Search: negative root

Question Expert Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

\(x^2+12x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

For \(x^2+12x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt

Correct Answer

A. \(0<\lambda<36\)

Step 1

Concept

For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).

Step 2

Why this answer is correct

The correct answer is A. \(0<\lambda<36\). For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).

Step 3

Exam Tip

दोनों ऋणात्मक जड़ों के लिए योग (-12) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(144-4\lambda>0\), इसलिए \(0<\lambda<36\)।

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

\(x^2+10x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

For \(x^2+10x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt

Correct Answer

B. \(0<\lambda<25\)

Step 1

Concept

For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).

Step 2

Why this answer is correct

The correct answer is B. \(0<\lambda<25\). For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).

Step 3

Exam Tip

दोनों ऋणात्मक जड़ों के लिए योग (-10) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(100-4\lambda>0\), इसलिए \(0<\lambda<25\)।

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

\(x^2+2x+\lambda=0\) की जड़ें वास्तविक और भिन्न हों तथा दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

For \(x^2+2x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt

Correct Answer

A. \(0<\lambda<1\)

Step 1

Concept

For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).

Step 2

Why this answer is correct

The correct answer is A. \(0<\lambda<1\). For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).

Step 3

Exam Tip

दोनों ऋणात्मक जड़ों के लिए योग (-2) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(4-4\lambda>0\), इसलिए \(0<\lambda<1\)।

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Question 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 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 Easy Mathematics Quadratic Equations Nature of Roots Class 10 Level 37

समीकरण \(3x^2+6x+3=0\) के मूल किस प्रकार के होंगे?

What type of roots will the equation \(3x^2+6x+3=0\) have?

Explanation opens after your attempt

Correct Answer

A. दो वास्तविक और समानtwo real and equal

Step 1

Concept

(D=6²-4(3)(3)=0). Hence both roots will be equal and real.

Step 2

Why this answer is correct

The correct answer is A. दो वास्तविक और समान / two real and equal. (D=6²-4(3)(3)=0). Hence both roots will be equal and real.

Step 3

Exam Tip

(D=6²-4(3)(3)=0) है। इसलिए दोनों मूल समान वास्तविक होंगे।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(x^2+14x+48=0\) के लिए सही हल कौनसा है?

Which is the correct solution for \(x^2+14x+48=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-6,-8)

Step 1

Concept

(x^-2+14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-6,-8). (x^-2+14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.

Step 3

Exam Tip

(x^-2+14x+48=(x+6)(x+8)), इसलिए (x=-6,-8) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद से ऋणात्मक मूल आ सकते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(5x^2+16x+3=0\) को हल करने पर मूल क्या होंगे?

What will be the roots after solving \(5x^2+16x+3=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=-3,-\frac{1}{5}\)

Step 1

Concept

(5x^-2+16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.

Step 2

Why this answer is correct

The correct answer is A. \(x=-3,-\frac{1}{5}\). (5x^-2+16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.

Step 3

Exam Tip

(5x^-2+16x+3=(5x+1)(x+3)), इसलिए मूल \(-\frac{1}{5}\) और (-3) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(12x^2+17x+5=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?

What roots are obtained by solving \(12x^2+17x+5=0\) by factorisation?

Explanation opens after your attempt

Correct Answer

A. \(x=-1,-\frac{5}{12}\)

Step 1

Concept

(12x^-2+17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.

Step 2

Why this answer is correct

The correct answer is A. \(x=-1,-\frac{5}{12}\). (12x^-2+17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.

Step 3

Exam Tip

(12x^-2+17x+5=(12x+5)(x+1)), इसलिए मूल \(-\frac{5}{12}\) और (-1) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(x^2+12x+32=0\) के लिए सही हल कौनसा है?

Which is the correct solution for \(x^2+12x+32=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-4,-8)

Step 1

Concept

(x^-2+12x+32=(x+4)(x+8)), so (x=-4,-8). In exams, a positive middle term and positive constant can give negative roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-4,-8). (x^-2+12x+32=(x+4)(x+8)), so (x=-4,-8). In exams, a positive middle term and positive constant can give negative roots.

Step 3

Exam Tip

(x^-2+12x+32=(x+4)(x+8)), इसलिए (x=-4,-8) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद से ऋणात्मक मूल आ सकते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(3x^2+11x+10=0\) को हल करने पर मूल क्या होंगे?

What will be the roots after solving \(3x^2+11x+10=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=-2,-\frac{5}{3}\)

Step 1

Concept

(3x^-2+11x+10=(3x+5)(x+2)), so the roots are \(-\frac{5}{3}\) and (-2). In exams, positive factors give negative roots.

