Question 118

If $$x=3+2\sqrt{2}$$, the value of $$x^{2}+\frac{1}{x^{2}}$$ is

Solution

$$x=3+2\sqrt{2}$$

then,

$$\frac{1}{x}=\frac{1}{3+2\sqrt{2}}$$ = $$\frac{1}{3+2\sqrt{2}}$$ X $$\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$$ = $$3-2\sqrt{2}$$

= $$3+2\sqrt{2}$$ + $$3-2\sqrt{2}$$ = 6

squaring on both sides

$$(x+\frac{1}{x})^{2}$$ = 36

$$x^{2}+\frac{1}{x^{2}}$$ + 2 = 36

$$x^{2}+\frac{1}{x^{2}}$$ = 36-2 = 34

so the answer is option D.

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