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80 results found for "negative" in Class 10.

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

यदि \(x^2+px+64=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+64=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \(12\sqrt{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.

Step 2

Why this answer is correct

The correct answer is A. \(12\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=64\) से \(r=4\sqrt{2}\) और \(p=3r=12\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।

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

यदि \(x^2+px+49=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+49=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{21\sqrt{2}}{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{21\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=49\) और \(p=3r=\frac{21\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।

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

यदि \(x^2+px+36=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+36=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \(9\sqrt{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.

Step 2

Why this answer is correct

The correct answer is A. \(9\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=36\) से \(r=3\sqrt{2}\) और \(p=3r=9\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।

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

यदि \(x^2+px+25=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+25=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{15\sqrt{2}}{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{15\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।

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

यदि \(x^2+px+16=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+16=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \(6\sqrt{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.

Step 2

Why this answer is correct

The correct answer is A. \(6\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=16\) से \(r=2\sqrt{2}\) और \(p=3r=6\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।

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

यदि \(x^2+px+9=0\) का एक मूल दूसरे का दुगुना है और दोनों ऋणात्मक हैं, तो (p) का सही सरल मान क्या है?

If one root of \(x^2+px+9=0\) is double the other and both are negative, what is the correct simplified value of (p)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{9\sqrt{2}}{2}\)

Step 1

Concept

\(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{9\sqrt{2}}{2}\). \(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 3

Exam Tip

\(\frac{9}{\sqrt{2}}\) को सरल करने पर \(\frac{9\sqrt{2}}{2}\) मिलता है। परीक्षा में हर को परिमेय बनाना न भूलें।

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

यदि \(x^2+px+9=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+9=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \(3\sqrt{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.

Step 2

Why this answer is correct

The correct answer is A. \(3\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=9\) और योग (-3r) है, इसलिए \(p=3r=3\sqrt{\frac{9}{2}}\) नहीं बल्कि \(r=\frac{3}{\sqrt{2}}\), अतः \(p=\frac{9}{\sqrt{2}}\) होता है। परीक्षा में ऐसे प्रश्नों में मानकर समीकरण बनाएं।

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

यदि \(x^2+4x+c=0\) की दोनों जड़ें वास्तविक और ऋणात्मक हैं, तो कौन-सी शर्त पर्याप्त और आवश्यक है?

If both roots of \(x^2+4x+c=0\) are real and negative, which condition is necessary and sufficient?

Explanation opens after your attempt
Correct Answer

A. \(0<c\le4\)

Step 1

Concept

The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, \(16-4c\ge0\), hence \(0<c\le4\).

Step 2

Why this answer is correct

The correct answer is A. \(0<c\le4\). The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, \(16-4c\ge0\), hence \(0<c\le4\).

Step 3

Exam Tip

योग (-4) पहले से ऋणात्मक है और गुणनफल धनात्मक चाहिए, इसलिए (c>0)। वास्तविक जड़ों के लिए \(16-4c\ge0\), अतः \(0<c\le4\)।

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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 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 \(x^2+px+18=0\) के मूलों का अनुपात (1:2) है और दोनों ऋणात्मक हैं तो (p) क्या होगा?

If roots of \(x^2+px+18=0\) are in the ratio (1:2) and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. (9)

Step 1

Concept

The roots are (-3) and (-6) because the product is (18). Their sum is (-9), so (p=9).

Step 2

Why this answer is correct

The correct answer is A. (9). The roots are (-3) and (-6) because the product is (18). Their sum is (-9), so (p=9).

Step 3

Exam Tip

मूल (-3) और (-6) होंगे क्योंकि गुणनफल (18) है। उनका योग (-9) है इसलिए (p=9) 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 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि \(x^2+px+12=0\) के मूलों का अनुपात (1:3) है और दोनों ऋणात्मक हैं तो (p) क्या होगा?

If roots of \(x^2+px+12=0\) are in the ratio (1:3) and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. (8)

Step 1

Concept

The roots are (-2) and (-6) because the product is (12). Their sum is (-8), so (-p=-8) gives (p=8).

Step 2

Why this answer is correct

The correct answer is A. (8). The roots are (-2) and (-6) because the product is (12). Their sum is (-8), so (-p=-8) gives (p=8).

Step 3

Exam Tip

मूल (-2) और (-6) होंगे क्योंकि गुणनफल (12) है। उनका योग (-8) है इसलिए (-p=-8) से (p=8) है।

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

यदि किसी द्विघात समीकरण के मूलों का गुणनफल ऋणात्मक है तो सही कथन कौन सा है?

If the product of roots of a quadratic equation is negative, which statement is correct?

Explanation opens after your attempt
Correct Answer

A. एक मूल धनात्मक और दूसरा ऋणात्मक हैOne root is positive and the other is negative

Step 1

Concept

A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.

Step 2

Why this answer is correct

The correct answer is A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative. A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.

Step 3

Exam Tip

ऋणात्मक गुणनफल तभी होता है जब मूलों के चिन्ह विपरीत हों। इसलिए एक मूल धनात्मक और दूसरा ऋणात्मक होगा।

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

यदि दो वास्तविक मूलों का गुणनफल धनात्मक और योग ऋणात्मक है तो दोनों मूल कैसे होंगे?

