Question

Rationalize the numerator.

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

Short answer

The solution is$\frac{-1}{\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}$

Step-by-step solution

Step 1. Apply the concept of rationalizing the numerator.

If a fraction has a numerator of the form$A+B\sqrt{C}$, we can rationalize the numerator by multiplying numerator and denominator by the conjugate radical $A-B\sqrt{C}$.

Step 2. Apply the concept of special multiplication.

If A and B are real numbers, then

$\left(A-B\right)\left(A+B\right)=A^2-B^2$

Step 3. Analyze the expression.

Given

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

Here the numerator is $\sqrt{x}-\sqrt{x+h}$, so multiply the numerator and denominator by$\sqrt{x}+\sqrt{x+h}$

Step 4. Rationalize the numerator.

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

So, the simplified expression after rationalizing numerator is given by:

localid="1643281202417" $\begin{array}{l}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}\cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}\\ \Rightarrow \frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{\left(\sqrt{x}-\sqrt{x+h}\right)\left(\sqrt{x}+\sqrt{x+h}\right)}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}\\ \Rightarrow \frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{{\left(\sqrt{x}\right)}^2-{\left(\sqrt{x+h}\right)}^2}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}\\ \Rightarrow \frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{x-x-h}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}\\ \Rightarrow \frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{-h}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}\\ \Rightarrow \frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}=\frac{-1}{\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}\end{array}$