Hard Mathematics Polynomials Class 10 Level 25

यदि \(x=\sqrt{7}+\sqrt{3}\) है तो \(x^2-10\) का मान क्या है?

If \(x=\sqrt{7}+\sqrt{3}\), what is the value of \(x^2-10\)?

Explanation opens after your attempt
Correct Answer

A. \(2\sqrt{21}\)

Step 1

Concept

\(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\). Therefore \(x^2-10=2\sqrt{21}\).

Step 2

Why this answer is correct

The correct answer is A. \(2\sqrt{21}\). \(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\). Therefore \(x^2-10=2\sqrt{21}\).

Step 3

Exam Tip

\(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\) है। इसलिए \(x^2-10=2\sqrt{21}\) होगा।

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(x=\sqrt{7}+\sqrt{3}\) है तो \(x^2-10\) का मान क्या है? / If \(x=\sqrt{7}+\sqrt{3}\), what is the value of \(x^2-10\)?

Correct Answer: A. \(2\sqrt{21}\). Explanation: \(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\) है। इसलिए \(x^2-10=2\sqrt{21}\) होगा। / \(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\). Therefore \(x^2-10=2\sqrt{21}\).

Which concept should I revise for this Mathematics MCQ?

\(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\). Therefore \(x^2-10=2\sqrt{21}\).

What exam hint can help solve this Mathematics question?

\(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\) है। इसलिए \(x^2-10=2\sqrt{21}\) होगा।

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