Chapter A.10, Problem 53AYU

To determine

To find: Rationalize the denominator of the expression $\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$.

Answer to Problem 53AYU

The rationalized form of $\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$ is $\frac{2x+h-2\sqrt{x+h}\sqrt{x}}{h}$.

Explanation of Solution

Given:

It is asked to find the rationalized form of the given expression $\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$.

Calculation:

When radicals occur in quotients, it is customary to rewrite the quotient so that the new denominator contains no radicals. This process is referred to as rationalizing the denominator.

The idea is to multiply by an appropriate expression so that the new denominator contains no radicals.

Here multiple and divide by $\sqrt{x+h}-\sqrt{x}$ to rationalize the given expression.

Consider $\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$.

Multiply and divide by $\sqrt{x+h}-\sqrt{x}$.

$\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\times \frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}-\sqrt{x}}=\frac{{\left(\sqrt{x+h}-\sqrt{x}\right)}^2}{{\left(\sqrt{x+h}\right)}^2-{\left(\sqrt{x}\right)}^2}= \frac{{\left(\sqrt{x+h}\right)}^2+{\left(\sqrt{x}\right)}^2-2\sqrt{x+h}\sqrt{x}}{x+h-x}=\frac{2x+h-2\sqrt{x+h}\sqrt{x}}{h}$