यदि \(a=\sqrt{2}+\sqrt{8}\), तो (a) किस प्रकार की संख्या है?
If \(a=\sqrt{2}+\sqrt{8}\), what type of number is (a)?
Explanation opens after your attempt
Correct Answer
B. अपरिमेयIrrational
Concept
\(\sqrt{8}=2\sqrt{2}\).
Why this answer is correct
So \(a=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\), and \(\sqrt{2}\) is irrational.
Exam Tip
Simplify like radical terms before deciding the type of number. चरण 1: \(\sqrt{8}=2\sqrt{2}\) होता है। चरण 2: इसलिए \(a=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\), और \(\sqrt{2}\) अपरिमेय है। चरण 3: समान मूल वाले पदों को पहले सरल करें।
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