If the product of two real roots is positive and their sum is negative, how will both roots be?

Explanation opens after your attempt
Correct Answer

B. दोनों ऋणात्मकBoth negative

Step 1

Concept

A positive product means both roots have the same sign. A negative sum means both roots are negative.

Step 2

Why this answer is correct

The correct answer is B. दोनों ऋणात्मक / Both negative. A positive product means both roots have the same sign. A negative sum means both roots are negative.

Step 3

Exam Tip

गुणनफल धनात्मक होने पर दोनों मूलों का चिन्ह समान होता है। योग ऋणात्मक होने से दोनों मूल ऋणात्मक होंगे।

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

यदि किसी द्विघात समीकरण के वास्तविक मूलों का गुणनफल ऋणात्मक है तो मूलों के चिन्ह कैसे होंगे?

If the product of real roots of a quadratic equation is negative then how are the signs of the roots?

Explanation opens after your attempt
Correct Answer

C. एक धनात्मक और एक ऋणात्मकOne positive and one negative

Step 1

Concept

A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.

Step 2

Why this answer is correct

The correct answer is C. एक धनात्मक और एक ऋणात्मक / One positive and one negative. A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.

Step 3

Exam Tip

ऋणात्मक गुणनफल तभी मिलता है जब एक मूल धनात्मक और दूसरा ऋणात्मक हो। \(\alpha\beta<0\) संकेतों की जांच का छोटा संकेत है।

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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 Expert Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29

किस विकल्प में मूलों का योग ऋणात्मक और गुणनफल ऋणात्मक होगा?

In which option will the sum of roots be negative and the product of roots be negative?

Explanation opens after your attempt
Correct Answer

A. \(x^2+5x-24=0\)

Step 1

Concept

In the first option, the sum is \(-\frac{b}{a}=-5\) and the product is \(\frac{c}{a}=-24\). So both are negative.

Step 2

Why this answer is correct

The correct answer is A. \(x^2+5x-24=0\). In the first option, the sum is \(-\frac{b}{a}=-5\) and the product is \(\frac{c}{a}=-24\). So both are negative.

Step 3

Exam Tip

पहले विकल्प में योग \(-\frac{b}{a}=-5\) और गुणनफल \(\frac{c}{a}=-24\) है। इसलिए दोनों ऋणात्मक हैं।

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

किस विकल्प में मूलों का योग धनात्मक और गुणनफल ऋणात्मक होगा?

In which option will the sum of roots be positive and the product of roots be negative?

Explanation opens after your attempt
Correct Answer

A. \(x^2-6x-16=0\)

Step 1

Concept

In the first option, the sum is \(-\frac{b}{a}=6\) and the product is \(\frac{c}{a}=-16\). So the sum is positive and the product is negative.

Step 2

Why this answer is correct

The correct answer is A. \(x^2-6x-16=0\). In the first option, the sum is \(-\frac{b}{a}=6\) and the product is \(\frac{c}{a}=-16\). So the sum is positive and the product is negative.

Step 3

Exam Tip

पहले विकल्प में योग \(-\frac{b}{a}=6\) और गुणनफल \(\frac{c}{a}=-16\) है। इसलिए योग धनात्मक और गुणनफल ऋणात्मक है।

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Question Hard Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30

किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?

In which equation is (x=0) one root and the other root negative?

Explanation opens after your attempt
Correct Answer

A. \(x^2+7x=0\)

Step 1

Concept

(x-2+7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.

Step 2

Why this answer is correct

The correct answer is A. \(x^2+7x=0\). (x-2+7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.

Step 3

Exam Tip

(x-2+7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।

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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 Hard Mathematics Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30

किस विकल्प में मूलों का योग ऋणात्मक और गुणनफल धनात्मक होगा?

In which option will the sum of roots be negative and the product of roots be positive?

Explanation opens after your attempt
Correct Answer

A. \(x^2+7x+12=0\)

Step 1

Concept

In the first option, the sum is \(-\frac{b}{a}=-7\) and the product is \(\frac{c}{a}=12\). So the sum is negative and the product is positive.

Step 2

Why this answer is correct

The correct answer is A. \(x^2+7x+12=0\). In the first option, the sum is \(-\frac{b}{a}=-7\) and the product is \(\frac{c}{a}=12\). So the sum is negative and the product is positive.

Step 3

Exam Tip

पहले विकल्प में योग \(-\frac{b}{a}=-7\) और गुणनफल \(\frac{c}{a}=12\) है। इसलिए योग ऋणात्मक और गुणनफल धनात्मक है।

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

किस समीकरण के मूलों का गुणनफल ऋणात्मक होगा?

Which equation will have a negative product of roots?

Explanation opens after your attempt
Correct Answer

A. \(2x^2+3x-5=0\)

Step 1

Concept

The product of roots is \(\frac{c}{a}\). In the first option, \(\frac{-5}{2}<0\), so the product is negative.

Step 2

Why this answer is correct

The correct answer is A. \(2x^2+3x-5=0\). The product of roots is \(\frac{c}{a}\). In the first option, \(\frac{-5}{2}<0\), so the product is negative.

Step 3

Exam Tip

मूलों का गुणनफल \(\frac{c}{a}\) है। पहले विकल्प में \(\frac{-5}{2}<0\), इसलिए गुणनफल ऋणात्मक है।

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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 Medium Mathematics Polynomials Geometrical meaning of the zeroes of a polynomial. Class 10 Level 22

यदि (p(2)) धनात्मक और (p(5)) ऋणात्मक है, तो ग्राफ के बारे में कौन सा कथन सबसे उचित है?

If (p(2)) is positive and (p(5)) is negative, which statement about the graph is most appropriate?

Explanation opens after your attempt
Correct Answer

A. ग्राफ (x=2) और (x=5) के बीच (x)-अक्ष को पार कर सकता हैThe graph may cross the (x)-axis between (x=2) and (x=5)

Step 1

Concept

A change of sign indicates a meeting with the (x)-axis in between. Tip: polynomial graphs are continuous.

Step 2

Why this answer is correct

The correct answer is A. ग्राफ (x=2) और (x=5) के बीच (x)-अक्ष को पार कर सकता है / The graph may cross the (x)-axis between (x=2) and (x=5). A change of sign indicates a meeting with the (x)-axis in between. Tip: polynomial graphs are continuous.

Step 3

Exam Tip

चिह्न बदलना बीच में (x)-अक्ष से मिलने का संकेत देता है। टिप: बहुपद के ग्राफ सतत होते हैं।

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Question Expert Science Unit 2: World of Living Life Processes Class 10 Level 23

किस स्थिति में पौधा रंध्र बंद करके भी स्वयं को बचाने की कोशिश करेगा और इसका भोजन निर्माण पर दुष्प्रभाव होगा?

In which situation will a plant try to protect itself by closing stomata and this will negatively affect food formation?

Explanation opens after your attempt
Correct Answer

A. अत्यधिक जल हानि की स्थिति मेंDuring excessive water loss

Step 1

Concept

Water vapour leaves when stomata remain open.

Step 2

Why this answer is correct

During high water loss the plant may close stomata.

Step 3

Exam Tip

This reduces carbon dioxide entry and photosynthesis. चरण 1: रंध्र खुले रहने पर जल वाष्प बाहर जाती है। चरण 2: जल हानि अधिक हो तो पौधा रंध्र बंद कर सकता है। चरण 3: इससे कार्बन डाइऑक्साइड प्रवेश कम होकर प्रकाश संश्लेषण घटता है।

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Question Medium Science Unit 2: World of Living Life Processes Class 10 Level 23

अंधेरे में रखे गए हरे पौधे में स्टार्च परीक्षण नकारात्मक आने का सबसे उचित कारण क्या है?

What is the most suitable reason for a negative starch test in a green plant kept in darkness?

Explanation opens after your attempt
Correct Answer

A. पौधे ने प्रकाश के बिना भोजन नहीं बनायाThe plant did not make food without light

Step 1

Concept

Light is necessary for photosynthesis.

Step 2

Why this answer is correct

In darkness food formation becomes very low or stops.

Step 3

Exam Tip

Therefore starch will not accumulate in the leaf. चरण 1: प्रकाश संश्लेषण के लिए प्रकाश जरूरी है। चरण 2: अंधेरे में भोजन निर्माण बहुत कम या बंद हो जाता है। चरण 3: इसलिए पत्ती में स्टार्च जमा नहीं होगा।

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Question Hard Science Unit 2: World of Living Control and Coordination Class 10 Level 18

जड़ में धनात्मक गुरुत्वानुवर्तन और ऋणात्मक प्रकाशानुवर्तन का संयुक्त अर्थ क्या है?

What is the combined meaning of positive geotropism and negative phototropism in roots?

Explanation opens after your attempt
Correct Answer

A. जड़ गुरुत्व की दिशा में और प्रकाश से दूर बढ़ती हैRoot grows towards gravity and away from light

Step 1

Concept

Positive geotropism means growth towards gravity.

Step 2

Why this answer is correct

Negative phototropism means growth away from light.

Step 3

Exam Tip

Roots usually grow downward into the soil. चरण 1: धनात्मक गुरुत्वानुवर्तन का अर्थ गुरुत्व की दिशा में वृद्धि है। चरण 2: ऋणात्मक प्रकाशानुवर्तन का अर्थ प्रकाश से दूर वृद्धि है। चरण 3: जड़ सामान्यतः नीचे मिट्टी की ओर बढ़ती है।

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Question Hard Science Unit 2: World of Living Control and Coordination Class 10 Level 17

किस परिस्थिति में नकारात्मक पुनर्भरण का सिद्धांत सबसे स्पष्ट होगा?

In which situation is the principle of negative feedback most clearly shown?

Explanation opens after your attempt
Correct Answer

A. रक्त शर्करा बढ़ने पर इंसुलिन बढ़ना और शर्करा सामान्य होने पर इंसुलिन घटनाInsulin rises when blood sugar increases and falls when sugar becomes normal

Step 1

Concept

In negative feedback the body reverses a change to maintain balance.

