Expert Mathematics Quadratic Equations Class 10 Level 30

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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Mathematics Answer, Explanation and Revision Hints

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?

Correct Answer: A. \(k\neq \frac{19}{9}\). Explanation: (x=3) रखने पर बायां पक्ष (9k-24+5=9k-19) होता है। मूल न होने के लिए \(9k-19\neq0\), इसलिए \(k\neq\frac{19}{9}\)। / Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).

Which concept should I revise for this Mathematics MCQ?

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

What exam hint can help solve this Mathematics question?

(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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