Question

What is the additive inverse of $\frac{1}{3-\sqrt{8}}$.

• A. $\sqrt{8}+$
• B. $-(\sqrt{8}+3)$
• C. $-(\sqrt{3}+8)$
• D. $\sqrt{3}+8$

Answer 1

The correct option is B $-(\sqrt{8}+3)$

Let $x$ be the additive inverse of given number
$\Rightarrow \frac{1}{3-\sqrt{8}}+x=0\Rightarrow x=-\frac{1}{3-\sqrt{8}}\Rightarrow x=\frac{1}{\sqrt{8}-3}$
Rationalising the number
$\Rightarrow x=\frac{1}{\sqrt{8}-3}\times \frac{\sqrt{8}+3}{\sqrt{8}+3}\Rightarrow x=\frac{\sqrt{8}+3}{{\left(\sqrt{8}\right)}^2-\left(3^2\right)}=\frac{\sqrt{8}+3}{8-9}=\frac{\sqrt{8}+3}{-1}\Rightarrow x=-(\sqrt{8}+3)$
So option $B$ is correct.