Step 2

Why this answer is correct

When blood sugar rises insulin helps reduce it.

Step 3

Exam Tip

When the level becomes normal insulin secretion decreases to maintain balance. चरण 1: नकारात्मक पुनर्भरण में शरीर बदलाव को उलटकर संतुलन लाता है। चरण 2: रक्त शर्करा बढ़ने पर इंसुलिन उसे कम करने में मदद करता है। चरण 3: सामान्य स्तर आने पर इंसुलिन स्राव घटता है जिससे संतुलन बना रहता है।

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Question Hard Science Unit 2: World of Living Control and Coordination Class 10 Level 17

किस स्थिति को ऋणात्मक प्रकाशानुवर्तन कहा जाएगा?

Which situation will be called negative phototropism?

Explanation opens after your attempt
Correct Answer

A. जड़ का प्रकाश से दूर बढ़नाRoot growing away from light

Step 1

Concept

Phototropism is a growth response to light.

Step 2

Why this answer is correct

Negative means growth away from the stimulus.

Step 3

Exam Tip

Growth of root away from light can be considered negative phototropism. चरण 1: प्रकाशानुवर्तन प्रकाश के प्रति वृद्धि प्रतिक्रिया है। चरण 2: ऋणात्मक का अर्थ उद्दीपन से दूर बढ़ना है। चरण 3: जड़ का प्रकाश से दूर बढ़ना ऋणात्मक प्रकाशानुवर्तन माना जा सकता है।

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Question Hard Science Unit 2: World of Living Control and Coordination Class 10 Level 17

नकारात्मक पुनर्भरण के आधार पर रक्त शर्करा सामान्य होने पर इंसुलिन स्राव में क्या परिवर्तन होना चाहिए?

According to negative feedback what should happen to insulin secretion when blood sugar becomes normal?

Explanation opens after your attempt
Correct Answer

A. स्राव घट जाना चाहिएSecretion should decrease

Step 1

Concept

Negative feedback maintains balance in the body.

Step 2

Why this answer is correct

When blood sugar becomes normal the need for extra insulin decreases.

Step 3

Exam Tip

Therefore insulin secretion should decrease. चरण 1: नकारात्मक पुनर्भरण शरीर में संतुलन बनाए रखता है। चरण 2: जब रक्त शर्करा सामान्य हो जाती है तो अधिक इंसुलिन की जरूरत घट जाती है। चरण 3: इसलिए इंसुलिन स्राव कम होना चाहिए।

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Question Hard Science Unit 2: World of Living Control and Coordination Class 10 Level 16

जड़ का धनात्मक गुरुत्वानुवर्तन और तने का ऋणात्मक गुरुत्वानुवर्तन पौधे के लिए साथ साथ कैसे लाभकारी हैं?

How are positive geotropism of root and negative geotropism of stem useful together for a plant?

Explanation opens after your attempt
Correct Answer

A. जड़ मिट्टी में जल खनिज लेती है और तना प्रकाश की ओर बढ़ता हैRoot gets water and minerals in soil and stem grows towards light

Step 1

Concept

Roots grow downward to get water and minerals.

Step 2

Why this answer is correct

Stem grows upward to get light.

Step 3

Exam Tip

Both directional growth responses are useful for plant life. चरण 1: जड़ें नीचे जाकर जल और खनिज प्राप्त करती हैं। चरण 2: तना ऊपर बढ़कर प्रकाश प्राप्त करता है। चरण 3: दोनों दिशात्मक वृद्धियां पौधे के जीवन के लिए उपयोगी हैं।

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Question Medium Science Unit 2: World of Living Control and Coordination Class 10 Level 17

किस स्थिति में ऋणात्मक गुरुत्वानुवर्तन दिखाई देगा?

In which situation will negative geotropism be seen?

Explanation opens after your attempt
Correct Answer

B. तने का ऊपर की ओर बढ़नाStem growing upward

Step 1

Concept

Negative tropism means growth away from the stimulus.

Step 2

Why this answer is correct

Gravity acts downward.

Step 3

Exam Tip

The stem grows upward so it shows negative geotropism. चरण 1: ऋणात्मक अनुवर्तन का अर्थ उद्दीपन के विपरीत दिशा में वृद्धि है। चरण 2: गुरुत्व नीचे की ओर कार्य करता है। चरण 3: तना ऊपर की ओर बढ़ता है इसलिए यह ऋणात्मक गुरुत्वानुवर्तन है।

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Question Medium Science Unit 2: World of Living Control and Coordination Class 10 Level 16

तना सामान्य रूप से ऋणात्मक गुरुत्वानुवर्तन क्यों दिखाता है?

Why does the stem generally show negative geotropism?

Explanation opens after your attempt
Correct Answer

A. क्योंकि वह गुरुत्व की विपरीत दिशा में ऊपर बढ़ता हैBecause it grows upward opposite to gravity

Step 1

Concept

Gravity acts downward.

Step 2

Why this answer is correct

The stem generally grows upward.

Step 3

Exam Tip

So the stem shows negative geotropism. चरण 1: गुरुत्व की दिशा नीचे की ओर होती है। चरण 2: तना सामान्य रूप से ऊपर की ओर बढ़ता है। चरण 3: इसलिए तना ऋणात्मक गुरुत्वानुवर्तन दिखाता है।

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Question Hard Social Science Chapter 2: Nationalism in India The First World War, Khilafat, and Non-Cooperation Class 10 Level 18

असहयोग आंदोलन को केवल नकारात्मक बहिष्कार कहना क्यों अधूरा होगा?

Why would it be incomplete to call the Non-Cooperation Movement only a negative boycott?

Explanation opens after your attempt
Correct Answer

A. क्योंकि इसमें स्वदेशी खादी और राष्ट्रीय शिक्षा जैसे रचनात्मक कार्य भी शामिल थेBecause it also included constructive work like swadeshi khadi and national education

Step 1

Concept

The movement boycotted colonial institutions.

Step 2

Why this answer is correct

It also promoted khadi swadeshi and national education.

Step 3

Exam Tip

Therefore it was a movement of both protest and construction. चरण 1: आंदोलन ने औपनिवेशिक संस्थाओं का बहिष्कार किया। चरण 2: साथ ही खादी स्वदेशी और राष्ट्रीय शिक्षा को बढ़ावा दिया। चरण 3: इसलिए यह विरोध और निर्माण दोनों का आंदोलन था।

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Question Expert Social Science Chapter 2: Nationalism in India The First World War, Khilafat, and Non-Cooperation Class 10 Level 18

क्यों कहा जा सकता है कि असहयोग आंदोलन केवल नकारात्मक बहिष्कार नहीं था?

Why can it be said that the Non-Cooperation Movement was not merely a negative boycott?

Explanation opens after your attempt
Correct Answer

A. क्योंकि इसमें स्वदेशी राष्ट्रीय शिक्षा और आत्मनिर्भरता जैसे रचनात्मक पक्ष भी थेBecause it also had constructive aspects like swadeshi national education and self-reliance

Step 1

Concept

The movement included boycott of foreign institutions.

Step 2

Why this answer is correct

At the same time khadi swadeshi and national education were promoted.

Step 3

Exam Tip

So it was both a programme of protest and construction. चरण 1: आंदोलन में विदेशी संस्थाओं का बहिष्कार था। चरण 2: साथ ही खादी स्वदेशी और राष्ट्रीय शिक्षा को बढ़ावा दिया गया। चरण 3: इसलिए यह विरोध और निर्माण दोनों का कार्यक्रम था।

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Question Expert Social Science Chapter 1: The Rise of Nationalism in Europe Nationalism and Imperialism Class 10 Level 2

बाल्कन के मामले में राष्ट्रवाद को सकारात्मक और नकारात्मक दोनों रूपों में क्यों देखा जा सकता है?

Why can nationalism be seen in both positive and negative forms in the case of the Balkans?

Explanation opens after your attempt
Correct Answer

A. इसने स्वतंत्रता की चेतना जगाई लेकिन आपसी संघर्ष भी बढ़ायाIt awakened freedom consciousness but also increased mutual conflict

Step 1

Concept

Nationalism inspires people toward freedom.

Step 2

Why this answer is correct

But when many claims clash, conflict increases.

Step 3

Exam Tip

The Balkans is a balanced example of this. चरण 1: राष्ट्रवाद लोगों को स्वतंत्रता के लिए प्रेरित करता है। चरण 2: लेकिन जब कई दावे टकराते हैं तो संघर्ष बढ़ता है। चरण 3: बाल्कन इसका संतुलित उदाहरण है।

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Question Medium Social Science Chapter 1: The Rise of Nationalism in Europe Nationalism and Imperialism Class 10 Level 3

निम्नलिखित में से कौन सा कथन राष्ट्रवाद के सकारात्मक और नकारात्मक दोनों रूपों को समझाता है?

Which statement explains both positive and negative forms of nationalism?

Explanation opens after your attempt
Correct Answer

B. राष्ट्रवाद एकता दे सकता है, पर संकीर्ण रूप में संघर्ष भी पैदा कर सकता हैNationalism can create unity, but in a narrow form it can also create conflict

Step 1

Concept

Nationalism can give people shared identity and unity.

Step 2

Why this answer is correct

But when it becomes narrow or aggressive, it can increase conflict.

Step 3

Exam Tip

Understand both sides in a balanced way for exams. चरण 1: राष्ट्रवाद लोगों को साझा पहचान और एकता दे सकता है। चरण 2: पर जब यह संकीर्ण या आक्रामक हो जाए तो संघर्ष बढ़ा सकता है। चरण 3: परीक्षा में दोनों पक्षों को संतुलित रूप से समझें।

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

समीकरण \(2x^2+4x+2=0\) के लिए विविक्तकर का चिन्ह क्या है?

What is the sign of the discriminant for \(2x^2+4x+2=0\)?

Explanation opens after your attempt
Correct Answer

A. शून्यzero

Step 1

Concept

(D=42-4(2)(2)=0). So this is the case of equal real roots.

