यदि (x=4) समीकरण \(ax^2-10x+8=0\) का मूल है तो (a) का मान क्या है?

If (x=4) is a root of \(ax^2-10x+8=0\), what is the value of (a)?

Explanation opens after your attempt
Correct Answer

A. (2)

Step 1

Concept

Putting (x=4) gives (16a-40+8=0), so (16a=32) and (a=2). Substitute when a coefficient is unknown.

Step 2

Why this answer is correct

The correct answer is A. (2). Putting (x=4) gives (16a-40+8=0), so (16a=32) and (a=2). Substitute when a coefficient is unknown.

Step 3

Exam Tip

(x=4) रखने पर (16a-40+8=0) इसलिए (16a=32) और (a=2)। अज्ञात गुणांक में प्रतिस्थापन करें।

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Mathematics Answer, Explanation and Revision Hints

यदि (x=4) समीकरण \(ax^2-10x+8=0\) का मूल है तो (a) का मान क्या है? / If (x=4) is a root of \(ax^2-10x+8=0\), what is the value of (a)?

Correct Answer: A. (2). Explanation: (x=4) रखने पर (16a-40+8=0) इसलिए (16a=32) और (a=2)। अज्ञात गुणांक में प्रतिस्थापन करें। / Putting (x=4) gives (16a-40+8=0), so (16a=32) and (a=2). Substitute when a coefficient is unknown.

Which concept should I revise for this Mathematics MCQ?

Putting (x=4) gives (16a-40+8=0), so (16a=32) and (a=2). Substitute when a coefficient is unknown.

What exam hint can help solve this Mathematics question?

(x=4) रखने पर (16a-40+8=0) इसलिए (16a=32) और (a=2)। अज्ञात गुणांक में प्रतिस्थापन करें।