Question 63

If $$\sqrt x - \frac{1}{\sqrt x} = 4, then x^2 + \frac{1}{x^2}$$ is equal to:

Solution

$$\sqrt x - \frac{1}{\sqrt x} = 4$$

Squaring both side,

$$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=4^2$$

$$x+\frac{1}{x}=16+2=18$$

Again squaring both side we get,

$$\left(x+\frac{1}{x}\right)^2=18^2$$

$$x^2+\frac{1}{x^2}+2=324$$

$$x^2+\frac{1}{x^2}=322$$

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