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Question
rationalize the numerator. assume that all variables represent positive real numbers.\\(\frac{sqrt{x + 5}-sqrt{x}}{5}\\)\\(\frac{sqrt{x + 5}-sqrt{x}}{5}=square\\) (simplify your answer.)
Step1: Multiply numerator and denominator by the conjugate of the numerator.
The conjugate of $\sqrt{x + 5}-\sqrt{x}$ is $\sqrt{x + 5}+\sqrt{x}$. So we multiply the fraction by $\frac{\sqrt{x + 5}+\sqrt{x}}{\sqrt{x + 5}+\sqrt{x}}$:
$$\frac{\sqrt{x + 5}-\sqrt{x}}{5}\times\frac{\sqrt{x + 5}+\sqrt{x}}{\sqrt{x + 5}+\sqrt{x}}=\frac{(\sqrt{x + 5}-\sqrt{x})(\sqrt{x + 5}+\sqrt{x})}{5(\sqrt{x + 5}+\sqrt{x})}$$
Step2: Apply the difference of squares formula $(a - b)(a + b)=a^2 - b^2$ to the numerator.
Here, $a=\sqrt{x + 5}$ and $b = \sqrt{x}$. So the numerator becomes $(\sqrt{x + 5})^2-(\sqrt{x})^2=(x + 5)-x$.
$$\frac{(x + 5)-x}{5(\sqrt{x + 5}+\sqrt{x})}$$
Step3: Simplify the numerator.
Simplify $(x + 5)-x$: $x+5 - x=5$.
$$\frac{5}{5(\sqrt{x + 5}+\sqrt{x})}$$
Step4: Cancel out the common factor of 5.
Cancel the 5 in the numerator and the 5 in the denominator:
$$\frac{1}{\sqrt{x + 5}+\sqrt{x}}$$
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$\frac{1}{\sqrt{x + 5}+\sqrt{x}}$