Step 2

Why this answer is correct

The correct answer is A. शून्य / zero. (D=42-4(2)(2)=0). So this is the case of equal real roots.

Step 3

Exam Tip

(D=42-4(2)(2)=0) है। इसलिए यह समान वास्तविक मूलों की स्थिति है।

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

समीकरण \(3x^2-2x+4=0\) के लिए विविक्तकर का चिन्ह क्या है?

What is the sign of the discriminant for \(3x^2-2x+4=0\)?

Explanation opens after your attempt
Correct Answer

A. ऋणात्मकnegative

Step 1

Concept

(D=(-2)2-4(3)(4)=-44<0). Hence the discriminant is negative.

Step 2

Why this answer is correct

The correct answer is A. ऋणात्मक / negative. (D=(-2)2-4(3)(4)=-44<0). Hence the discriminant is negative.

Step 3

Exam Tip

(D=(-2)2-4(3)(4)=-44<0) है। अतः विविक्तकर ऋणात्मक है।

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

समीकरण \(2x^2-7x+3=0\) के मूलों की प्रकृति क्या है?

What is the nature of roots of \(2x^2-7x+3=0\)?

Explanation opens after your attempt
Correct Answer

A. दो वास्तविक और असमानtwo real and distinct

Step 1

Concept

(D=(-7)2-4(2)(3)=25>0). Hence the roots are real and distinct.

Step 2

Why this answer is correct

The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=(-7)2-4(2)(3)=25>0). Hence the roots are real and distinct.

Step 3

Exam Tip

(D=(-7)2-4(2)(3)=25>0) है। इसलिए मूल वास्तविक और असमान हैं।

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

समीकरण \(x^2+4x+p=0\) के वास्तविक मूल न होने के लिए कौन सी शर्त सही है?

For \(x^2+4x+p=0\) to have no real roots, which condition is correct?

Explanation opens after your attempt
Correct Answer

A. (p>4)

Step 1

Concept

For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.

Step 2

Why this answer is correct

The correct answer is A. (p>4). For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.

Step 3

Exam Tip

वास्तविक मूल न होने के लिए (D<0), इसलिए (16-4p<0) से (p>4)। ऋणात्मक विविक्तकर पर वास्तविक मूल नहीं होते।

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

यदि (D=-7) हो तो मूलों की प्रकृति क्या होगी?

If (D=-7), what will be the nature of roots?

Explanation opens after your attempt
Correct Answer

A. वास्तविक मूल नहींno real roots

Step 1

Concept

(-7<0), so there will be no real roots. Never treat negative (D) as equal roots.

Step 2

Why this answer is correct

The correct answer is A. वास्तविक मूल नहीं / no real roots. (-7<0), so there will be no real roots. Never treat negative (D) as equal roots.

Step 3

Exam Tip

(-7<0) है, इसलिए वास्तविक मूल नहीं होंगे। ऋणात्मक (D) को कभी समान मूल न मानें।

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

किस समीकरण के वास्तविक मूल नहीं होंगे?

Which equation will have no real roots?

Explanation opens after your attempt
Correct Answer

A. \(x^2+3x+5=0\)

Step 1

Concept

For the first equation, (D=32-4(1)(5)=-11<0). A negative discriminant gives no real roots.

Step 2

Why this answer is correct

The correct answer is A. \(x^2+3x+5=0\). For the first equation, (D=32-4(1)(5)=-11<0). A negative discriminant gives no real roots.

Step 3

Exam Tip

पहले समीकरण में (D=32-4(1)(5)=-11<0) है। ऋणात्मक विविक्तकर वास्तविक मूल नहीं देता।

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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 Easy Mathematics Quadratic Equations Nature of Roots Class 10 Level 37

समीकरण \(x^2-4x+4=0\) के मूलों की प्रकृति पहचानिए।

Identify the nature of roots of the equation \(x^2-4x+4=0\).

Explanation opens after your attempt
Correct Answer

A. दो वास्तविक और समानtwo real and equal

Step 1

Concept

Here (D=(-4)2-4(1)(4)=0). (D=0) means equal real roots.

Step 2

Why this answer is correct

The correct answer is A. दो वास्तविक और समान / two real and equal. Here (D=(-4)2-4(1)(4)=0). (D=0) means equal real roots.

Step 3

Exam Tip

यहां (D=(-4)2-4(1)(4)=0) है। (D=0) समान वास्तविक मूल बताता है।

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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

\(x^2-20x+96=0\) में कौनसी संख्या जोड़ी मध्य पद बनाने में मदद करेगी?

Which number pair helps in making the middle term in \(x^2-20x+96=0\)?

Explanation opens after your attempt
Correct Answer

A. (-8) और (-12)(-8) and (-12)

Step 1

Concept

(-8+(-12)=-20) and ((-8)(-12)=96), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

Step 2

Why this answer is correct

The correct answer is A. (-8) और (-12) / (-8) and (-12). (-8+(-12)=-20) and ((-8)(-12)=96), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

Step 3

Exam Tip

(-8+(-12)=-20) और ((-8)(-12)=96), इसलिए यह जोड़ी सही है। परीक्षा में (c) धनात्मक और (b) ऋणात्मक हो तो दोनों संख्याएं ऋणात्मक होती 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

द्विघात सूत्र से \(5x^2-10x-3=0\) के लिए (D) का मान क्या है?

