Question 159

If $$\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b} - \sqrt{a-2b}}=\frac{\sqrt{3}}{1}$$, find the value of $$\frac{a}{b}$$

Solution

here in this question $$\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b-}\sqrt{a-2b}}=\frac{\sqrt{3}}{1}$$

using componendo and dividendo, we will get

$$\frac{\sqrt(a+2b)}{\sqrt(a-2b)} = \frac{\sqrt3 + 1 }{\sqrt3 - 1}$$

now on squaring both side and solving, we will get

16 b = 4a$$\surd3$$

$$\frac{a}{b}$$ = $$\frac{4}{\surd3}$$

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