Expert Mathematics Quadratic Equations Class 10 Level 28

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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Mathematics Answer, Explanation and Revision Hints

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?

Correct Answer: A. \(k\neq \frac{11}{4}\). Explanation: (x=2) रखने पर बायां पक्ष (4k-14+3=4k-11) होता है। मूल न होने के लिए \(4k-11\neq0\), इसलिए \(k\neq\frac{11}{4}\)। / Putting (x=2), the left side becomes (4k-14+3=4k-11). For it not to be a root, \(4k-11\neq0\), so \(k\neq\frac{11}{4}\).

Which concept should I revise for this Mathematics MCQ?

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

What exam hint can help solve this Mathematics question?

(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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