Question

$\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $=$ $a+b$ $\sqrt{3}$; then what is the value of $a+b$ ?

• A. $-3$
• B. $1$
• C. $3$
• D. $-1$

Answer 1

The correct option is B $1$

Let $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$
$\Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1{)}^2}{2}=\frac{3+1-2\sqrt{3}}{2}$
$=2-\sqrt{3}=a+b\sqrt{3}$
$\Rightarrow \therefore a=2,b=-1$
$\therefore a+b=2-1=1$
Hence, option 'B' is correct.