Question 129

If x = 2 + √3 , then the value, $$\sqrt{x} + \frac{1}{\sqrt{x}}$$

Solution

$$x=2+\sqrt{3}$$

$$\frac{1}{x}=2-\sqrt{3}$$

$$(\sqrt{x} + \frac{1}{\sqrt{x}})^{2}$$ = $$x+\frac{1}{x}$$ + 2

$$(\sqrt{x} + \frac{1}{\sqrt{x}})^{2}$$ = 4 + 2 = 6

$$\sqrt{x} + \frac{1}{\sqrt{x}}$$ = $$\sqrt{6}$$

so the answer is option B.

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