यदि \(z=\sqrt{31}-\sqrt{22}\) है तो \(z^2\) का मान कौन-सा है?

If \(z=\sqrt{31}-\sqrt{22}\), which is the value of \(z^2\)?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

D. \(53-2\sqrt{682}\)

Step 1

Concept

(\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}). Keep the middle term negative.

Step 2

Why this answer is correct

The correct answer is D. \(53-2\sqrt{682}\). (\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}). Keep the middle term negative.

Step 3

Exam Tip

(\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}) है। मध्य पद का चिन्ह ऋणात्मक रखें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(z=\sqrt{31}-\sqrt{22}\) है तो \(z^2\) का मान कौन-सा है? / If \(z=\sqrt{31}-\sqrt{22}\), which is the value of \(z^2\)?

Correct Answer: D. \(53-2\sqrt{682}\). Explanation: (\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}) है। मध्य पद का चिन्ह ऋणात्मक रखें। / (\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}). Keep the middle term negative.

Which concept should I revise for this Mathematics MCQ?

(\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}). Keep the middle term negative.

What exam hint can help solve this Mathematics question?

(\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}) है। मध्य पद का चिन्ह ऋणात्मक रखें।