Question 66

If $$\sqrt x - \frac{1}{\sqrt x} = 3\sqrt2, then x^2 + \frac{1}{x^2}$$ is equal to:

Solution

squaring both sides,

$$x+\frac{1}{x}-2=18$$ $$[(a-b)^{2}=a^{2}+b^{2}-2ab]$$

$$\Rightarrow$$ $$x+\frac{1}{x}=20$$

squaring again both sides,
$$\Rightarrow$$ $$x^{2}+\frac{1}{x^{2}}+2=20^{2}$$

$$\Rightarrow$$ $$x^{2}+\frac{1}{x^{2}}=400-2$$ = 398

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