यदि \(\sqrt{12}+\sqrt{27}\) को पहले सरल किया जाए, तो इसका वर्ग क्या होगा?
If \(\sqrt{12}+\sqrt{27}\) is simplified first, what will its square be?
Explanation opens after your attempt
Correct Answer
A. (75)
Concept
\(\sqrt{12}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\), so the square is (75). In exams add like radicals first.
Why this answer is correct
The correct answer is A. (75). \(\sqrt{12}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\), so the square is (75). In exams add like radicals first.
Exam Tip
\(\sqrt{12}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\), इसलिए वर्ग (75) है। परीक्षा में समान मूलों को पहले जोड़ें।
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