता है। वर्ग समीकरण में धनात्मक और ऋणात्मक दोनों मूल देखें।
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Question 6/6
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(2x^2-8=0\) के मूल कौन से हैं?
What are the roots of \(2x^2-8=0\)?
Explanation opens after your attempt
Correct Answer
A. (2) और (-2)/(2) and (-2)
Step 1
Concept
From \(2x^2-8=0\) we get \(x^2=4\) so \(x=\pm 2\). Take both signs while finding square roots.
Step 2
Why this answer is correct
The correct answer is A. (2) और (-2) / (2) and (-2). From \(2x^2-8=0\) we get \(x^2=4\) so \(x=\pm 2\). Take both signs while finding square roots.
Step 3
Exam Tip
\(2x^2-8=0\) से \(x^2=4\) मिलता है इसलिए \(x=\pm 2\)। वर्गमूल लेते समय दोनों चिन्ह लें।
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