Using the quadratic formula setup, what is the value of (D) for \(5x^2-10x-3=0\)?

Explanation opens after your attempt
Correct Answer

A. (160)

Step 1

Concept

Here (D=(-10)2-4(5)(-3)=160). In exams, a negative (c) makes the second term add.

Step 2

Why this answer is correct

The correct answer is A. (160). Here (D=(-10)2-4(5)(-3)=160). In exams, a negative (c) makes the second term add.

Step 3

Exam Tip

यहां (D=(-10)2-4(5)(-3)=160) है। परीक्षा में ऋणात्मक (c) के कारण दूसरा पद जुड़ता है।

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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

\(x^2-18x+80=0\) में कौनसी संख्या जोड़ी मध्य पद बनाने में मदद करेगी?

Which number pair helps in making the middle term in \(x^2-18x+80=0\)?

Explanation opens after your attempt
Correct Answer

A. (-8) और (-10)(-8) and (-10)

Step 1

Concept

(-8+(-10)=-18) and ((-8)(-10)=80), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

Step 2

Why this answer is correct

The correct answer is A. (-8) और (-10) / (-8) and (-10). (-8+(-10)=-18) and ((-8)(-10)=80), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

Step 3

Exam Tip

(-8+(-10)=-18) और ((-8)(-10)=80), इसलिए यह जोड़ी सही है। परीक्षा में (c) धनात्मक और (b) ऋणात्मक हो तो दोनों संख्याएं ऋणात्मक होती 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

द्विघात सूत्र से \(3x^2-6x-2=0\) के लिए (D) का मान क्या है?

Using the quadratic formula setup, what is the value of (D) for \(3x^2-6x-2=0\)?

Explanation opens after your attempt
Correct Answer

A. (60)

Step 1

Concept

Here (D=(-6)2-4(3)(-2)=60). In exams, a negative (c) makes the second term add.

Step 2

Why this answer is correct

The correct answer is A. (60). Here (D=(-6)2-4(3)(-2)=60). In exams, a negative (c) makes the second term add.

Step 3

Exam Tip

यहां (D=(-6)2-4(3)(-2)=60) है। परीक्षा में ऋणात्मक (c) के कारण दूसरा पद जुड़ता है।

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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

\(x^2-14x+45=0\) में कौनसी संख्या जोड़ी मध्य पद बनाने में मदद करेगी?

Which number pair helps in making the middle term in \(x^2-14x+45=0\)?

Explanation opens after your attempt
Correct Answer

A. (-5) और (-9)(-5) and (-9)

Step 1

Concept

(-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.

Step 2

Why this answer is correct

The correct answer is A. (-5) और (-9) / (-5) and (-9). (-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.

Step 3

Exam Tip

(-5+(-9)=-14) और ((-5)(-9)=45), इसलिए यह जोड़ी सही है। परीक्षा में (c) धनात्मक और (b) ऋणात्मक हो तो दोनों संख्याएं ऋणात्मक हो सकती 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 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(4x^2-8x-1=0\) का विविक्तकर (D) क्या है?

What is the discriminant (D) of \(4x^2-8x-1=0\)?

Explanation opens after your attempt
Correct Answer

A. (80)

Step 1

Concept

Here (D=(-8)2-4(4)(-1)=80). In exams, a negative (c) increases (D).

Step 2

Why this answer is correct

The correct answer is A. (80). Here (D=(-8)2-4(4)(-1)=80). In exams, a negative (c) increases (D).

Step 3

Exam Tip

यहां (D=(-8)2-4(4)(-1)=80) है। परीक्षा में ऋणात्मक (c) से (D) बढ़ जाता है।

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

द्विघात सूत्र से \(2x^2-4x-3=0\) के लिए (D) का मान क्या है?

Using the quadratic formula setup, what is the value of (D) for \(2x^2-4x-3=0\)?

Explanation opens after your attempt
Correct Answer

A. (40)

Step 1

Concept

Here (D=(-4)2-4(2)(-3)=40). In exams, a negative (c) makes the term add.

Step 2

Why this answer is correct

The correct answer is A. (40). Here (D=(-4)2-4(2)(-3)=40). In exams, a negative (c) makes the term add.

Step 3

Exam Tip

यहां (D=(-4)2-4(2)(-3)=40) है। परीक्षा में ऋणात्मक (c) के कारण जोड़ बनता है।

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

द्विघात सूत्र से \(x^2-4x-5=0\) के मूल क्या मिलेंगे?

Using the quadratic formula, what roots are obtained for \(x^2-4x-5=0\)?

Explanation opens after your attempt
Correct Answer

A. (x=5,-1)

Step 1

Concept

Here (D=(-4)2-4(1)(-5)=36), so \(x=\frac{4\pm6}{2}\). In exams, do not forget the negative sign of (c) while using the formula.

Step 2

Why this answer is correct

The correct answer is A. (x=5,-1). Here (D=(-4)2-4(1)(-5)=36), so \(x=\frac{4\pm6}{2}\). In exams, do not forget the negative sign of (c) while using the formula.