Step 2

Why this answer is correct

The correct answer is A. \(x=-2,-\frac{5}{3}\). (3x^-2+11x+10=(3x+5)(x+2)), so the roots are \(-\frac{5}{3}\) and (-2). In exams, positive factors give negative roots.

Step 3

Exam Tip

(3x^-2+11x+10=(3x+5)(x+2)), इसलिए मूल \(-\frac{5}{3}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(3x^2+8x+4=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?

What roots are obtained by solving \(3x^2+8x+4=0\) by factorisation?

Explanation opens after your attempt

Correct Answer

A. \(x=-2,-\frac{2}{3}\)

Step 1

Concept

(3x^-2+8x+4=(3x+2)(x+2)), so the roots are \(-\frac{2}{3}\) and (-2). In exams, positive factors give negative roots.

Step 2

Why this answer is correct

The correct answer is A. \(x=-2,-\frac{2}{3}\). (3x^-2+8x+4=(3x+2)(x+2)), so the roots are \(-\frac{2}{3}\) and (-2). In exams, positive factors give negative roots.

Step 3

Exam Tip

(3x^-2+8x+4=(3x+2)(x+2)), इसलिए मूल \(-\frac{2}{3}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(x^2+10x+21=0\) के लिए सही हल कौनसा है?

Which is the correct solution for \(x^2+10x+21=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-3,-7)

Step 1

Concept

(x^-2+10x+21=(x+3)(x+7)), so (x=-3,-7). In exams, a positive middle term and positive constant can give both negative roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-3,-7). (x^-2+10x+21=(x+3)(x+7)), so (x=-3,-7). In exams, a positive middle term and positive constant can give both negative roots.

Step 3

Exam Tip

(x^-2+10x+21=(x+3)(x+7)), इसलिए (x=-3,-7) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद पर दोनों मूल ऋणात्मक हो सकते हैं।

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Question Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(2x^2+7x+6=0\) को हल करने पर मूल क्या होंगे?

What will be the roots after solving \(2x^2+7x+6=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=-\frac{3}{2},-2\)

Step 1

Concept

(2x^-2+7x+6=(2x+3)(x+2)), so \(x=-\frac{3}{2}\) and (-2). In exams, positive factors give negative roots.

Step 2

Why this answer is correct

The correct answer is A. \(x=-\frac{3}{2},-2\). (2x^-2+7x+6=(2x+3)(x+2)), so \(x=-\frac{3}{2}\) and (-2). In exams, positive factors give negative roots.

Step 3

Exam Tip

(2x^-2+7x+6=(2x+3)(x+2)), इसलिए \(x=-\frac{3}{2}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।

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Question Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(x^2+9x+20=0\) के मूल कौनसे हैं?

Which are the roots of \(x^2+9x+20=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-4,-5)

Step 1

Concept

(x^-2+9x+20=(x+4)(x+5)), so the roots are (-4) and (-5). In exams, a positive middle term can give negative roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-4,-5). (x^-2+9x+20=(x+4)(x+5)), so the roots are (-4) and (-5). In exams, a positive middle term can give negative roots.

Step 3

Exam Tip

(x^-2+9x+20=(x+4)(x+5)), इसलिए मूल (-4) और (-5) हैं। परीक्षा में धनात्मक मध्य पद से ऋणात्मक मूल मिल सकते हैं।

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Question Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(x^2+8x+12=0\) के मूल कौनसे हैं?

Which are the roots of \(x^2+8x+12=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-2,-6)

Step 1

Concept

(x^-2+8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-2,-6). (x^-2+8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.

Step 3

Exam Tip

(x^-2+8x+12=(x+2)(x+6)), इसलिए मूल (-2) और (-6) हैं। परीक्षा में धनात्मक मध्य पद से ऋणात्मक मूल मिल सकते हैं।

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Question Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(x^2+6x+8=0\) के लिए सही मूल कौनसे हैं?

Which are the correct roots for \(x^2+6x+8=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-2,-4)

Step 1

Concept

(x^-2+6x+8=(x+2)(x+4)), so (x=-2) and (x=-4). In exams, a positive middle term can give negative roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-2,-4). (x^-2+6x+8=(x+2)(x+4)), so (x=-2) and (x=-4). In exams, a positive middle term can give negative roots.

Step 3

Exam Tip

(x^-2+6x+8=(x+2)(x+4)), इसलिए (x=-2) और (x=-4) हैं। परीक्षा में धनात्मक मध्य पद के साथ ऋणात्मक मूल आ सकते हैं।

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Question 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 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 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 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

क्या (x=-3) समीकरण \(2x^2+7x+3=0\) का मूल है?

Is (x=-3) a root of \(2x^2+7x+3=0\)?

