Question 52

If $$\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{6}, then x^2 + \frac{1}{x^2}$$ is equal to:

Solution

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

Squaring both the sides , we get,

$$x + \frac{1}{x} + 2 =6$$

x +$$\frac{1}{x}$$ = 4

Squaring both the sides , we get ,

$$x^2 + \frac{1}{x^2} + 2 =16$$

$$x^2 + \frac{1}{x^2}$$=14

So, the answer would be option b)14.

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