Expert Mathematics Polynomials Class 10 Level 26

किस विकल्प में \(\sqrt{3}\) और \(\sqrt{12}\) का योग परिमेय गुणांक वाले सरल अपरिमेय रूप में सही लिखा गया है?

In which option is the sum of \(\sqrt{3}\) and \(\sqrt{12}\) correctly written as a simple irrational form with rational coefficient?

Explanation opens after your attempt
Correct Answer

A. \(3\sqrt{3}\)

Step 1

Concept

\(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\). In exams make radicals like terms before adding.

Step 2

Why this answer is correct

The correct answer is A. \(3\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\). In exams make radicals like terms before adding.

Step 3

Exam Tip

\(\sqrt{12}=2\sqrt{3}\), इसलिए \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\) है। परीक्षा में मूलों को जोड़ने से पहले समान मूल बनाएं।

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Mathematics Answer, Explanation and Revision Hints

किस विकल्प में \(\sqrt{3}\) और \(\sqrt{12}\) का योग परिमेय गुणांक वाले सरल अपरिमेय रूप में सही लिखा गया है? / In which option is the sum of \(\sqrt{3}\) and \(\sqrt{12}\) correctly written as a simple irrational form with rational coefficient?

Correct Answer: A. \(3\sqrt{3}\). Explanation: \(\sqrt{12}=2\sqrt{3}\), इसलिए \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\) है। परीक्षा में मूलों को जोड़ने से पहले समान मूल बनाएं। / \(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\). In exams make radicals like terms before adding.

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\). In exams make radicals like terms before adding.

What exam hint can help solve this Mathematics question?

\(\sqrt{12}=2\sqrt{3}\), इसलिए \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\) है। परीक्षा में मूलों को जोड़ने से पहले समान मूल बनाएं।

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