Chapter 11.9, Problem 29WE

To determine

Rationalize: The denominator of given fraction.

Answer to Problem 29WE

Rationalized form of denominator of fraction $\frac{\sqrt{3}-4}{\sqrt{7}+2}$ is $\frac{\sqrt{21}-2\sqrt{3}-4\sqrt{7}-8}{3}$ .

Explanation of Solution

Given information:

The expression $\frac{\sqrt{3}-4}{\sqrt{7}+2}$ .

Formula used:

Rationalizing the denominator is to eliminate the radical terms like square root and cube root from denominator,

To rationalize the denominator, we multiply the conjugate to denominator to both numerator and denominator.

  $\left(a+b\right)\left(a-b\right)=a^2-b^2$

  $\frac{\sqrt{3}-4}{\sqrt{7}+2}$

Recall that to rationalize the denominator, we multiply the conjugate of denominator to both numerator and denominator.

Conjugate of $\sqrt{7}+2$ is $\sqrt{7}-2$ ,

Multiplying $\sqrt{7}-2$ to both numerator and denominator.

  $=\frac{\sqrt{3}-4}{\sqrt{7}+2}\times \frac{\sqrt{7}-2}{\sqrt{7}-2}$

Using identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in denominator,

  $\begin{array}{l}=\frac{\left(\sqrt{3}-4\right)\left(\sqrt{7}-2\right)}{{\left(\sqrt{7}\right)}^2-2^2}\\ =\frac{\sqrt{3}\left(\sqrt{7}-2\right)-4\left(\sqrt{7}-2\right)}{7-4}\\ =\frac{\sqrt{3}\cdot \sqrt{7}-\sqrt{3}\cdot 2-4\cdot \sqrt{7}+4\cdot 2}{7-4}\\ =\boxed{\frac{\sqrt{21}-2\sqrt{3}-4\sqrt{7}+8}{3}}\end{array}$

Rationalized form of denominator of fraction $\frac{\sqrt{3}-4}{\sqrt{7}+2}$ is $\frac{\sqrt{21}-2\sqrt{3}-4\sqrt{7}-8}{3}$ .