यदि \(x^2+ax+a=0\) का एक मूल (1) है तो दूसरा मूल क्या होगा?

If one root of \(x^2+ax+a=0\) is (1), what will be the other root?

Explanation opens after your attempt
Correct Answer

A. \(-\frac{1}{2}\)

Step 1

Concept

Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).

Step 2

Why this answer is correct

The correct answer is A. \(-\frac{1}{2}\). Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).

Step 3

Exam Tip

(x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा।

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(x^2+ax+a=0\) का एक मूल (1) है तो दूसरा मूल क्या होगा? / If one root of \(x^2+ax+a=0\) is (1), what will be the other root?

Correct Answer: A. \(-\frac{1}{2}\). Explanation: (x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा। / Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).

Which concept should I revise for this Mathematics MCQ?

Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).

What exam hint can help solve this Mathematics question?

(x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा।