Question

If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$, then the values of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are _________.

• A. $a=2,b=1$
• B. $a=2,b=-1$
• C. $a=-2,b=1$
• D. $a=4,b=-1$

Answer 1

The correct option is B

$a=2,b=-1$

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$

$\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=a+b\sqrt{3}$

$\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}{)}^2-(1{)}^2}=a+b\sqrt{3}$

$\frac{(\sqrt{3}{)}^2-2\sqrt{3}+(1{)}^2}{3-1}=a+b\sqrt{3}$

$\frac{3-2\sqrt{3}+1}{2}=a+b\sqrt{3}$

$\frac{4-2\sqrt{3}}{2}=a+b\sqrt{3}$

$\text{/}2\left(\frac{2-\sqrt{3}}{\text{/}2}\right)=a+b\sqrt{3}$

Hence, $a=2$ and $b=-1$.