यदि \(x-\frac{1}{x}=4\) हो तो \(x^2+\frac{1}{x^2}\) का मान क्या होगा?

If \(x-\frac{1}{x}=4\), what is the value of \(x^2+\frac{1}{x^2}\)?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. (18)

Step 1

Concept

Squaring gives \(x^2-2+\frac{1}{x^2}=16\), so \(x^2+\frac{1}{x^2}=18\). Exam tip: add (2) in the subtraction identity.

Step 2

Why this answer is correct

The correct answer is B. (18). Squaring gives \(x^2-2+\frac{1}{x^2}=16\), so \(x^2+\frac{1}{x^2}=18\). Exam tip: add (2) in the subtraction identity.

Step 3

Exam Tip

वर्ग करने पर \(x^2-2+\frac{1}{x^2}=16\) है इसलिए \(x^2+\frac{1}{x^2}=18\)। परीक्षा में घटाव पहचान में (2) जोड़ें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(x-\frac{1}{x}=4\) हो तो \(x^2+\frac{1}{x^2}\) का मान क्या होगा? / If \(x-\frac{1}{x}=4\), what is the value of \(x^2+\frac{1}{x^2}\)?

Correct Answer: B. (18). Explanation: वर्ग करने पर \(x^2-2+\frac{1}{x^2}=16\) है इसलिए \(x^2+\frac{1}{x^2}=18\)। परीक्षा में घटाव पहचान में (2) जोड़ें। / Squaring gives \(x^2-2+\frac{1}{x^2}=16\), so \(x^2+\frac{1}{x^2}=18\). Exam tip: add (2) in the subtraction identity.

Which concept should I revise for this Mathematics MCQ?

Squaring gives \(x^2-2+\frac{1}{x^2}=16\), so \(x^2+\frac{1}{x^2}=18\). Exam tip: add (2) in the subtraction identity.

What exam hint can help solve this Mathematics question?

वर्ग करने पर \(x^2-2+\frac{1}{x^2}=16\) है इसलिए \(x^2+\frac{1}{x^2}=18\)। परीक्षा में घटाव पहचान में (2) जोड़ें।