Question

Which one of the following is the rationalize form of $\frac{11}{\sqrt{17}-\sqrt{6}}$ ?

• A. $\sqrt{17}-\sqrt{6}$
• B. $\sqrt{17}+2\sqrt{6}$
• C. $\sqrt{17}+\sqrt{6}$
• D. $11\sqrt{11}-\sqrt{6}$

Answer 1

Answer: (C)

$\frac{11}{\sqrt{17}-\sqrt{6}}\times \frac{\sqrt{17}+\sqrt{6}}{\sqrt{17}+\sqrt{6}}$
Using $(a+b)(a-b)=a^2-b^2$, we get
$\frac{11(\sqrt{17}+\sqrt{6})}{(\sqrt{17}{)}^2-(\sqrt{6}{)}^2}=\frac{11(\sqrt{17}+\sqrt{6})}{17-6}<br/>=\frac{11(\sqrt{17}+\sqrt{6})}{11}$
$=\sqrt{17}+\sqrt{6}$