Question 69

If $$\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$$, then the value of $$x^2 + \frac{1}{x^2}$$ is:

Solution

$$\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$$

$$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=\left(\sqrt{7}\right)^2$$

$$x+\frac{1}{x}-2=7$$

$$x+\frac{1}{x}=9$$

$$\left(x+\frac{1}{x}\right)^2=9^2$$

$$x^2+\frac{1}{x^2}+2=81$$

$$x^2+\frac{1}{x^2}=79$$

Hence, the correct answer is Option C

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