Question 66
If $$x - \frac{1}{x} = 13,$$ then the value of $$x^2 + \frac{1}{x^2}$$ is:
Solution
Given, $$x-\frac{1}{x}=13$$
$$\Rightarrow$$ $$\left(x-\frac{1}{x}\right)^2=13^2$$
$$\Rightarrow$$ $$x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=169$$
$$\Rightarrow$$ $$x^2+\frac{1}{x^2}-2=169$$
$$\Rightarrow$$ $$x^2+\frac{1}{x^2}=171$$
Hence, the correct answer is Option B
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