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Which option gives the correct comparison between (\sqrt{3}+\sqrt{6}) and (\sqrt{12})?
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Answer and explanation
Correct answer: (\sqrt{3}+\sqrt{6}>\sqrt{12})
Step 1: All terms are positive and (\sqrt{6}>0). Step 2: Since (\sqrt{12}=2\sqrt{3}) and (\sqrt{6}>\sqrt{3}), the sum (\sqrt{3}+\sqrt{6}) is greater than (2\sqrt{3}). Step 3: For comparison, convert what you can and use positivity.
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What is the correct answer to this question?
(\sqrt{3}+\sqrt{6}>\sqrt{12})
Why is this the correct answer?
Step 1: All terms are positive and (\sqrt{6}>0). Step 2: Since (\sqrt{12}=2\sqrt{3}) and (\sqrt{6}>\sqrt{3}), the sum (\sqrt{3}+\sqrt{6}) is greater than (2\sqrt{3}). Step 3: For comparison, convert what you can and use positivity.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Irrational numbers.
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