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equal-negative-roots MCQ Questions for Class 10

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12 questions tagged with equal-negative-roots.

Question 1/12 Expert Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि \(x^2+bx+49=0\) की जड़ें समान और ऋणात्मक हैं, तो (b) का मान क्या होगा?

If \(x^2+bx+49=0\) has equal and negative roots, what is the value of (b)?

Explanation opens after your attempt
Correct Answer

B. (14)

Step 1

Concept

For equal roots, \(b^2-196=0\), so \(b=\pm14\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=14).

Step 2

Why this answer is correct

The correct answer is B. (14). For equal roots, \(b^2-196=0\), so \(b=\pm14\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=14).

Step 3

Exam Tip

समान जड़ों के लिए \(b^2-196=0\), इसलिए \(b=\pm14\)। समान जड़ \(-\frac{b}{2}\) ऋणात्मक होनी चाहिए, अतः (b=14)।

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Question 2/12 Expert Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि \(x^2+bx+25=0\) की जड़ें समान और ऋणात्मक हैं, तो (b) का मान क्या होगा?

If \(x^2+bx+25=0\) has equal and negative roots, what is the value of (b)?

Explanation opens after your attempt
Correct Answer

A. (10)

Step 1

Concept

For equal roots, \(b^2-100=0\), so \(b=\pm10\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=10).

Step 2

Why this answer is correct

The correct answer is A. (10). For equal roots, \(b^2-100=0\), so \(b=\pm10\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=10).

Step 3

Exam Tip

समान जड़ों के लिए \(b^2-100=0\), इसलिए \(b=\pm10\)। समान जड़ \(-\frac{b}{2}\) ऋणात्मक होनी चाहिए, अतः (b=10)।

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Question 3/12 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

\(2x^2+kx+2=0\) की जड़ें समान और ऋणात्मक हों, तो (k) का मान क्या होगा?

If \(2x^2+kx+2=0\) has equal and negative roots, what is (k)?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

For equal roots, \(k^2-16=0\), so \(k=\pm4\). The equal root is \(-\frac{k}{4}\), which is negative only when (k=4).

Step 2

Why this answer is correct

The correct answer is A. (4). For equal roots, \(k^2-16=0\), so \(k=\pm4\). The equal root is \(-\frac{k}{4}\), which is negative only when (k=4).

Step 3

Exam Tip

समान जड़ों के लिए \(k^2-16=0\), इसलिए \(k=\pm4\)। समान जड़ \(-\frac{k}{4}\) है, जो ऋणात्मक तभी होगी जब (k=4)।

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Question 4/12 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि \(x^2+kx+49=0\) के मूल एक दूसरे के बराबर और ऋणात्मक हैं तो (k) क्या होगा?

If roots of \(x^2+kx+49=0\) are equal and negative, what is (k)?

Explanation opens after your attempt
Correct Answer

A. (14)

Step 1

Concept

The equal negative roots are (-7) and (-7). Their sum is (-14), so (k=14).

Step 2

Why this answer is correct

The correct answer is A. (14). The equal negative roots are (-7) and (-7). Their sum is (-14), so (k=14).

Step 3

Exam Tip

बराबर ऋणात्मक मूल (-7) और (-7) हैं। योग (-14) है इसलिए (k=14) है।

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Question 5/12 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि समीकरण \(x^2+px+25=0\) के मूल बराबर हैं और दोनों ऋणात्मक हैं तो (p) का मान क्या होगा?

If the roots of \(x^2+px+25=0\) are equal and both negative, what is the value of (p)?

Explanation opens after your attempt
Correct Answer

A. (10)

Step 1

Concept

The equal negative roots are (-5) and (-5) because the product is (25). The sum is (-10), so (p=10).

Step 2

Why this answer is correct

The correct answer is A. (10). The equal negative roots are (-5) and (-5) because the product is (25). The sum is (-10), so (p=10).

Step 3

Exam Tip

बराबर ऋणात्मक मूल (-5) और (-5) होंगे क्योंकि गुणनफल (25) है। योग (-10) है इसलिए (p=10) है।

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Question 6/12 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि \(x^2+kx+36=0\) के मूल एक दूसरे के बराबर और ऋणात्मक हैं तो (k) क्या होगा?

If roots of \(x^2+kx+36=0\) are equal and negative, what is (k)?

Explanation opens after your attempt
Correct Answer

A. (12)

Step 1

Concept

The equal negative roots are (-6) and (-6). Their sum is (-12), so (-k=-12) gives (k=12).

