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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