Explanation opens after your attempt

Correct Answer

A. हाँYes

Step 1

Concept

Putting (x=-3) gives (18-21+3=0). Watch the signs carefully when substituting a negative root.

Step 2

Why this answer is correct

The correct answer is A. हाँ / Yes. Putting (x=-3) gives (18-21+3=0). Watch the signs carefully when substituting a negative root.

Step 3

Exam Tip

(x=-3) रखने पर (18-21+3=0) मिलता है। ऋणात्मक मूल रखते समय चिन्हों को ध्यान से देखें।

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Question Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

क्या (x=-2) समीकरण \(3x^2+2x-8=0\) का मूल है?

Is (x=-2) a root of \(3x^2+2x-8=0\)?

Explanation opens after your attempt

Correct Answer

A. हाँYes

Step 1

Concept

Putting (x=-2) gives (12-4-8=0). Handle signs carefully when substituting a negative root.

Step 2

Why this answer is correct

The correct answer is A. हाँ / Yes. Putting (x=-2) gives (12-4-8=0). Handle signs carefully when substituting a negative root.

Step 3

Exam Tip

(x=-2) रखने पर (12-4-8=0) मिलता है। ऋणात्मक मूल रखते समय चिन्हों को ध्यान से संभालें।

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Question Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

क्या (x=-1) समीकरण \(2x^2+5x+3=0\) का मूल है?

Is (x=-1) a root of \(2x^2+5x+3=0\)?

Explanation opens after your attempt

Correct Answer

A. हाँYes

Step 1

Concept

Putting (x=-1) gives (2-5+3=0). Check signs carefully for a negative root.

Step 2

Why this answer is correct

The correct answer is A. हाँ / Yes. Putting (x=-1) gives (2-5+3=0). Check signs carefully for a negative root.

Step 3

Exam Tip

(x=-1) रखने पर (2-5+3=0) मिलता है। ऋणात्मक मूल में चिन्हों की जांच ध्यान से करें।

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Question Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

यदि (x=-4) समीकरण \(2x^2+px-8=0\) का मूल है तो (p) का मान क्या है?

If (x=-4) is a root of \(2x^2+px-8=0\), what is the value of (p)?

Explanation opens after your attempt

Correct Answer

A. (6)

Step 1

Concept

Putting (x=-4) gives (32-4p-8=0), so (p=6). Check signs carefully with a negative root.

Step 2

Why this answer is correct

The correct answer is A. (6). Putting (x=-4) gives (32-4p-8=0), so (p=6). Check signs carefully with a negative root.

Step 3

Exam Tip

(x=-4) रखने पर (32-4p-8=0) इसलिए (p=6) मिलता है। ऋणात्मक मूल में चिन्हों की जांच सावधानी से करें।

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Question Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

क्या (x=-1) समीकरण \(2x^2+x-1=0\) का मूल है?

Is (x=-1) a root of \(2x^2+x-1=0\)?

Explanation opens after your attempt

Correct Answer

A. हाँYes

Step 1

Concept

Putting (x=-1) gives (2-1-1=0). Sign checking is important with a negative root.

Step 2

Why this answer is correct

The correct answer is A. हाँ / Yes. Putting (x=-1) gives (2-1-1=0). Sign checking is important with a negative root.

Step 3

Exam Tip

(x=-1) रखने पर (2-1-1=0) मिलता है। ऋणात्मक मूल में चिन्हों की जांच जरूरी है।

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Question Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि किसी समीकरण के मूल (2) और (-5) हैं तो उनका योग क्या है?

If the roots of an equation are (2) and (-5), what is their sum?

Explanation opens after your attempt

Correct Answer

B. (-3)

Step 1

Concept

The sum is (2+(-5)=-3). Do not forget the sign while adding a negative root.

Step 2

Why this answer is correct

The correct answer is B. (-3). The sum is (2+(-5)=-3). Do not forget the sign while adding a negative root.

Step 3

Exam Tip

योग (2+(-5)=-3) है। ऋणात्मक मूल जोड़ते समय चिन्ह न भूलें।

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Question Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि (-1) समीकरण \(x^2+rx-12=0\) का मूल है तो (r) का मान क्या है?

If (-1) is a root of \(x^2+rx-12=0\), what is the value of (r)?

Explanation opens after your attempt

Correct Answer

A. (-11)

Step 1

Concept

Putting (x=-1) gives (1-r-12=0), so (r=-11). Check the sign of each term while using a negative root.

Step 2

Why this answer is correct

The correct answer is A. (-11). Putting (x=-1) gives (1-r-12=0), so (r=-11). Check the sign of each term while using a negative root.

