Question

If $x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$ then $x^2$ = ___

• A. $161+72\sqrt{5}$
• B. $9-4\sqrt{2}$
• C. 161
• D. $161-72\sqrt{5}$

Answer 1

The correct option is D

$161-72\sqrt{5}$

$x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$

$=\frac{\sqrt{5}-2}{\sqrt{5}+2}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}$

$=\frac{(\sqrt{5}-2{)}^2}{(\sqrt{5}{)}^2-(2{)}^2}$ (The numerator is of the form $(a-b{)}^2$ and the denominator is of the form $a^2-b^2$)

$=\frac{(\sqrt{5}{)}^2-2(2\sqrt{5})+(2{)}^2}{5-4}$

$=\frac{5-4\sqrt{5}+4}{1}$

$=9-4\sqrt{5}$

$X^2=(9-4\sqrt{5}{)}^2$

$=(9{)}^2-2(9\left)\right(4\sqrt{5})+(4\sqrt{5}{)}^2$

$=81-72\sqrt{5}+80$

$=161-72\sqrt{5}$