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# If x = 3 - $$\sqrt{3}$$, find x2 + $$\frac{36}{x^2}$$

### Question

If x = 3 - $$\sqrt{3}$$, find x2 + $$\frac{36}{x^2}$$

A) 9

B) 18

C) 24

D) 27

### Explanation:

x = 3 - $$\sqrt{3}$$
x2 = (3 - $$\sqrt{3}$$)2
= 9 + 3 - 6$$\sqrt{34}$$
= 12 - 6$$\sqrt{3}$$
= 6(2 - $$\sqrt{3}$$)
∴ x2 + $$\frac{36}{x^2}$$ = 6(2 - $$\sqrt{3}$$) + $$\frac{36}{6(2 - \sqrt{3})}$$
6(2 - $$\sqrt{3}$$) + $$\frac{6}{2 - \sqrt{3}}$$ = 6(- $$\sqrt{3}$$) + $$\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}$$
= 6(2 - $$\sqrt{3}$$) + $$\frac{6(2 + \sqrt{3})}{4 - 3}$$
6(2 - $$\sqrt{3}$$) + 6(2 + $$\sqrt{3}$$) = 12 + 12
= 24

## Dicussion (1)

• x = 3 - $$\sqrt{3}$$
x2 = (3 - $$\sqrt{3}$$)2
= 9 + 3 - 6$$\sqrt{34}$$
= 12 - 6$$\sqrt{3}$$
= 6(2 - $$\sqrt{3}$$)
∴ x2 + $$\frac{36}{x^2}$$ = 6(2 - $$\sqrt{3}$$) + $$\frac{36}{6(2 - \sqrt{3})}$$
6(2 - $$\sqrt{3}$$) + $$\frac{6}{2 - \sqrt{3}}$$ = 6(- $$\sqrt{3}$$) + $$\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}$$
= 6(2 - $$\sqrt{3}$$) + $$\frac{6(2 + \sqrt{3})}{4 - 3}$$
6(2 - $$\sqrt{3}$$) + 6(2 + $$\sqrt{3}$$) = 12 + 12
= 24