Question

$\text{If}a=\frac{2-\sqrt{5}}{2+\sqrt{5}}$ and $b=\frac{2+\sqrt{5}}{2-\sqrt{5}}$, then $(a+b{)}^2=$_______.

• A. 324
• B. 256
• C. 81
• D. - 18

Answer 1

The correct option is A

324

Solving for $a$ :
$a=\frac{2-\sqrt{5}}{2+\sqrt{5}}\times \frac{2-\sqrt{5}}{2-\sqrt{5}}$

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

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

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

$a=$ $4\sqrt{5}-9$ ...... (i)

Solving for $b$ :
$b=\frac{2+\sqrt{5}}{2-\sqrt{5}}\times \frac{2+\sqrt{5}}{2+\sqrt{5}}$

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

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

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

$b=$ $-(9+4\sqrt{5})$ ...... (ii)

$a+b=4\sqrt{5}-9-9-4\sqrt{5}$

$=-18$

$\Rightarrow$ $(a+b{)}^2=(-18{)}^2$

$=$ $324$