\(\sqrt{32}+\sqrt{128}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt{32}+\sqrt{128}\)?

Explanation opens after your attempt
Correct Answer

A. \(12\sqrt{2}\)

Step 1

Concept

\(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\).

Step 2

Why this answer is correct

\(4\sqrt{2}+8\sqrt{2}=12\sqrt{2}\).

Step 3

Exam Tip

Radicals can be added only when they become like radicals. चरण 1: \(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\)। चरण 2: \(4\sqrt{2}+8\sqrt{2}=12\sqrt{2}\)। चरण 3: समान वर्गमूल बनने पर ही उन्हें जोड़ा जा सकता है।

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Mathematics Answer, Explanation and Revision Hints

\(\sqrt{32}+\sqrt{128}\) का सरल रूप क्या है? / What is the simplified form of \(\sqrt{32}+\sqrt{128}\)?

Correct Answer: A. \(12\sqrt{2}\). Explanation: चरण 1: \(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\)। चरण 2: \(4\sqrt{2}+8\sqrt{2}=12\sqrt{2}\)। चरण 3: समान वर्गमूल बनने पर ही उन्हें जोड़ा जा सकता है। / Step 1: \(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\). Step 2: \(4\sqrt{2}+8\sqrt{2}=12\sqrt{2}\). Step 3: Radicals can be added only when they become like radicals.

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\).

What exam hint can help solve this Mathematics question?

Radicals can be added only when they become like radicals. चरण 1: \(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\)। चरण 2: \(4\sqrt{2}+8\sqrt{2}=12\sqrt{2}\)। चरण 3: समान वर्गमूल बनने पर ही उन्हें जोड़ा जा सकता है।