Question 71

If $$\sqrt x - \frac{1}{\sqrt x} = \sqrt6$$, then $$x^2 + \frac{1}{x^2}$$ is equal to:

Solution

$$\sqrt x - \frac{1}{\sqrt x} = \sqrt6$$

Squaring both side,

$$\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2-2=\left(\sqrt{6}\right)^2$$

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

$$x+\frac{1}{x}=8$$

Again squaring both side,

$$\left(x+\frac{1}{x}\right)^2=8^2$$

$$\left(x^2+\frac{1}{x^2}+2=64\right)$$

$$\left(x^2+\frac{1}{x^2}=62\right)$$

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