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2 results found for "surd algebra" in Class 10.

Question Expert Mathematics Chapter 1: Real Numbers 5: Irrational numbers Class 10 Level 14

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

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

Explanation opens after your attempt
Correct Answer

A. \(7+4\sqrt{7}\)

Step 1

Concept

((x-2)=\sqrt{7}) and ((x+2)=\sqrt{7}+4).

Step 2

Why this answer is correct

The product is (\sqrt{7}\(\sqrt{7}+4\)=7+4\sqrt{7}).

Step 3

Exam Tip

Before applying an identity directly, substitute the given value of (x) carefully. चरण 1: ((x-2)=\sqrt{7}) और ((x+2)=\sqrt{7}+4)। चरण 2: गुणन (\sqrt{7}\(\sqrt{7}+4\)=7+4\sqrt{7}) है। चरण 3: सीधे सूत्र लगाने से पहले (x) का दिया हुआ मान ध्यान से रखें।

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Question Expert Mathematics Chapter 1: Real Numbers 5: Irrational numbers Class 10 Level 13

यदि \(a=1+\sqrt{5}\), तो \(a^2-2a\) का मान क्या है?

If \(a=1+\sqrt{5}\), what is the value of \(a^2-2a\)?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

(a-2-2a=a(a-2)).

Step 2

Why this answer is correct

\(a-2=\sqrt{5}-1\), so (a(a-2)=\(1+\sqrt{5}\)\(\sqrt{5}-1\)=4).

Step 3

Exam Tip

Recognizing the hidden conjugate form is a quick method. चरण 1: (a-2-2a=a(a-2)) है। चरण 2: \(a-2=\sqrt{5}-1\), इसलिए (a(a-2)=\(1+\sqrt{5}\)\(\sqrt{5}-1\)=4)। चरण 3: छिपे हुए संयुग्मी रूप को पहचानना तेज तरीका है।

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