Step 3

Exam Tip

यहां (D=(-4)2-4(1)(-5)=36), इसलिए \(x=\frac{4\pm6}{2}\) मिलता है। परीक्षा में सूत्र लगाते समय (c) का ऋण चिन्ह न भूलें।

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

\(x^2+36=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?

What is the correct statement about the real roots of \(x^2+36=0\)?

Explanation opens after your attempt
Correct Answer

A. कोई वास्तविक मूल नहींNo real roots

Step 1

Concept

\(x^2=-36\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).

Step 2

Why this answer is correct

The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. \(x^2=-36\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).

Step 3

Exam Tip

\(x^2=-36\) वास्तविक संख्याओं में संभव नहीं है। परीक्षा में \(x^2\) का मान वास्तविक (x) के लिए ऋणात्मक नहीं होता।

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

\(x^2+6x-55=0\) के सही गुणनखंड कौनसे हैं?

What are the correct factors of \(x^2+6x-55=0\)?

Explanation opens after your attempt
Correct Answer

A. ((x+11)(x-5))

Step 1

Concept

(11+(-5)=6) and \(11\times(-5)=-55\), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.

Step 2

Why this answer is correct

The correct answer is A. ((x+11)(x-5)). (11+(-5)=6) and \(11\times(-5)=-55\), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.

Step 3

Exam Tip

(11+(-5)=6) और \(11\times(-5)=-55\), इसलिए ((x+11)(x-5)) सही है। परीक्षा में एक चिन्ह धनात्मक और एक ऋणात्मक रखें।

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

\(x^2-19x+90=0\) के गुणनखंड कौनसे होंगे?

What will be the factors of \(x^2-19x+90=0\)?

Explanation opens after your attempt
Correct Answer

A. ((x-9)(x-10))

Step 1

Concept

(9+10=19) and \(9\times10=90\), so ((x-9)(x-10)) is correct. In exams, take both negative signs for a negative middle term.

Step 2

Why this answer is correct

The correct answer is A. ((x-9)(x-10)). (9+10=19) and \(9\times10=90\), so ((x-9)(x-10)) is correct. In exams, take both negative signs for a negative middle term.

Step 3

Exam Tip

(9+10=19) और \(9\times10=90\), इसलिए ((x-9)(x-10)) सही है। परीक्षा में ऋणात्मक मध्य पद के लिए दोनों ऋणात्मक चिन्ह लें।

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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 36

यदि किसी द्विघात समीकरण में (D=-9) हो, तो कौनसा निष्कर्ष सही है?

If a quadratic equation has (D=-9), which conclusion is correct?

Explanation opens after your attempt
Correct Answer

A. वास्तविक मूल नहीं होंगेThere will be no real roots

Step 1

Concept

When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.

Step 2

Why this answer is correct

The correct answer is A. वास्तविक मूल नहीं होंगे / There will be no real roots. When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.

Step 3

Exam Tip

(D<0) होने पर वास्तविक वर्गमूल नहीं मिलता। परीक्षा में ऋणात्मक विविक्तकर का अर्थ याद रखें।

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

\(x^2+25=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?

What is the correct statement about the real roots of \(x^2+25=0\)?

Explanation opens after your attempt
Correct Answer

A. कोई वास्तविक मूल नहींNo real roots

Step 1

Concept

\(x^2=-25\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).

Step 2

Why this answer is correct

The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. \(x^2=-25\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).

Step 3

Exam Tip

\(x^2=-25\) वास्तविक संख्याओं में संभव नहीं है। परीक्षा में \(x^2\) का मान वास्तविक (x) के लिए ऋणात्मक नहीं होता।

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

\(x^2+5x-14=0\) के सही गुणनखंड कौनसे हैं?

What are the correct factors of \(x^2+5x-14=0\)?

Explanation opens after your attempt
Correct Answer

A. ((x+7)(x-2))

Step 1

Concept

(7+(-2)=5) and \(7\times(-2)=-14\), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.

Step 2

Why this answer is correct

The correct answer is A. ((x+7)(x-2)). (7+(-2)=5) and \(7\times(-2)=-14\), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.

Step 3

Exam Tip

(7+(-2)=5) और \(7\times(-2)=-14\), इसलिए ((x+7)(x-2)) सही है। परीक्षा में एक चिन्ह धनात्मक और एक ऋणात्मक रखें।

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

\(x^2-13x+40=0\) के गुणनखंड कौनसे होंगे?

What will be the factors of \(x^2-13x+40=0\)?

Explanation opens after your attempt
Correct Answer

A. ((x-5)(x-8))

Step 1

Concept

(5+8=13) and \(5\times8=40\), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.

Step 2

Why this answer is correct

The correct answer is A. ((x-5)(x-8)). (5+8=13) and \(5\times8=40\), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.

Step 3

Exam Tip

(5+8=13) और \(5\times8=40\), इसलिए ((x-5)(x-8)) सही है। परीक्षा में ऋणात्मक मध्य पद के लिए दोनों ऋणात्मक चिन्ह लें।

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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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