Step 2

Why this answer is correct

The correct answer is A. (12). The equal negative roots are (-6) and (-6). Their sum is (-12), so (-k=-12) gives (k=12).

Step 3

Exam Tip

बराबर ऋणात्मक मूल (-6) और (-6) हैं। योग (-12) है इसलिए (-k=-12) से (k=12) है।

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Question 7/12 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि समीकरण \(x^2+px+16=0\) के मूल बराबर हैं और दोनों ऋणात्मक हैं तो (p) का मान क्या होगा?

If the roots of \(x^2+px+16=0\) are equal and both negative, what is the value of (p)?

Explanation opens after your attempt
Correct Answer

A. (8)

Step 1

Concept

For equal roots, the roots are (-4) and (-4) because the product is (16). The sum is (-8), so (-p=-8) gives (p=8).

Step 2

Why this answer is correct

The correct answer is A. (8). For equal roots, the roots are (-4) and (-4) because the product is (16). The sum is (-8), so (-p=-8) gives (p=8).

Step 3

Exam Tip

बराबर मूलों के लिए मूल (-4) और (-4) होंगे क्योंकि गुणनफल (16) है। योग (-8) है इसलिए (-p=-8) से (p=8) है।

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Question 8/12 Expert Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30

समीकरण \(x^2+px+49=0\) के मूल समान और ऋणात्मक हैं। (p) का मान क्या होगा?

The roots of \(x^2+px+49=0\) are equal and negative. What is the value of (p)?

Explanation opens after your attempt
Correct Answer

A. (14)

Step 1

Concept

For equal roots, \(p^2-196=0\) gives \(p=\pm14\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=14).

Step 2

Why this answer is correct

The correct answer is A. (14). For equal roots, \(p^2-196=0\) gives \(p=\pm14\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=14).

Step 3

Exam Tip

समान मूलों के लिए \(p^2-196=0\) से \(p=\pm14\) मिलता है। ऋणात्मक समान मूल के लिए \(-\frac{p}{2}<0\), इसलिए (p=14)।

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Question 9/12 Hard Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30

किस समीकरण में मूल बराबर और ऋणात्मक होंगे?

Which equation will have equal and negative roots?

Explanation opens after your attempt
Correct Answer

A. \(x^2+16x+64=0\)

Step 1

Concept

(x-2+16x+64=(x+8)2), so both roots are (-8). Both roots are equal and negative.

Step 2

Why this answer is correct

The correct answer is A. \(x^2+16x+64=0\). (x-2+16x+64=(x+8)2), so both roots are (-8). Both roots are equal and negative.

Step 3

Exam Tip

(x-2+16x+64=(x+8)2), इसलिए दोनों मूल (-8) हैं। दोनों मूल बराबर और ऋणात्मक हैं।

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Question 10/12 Hard Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29

समीकरण \(x^2+px+9=0\) के मूल समान और ऋणात्मक हैं। (p) का मान क्या होगा?

The roots of \(x^2+px+9=0\) are equal and negative. What is the value of (p)?

Explanation opens after your attempt
Correct Answer

A. (6)

Step 1

Concept

For equal roots, \(p^2-36=0\) gives \(p=\pm6\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=6).

Step 2

Why this answer is correct

The correct answer is A. (6). For equal roots, \(p^2-36=0\) gives \(p=\pm6\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=6).

Step 3

Exam Tip

समान मूलों के लिए \(p^2-36=0\) से \(p=\pm6\) मिलता है। ऋणात्मक समान मूल के लिए \(-\frac{p}{2}<0\), इसलिए (p=6)।

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Question 11/12 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?

Explanation opens after your attempt
Correct Answer

A. (18)

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 12/12 Medium Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28

समीकरण \(x^2+kx+25=0\) में यदि मूल समान और ऋणात्मक हैं, तो (k) का संभव मान कौन-सा है?

In \(x^2+kx+25=0\), if the roots are equal and negative, which possible value of (k) is correct?

Explanation opens after your attempt
Correct Answer

A. (10)

Step 1

Concept

For equal roots, (D=0) gives \(k^2=100\), and for equal negative roots \(-\frac{k}{2}<0\) is needed. Hence (k=10) is correct.

Step 2

Why this answer is correct

The correct answer is A. (10). For equal roots, (D=0) gives \(k^2=100\), and for equal negative roots \(-\frac{k}{2}<0\) is needed. Hence (k=10) is correct.

Step 3

Exam Tip

समान मूलों के लिए (D=0) से \(k^2=100\), और ऋणात्मक समान मूल के लिए \(-\frac{k}{2}<0\) चाहिए। इसलिए (k=10) सही है।

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