Concept-wise Practice

not-root MCQ Questions for Class 10

not-root se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

7 questions tagged with not-root.

यदि (x=3) समीकरण \(kx^2-8x+5=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?

If (x=3) is not a root of \(kx^2-8x+5=0\), what condition must (k) satisfy?

Explanation opens after your attempt
Correct Answer

A. \(k\neq \frac{19}{9}\)

Step 1

Concept

Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).

Step 2

Why this answer is correct

The correct answer is A. \(k\neq \frac{19}{9}\). Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).

Step 3

Exam Tip

(x=3) रखने पर बायां पक्ष (9k-24+5=9k-19) होता है। मूल न होने के लिए \(9k-19\neq0\), इसलिए \(k\neq\frac{19}{9}\)।

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यदि (x=-2) समीकरण \(kx^2+5x+6=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?

If (x=-2) is not a root of \(kx^2+5x+6=0\), what condition must (k) satisfy?

Explanation opens after your attempt
Correct Answer

A. \(k\neq1\)

Step 1

Concept

Putting (x=-2), the left side becomes (4k-10+6=4k-4). For it not to be a root, \(4k-4\neq0\), so \(k\neq1\).

Step 2

Why this answer is correct

The correct answer is A. \(k\neq1\). Putting (x=-2), the left side becomes (4k-10+6=4k-4). For it not to be a root, \(4k-4\neq0\), so \(k\neq1\).

Step 3

Exam Tip

(x=-2) रखने पर बायां पक्ष (4k-10+6=4k-4) होता है। मूल न होने के लिए \(4k-4\neq0\), इसलिए \(k\neq1\)।

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यदि (x=2) समीकरण \(kx^2-7x+3=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?

If (x=2) is not a root of \(kx^2-7x+3=0\), what condition must (k) satisfy?

Explanation opens after your attempt
Correct Answer

A. \(k\neq \frac{11}{4}\)

Step 1

Concept

Putting (x=2), the left side becomes (4k-14+3=4k-11). For it not to be a root, \(4k-11\neq0\), so \(k\neq\frac{11}{4}\).

Step 2

Why this answer is correct

The correct answer is A. \(k\neq \frac{11}{4}\). Putting (x=2), the left side becomes (4k-14+3=4k-11). For it not to be a root, \(4k-11\neq0\), so \(k\neq\frac{11}{4}\).

Step 3

Exam Tip

(x=2) रखने पर बायां पक्ष (4k-14+3=4k-11) होता है। मूल न होने के लिए \(4k-11\neq0\), इसलिए \(k\neq\frac{11}{4}\)।

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यदि (x=-1) समीकरण \(kx^2+3x+2=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?

If (x=-1) is not a root of \(kx^2+3x+2=0\), what condition must (k) satisfy?

Explanation opens after your attempt
Correct Answer

A. \(k\neq1\)

Step 1

Concept

Putting (x=-1), the left side becomes (k-3+2=k-1). For it not to be a root, \(k-1\neq0\), so \(k\neq1\).

Step 2

Why this answer is correct

The correct answer is A. \(k\neq1\). Putting (x=-1), the left side becomes (k-3+2=k-1). For it not to be a root, \(k-1\neq0\), so \(k\neq1\).

Step 3

Exam Tip

(x=-1) रखने पर बायां पक्ष (k-3+2=k-1) होता है। मूल न होने के लिए \(k-1\neq0\), इसलिए \(k\neq1\)।

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यदि (x=1) समीकरण \(kx^2-5x+2=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?

If (x=1) is not a root of \(kx^2-5x+2=0\), what condition must (k) satisfy?

Explanation opens after your attempt
Correct Answer

A. \(k\neq3\)

Step 1

Concept

Putting (x=1), the left side becomes (k-5+2=k-3). For it not to be a root, \(k-3\neq0\), so \(k\neq3\).

Step 2

Why this answer is correct

The correct answer is A. \(k\neq3\). Putting (x=1), the left side becomes (k-5+2=k-3). For it not to be a root, \(k-3\neq0\), so \(k\neq3\).

Step 3

Exam Tip

(x=1) रखने पर बायां पक्ष (k-5+2=k-3) होता है। मूल न होने के लिए \(k-3\neq0\), इसलिए \(k\neq3\)।

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किस समीकरण में (x=-2) मूल नहीं है?

In which equation is (x=-2) not a root?

Explanation opens after your attempt
Correct Answer

D. \(x^2-2x+4=0\)

Step 1

Concept

Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.

Step 2

Why this answer is correct

The correct answer is D. \(x^2-2x+4=0\). Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.

Step 3

Exam Tip

(x=-2) रखने पर \(4+4+4=12\neq0\) मिलता है। मूल न होने की जांच भी प्रतिस्थापन से करें।

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किस समीकरण में (x=1) मूल नहीं है?

In which equation is (x=1) not a root?

Explanation opens after your attempt
Correct Answer

D. \(x^2+x+1=0\)

Step 1

Concept

Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.

Step 2

Why this answer is correct

The correct answer is D. \(x^2+x+1=0\). Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.

Step 3

Exam Tip

(x=1) रखने पर \(1+1+1=3\neq 0\) मिलता है। किसी विकल्प में मूल न होने की जांच भी प्रतिस्थापन से करें।

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