Step 3

Exam Tip

(x=-1) रखने पर (1-r-12=0) इसलिए (r=-11)। ऋणात्मक मूल रखते समय प्रत्येक पद का चिन्ह देखें।

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Question 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 Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि (-2) समीकरण \(x^2+kx+4=0\) का मूल है तो (k) का मान क्या है?

If (-2) is a root of \(x^2+kx+4=0\) then what is the value of (k)?

Explanation opens after your attempt

Correct Answer

B. (4)

Step 1

Concept

Putting (x=-2) gives (4-2k+4=0) so (k=4). Be careful with signs while substituting a negative root.

Step 2

Why this answer is correct

The correct answer is B. (4). Putting (x=-2) gives (4-2k+4=0) so (k=4). Be careful with signs while substituting a negative root.

Step 3

Exam Tip

(x=-2) रखने पर (4-2k+4=0) इसलिए (k=4)। ऋणात्मक मूल रखते समय चिन्ह सावधानी से लगाएं।

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Question 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 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 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 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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Question Easy Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29

\(x^2+6x+9=0\) में (x=-3) रखने पर क्या होगा?

What happens when (x=-3) is put in \(x^2+6x+9=0\)?

Explanation opens after your attempt

Correct Answer

A. समीकरण संतुष्ट होता हैThe equation is satisfied

Step 1

Concept

((-3)²+6(-3)+9=0). Hence (x=-3) is a solution.

Step 2

Why this answer is correct

The correct answer is A. समीकरण संतुष्ट होता है / The equation is satisfied. ((-3)²+6(-3)+9=0). Hence (x=-3) is a solution.

Step 3

Exam Tip

((-3)²+6(-3)+9=0) है। अतः (x=-3) हल है।

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Question Easy Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28

\(x^2+4x+4=0\) में (x=-2) रखने पर क्या होगा?

What happens when (x=-2) is put in \(x^2+4x+4=0\)?

Explanation opens after your attempt

Correct Answer

A. समीकरण संतुष्ट होता हैThe equation is satisfied

Step 1

Concept

((-2)²+4(-2)+4=0). Hence (x=-2) is a solution.

Step 2

Why this answer is correct

The correct answer is A. समीकरण संतुष्ट होता है / The equation is satisfied. ((-2)²+4(-2)+4=0). Hence (x=-2) is a solution.

Step 3

Exam Tip

((-2)²+4(-2)+4=0) है। अतः (x=-2) हल है।

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Question Hard Mathematics Real Numbers 5: Irrational numbers Class 10 Level 14

यदि (x) अपरिमेय है और \(x^2=7\), तो (x) का संभावित मान कौन-सा है?

If (x) is irrational and \(x^2=7\), which can be the value of (x)?

Explanation opens after your attempt

Correct Answer

B. \(\sqrt{7}\)

Step 1

Concept

From \(x^2=7\), \(x=\sqrt{7}\) or \(x=-\sqrt{7}\).

Step 2

Why this answer is correct

Among the options, \(\sqrt{7}\) is present and it is irrational.

Step 3

Exam Tip

Remember both positive and negative roots, then match the given options. चरण 1: \(x^2=7\) से \(x=\sqrt{7}\) या \(x=-\sqrt{7}\) हो सकता है। चरण 2: दिए गए विकल्पों में \(\sqrt{7}\) है, जो अपरिमेय है। चरण 3: वर्ग समीकरण में धन और ऋण दोनों मूल याद रखें, पर विकल्प के अनुसार चुनें।

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\(x^2+12x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

Concept

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Exam Tip

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

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\(x^2+10x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

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Exam Tip

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

Concept

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Exam Tip

\(x^2+2x+\lambda=0\) की जड़ें वास्तविक और भिन्न हों तथा दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

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

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किस समीकरण में मूल बराबर और ऋणात्मक होंगे?

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समीकरण \(3x^2+6x+3=0\) के मूल किस प्रकार के होंगे?

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\(x^2+14x+48=0\) के लिए सही हल कौनसा है?

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\(5x^2+16x+3=0\) को हल करने पर मूल क्या होंगे?

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\(12x^2+17x+5=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?

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\(x^2+12x+32=0\) के लिए सही हल कौनसा है?

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\(3x^2+11x+10=0\) को हल करने पर मूल क्या होंगे?

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Exam Tip

\(3x^2+8x+4=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?

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\(x^2+10x+21=0\) के लिए सही हल कौनसा है?

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\(2x^2+7x+6=0\) को हल करने पर मूल क्या होंगे?

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\(x^2+9x+20=0\) के मूल कौनसे हैं?

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\(x^2+8x+12=0\) के मूल कौनसे हैं?

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Exam Tip

\(x^2+6x+8=0\) के लिए सही मूल कौनसे हैं?

Concept

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Exam Tip

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

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