## Precalculus (6th Edition) Blitzer

The rationalized form of the expression $\frac{\sqrt{x+7}-\sqrt{x}}{7}$ is $\frac{1}{\left( \sqrt{x+7}+\sqrt{x} \right)}$.
Consider the provided expression, $\frac{\sqrt{x+7}-\sqrt{x}}{7}$ Multiply the numerator and denominator by $\sqrt{x+7}+\sqrt{x}$ so that the radicand in the numerator is a perfect square. Therefore, $\frac{\sqrt{x+7}-\sqrt{x}}{7}=\frac{\sqrt{x+7}-\sqrt{x}}{7}\cdot \frac{\sqrt{x+7}+\sqrt{x}}{\sqrt{x+7}+\sqrt{x}}$ Apply the difference of squares property in the numerator, $\left( \sqrt{x+7}+\sqrt{x} \right)\left( \sqrt{x+7}-\sqrt{x} \right)={{\left( \sqrt{x+7} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}$ Apply the power of a power property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ in the expression ${{\left( \sqrt{x+7} \right)}^{2}},{{\left( \sqrt{x} \right)}^{2}}$. \begin{align} & {{\left( \sqrt{x+7} \right)}^{2}}={{\left( x+7 \right)}^{2\cdot \frac{1}{2}}} \\ & ={{\left( x+7 \right)}^{1}} \\ & =x+7 \end{align} And, \begin{align} & {{\left( \sqrt{x} \right)}^{2}}={{\left( x \right)}^{2\cdot \frac{1}{2}}} \\ & ={{\left( x \right)}^{1}} \\ & =x \end{align} Rewrite the original expression as: \begin{align} & \frac{\sqrt{x+7}-\sqrt{x}}{7}=\frac{\sqrt{x+7}-\sqrt{x}}{7}\cdot \frac{\sqrt{x+7}+\sqrt{x}}{\sqrt{x+7}+\sqrt{x}} \\ & =\frac{{{\left( \sqrt{x+7} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \\ & =\frac{x+7-x}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \\ & =\frac{7}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \end{align} Further simplify, \begin{align} & \frac{\sqrt{x+7}-\sqrt{x}}{7}=\frac{7}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \\ & =\frac{1}{\left( \sqrt{x+7}+\sqrt{x} \right)} \end{align} Therefore, the rationalized form of the expression $\frac{\sqrt{x+7}-\sqrt{x}}{7}$ is $\frac{1}{\left( \sqrt{x+7}+\sqrt{x} \right)}$.