Rationalize the numerator
$$\frac{\sqrt{x+4} -2}{x }$$
answer:
$$\frac{1}{\sqrt{x+4} +2} $$
$\frac{\sqrt{x+h} - \sqrt{2}}{x}$ answer:$\frac{1} {\sqrt{x+h} + \sqrt{x}}$
Please show me the steps. I feel like the answer is easy, but its really making me go mad.
$\endgroup$ 43 Answers
$\begingroup$We have,
$$\frac{\sqrt{x+4}-2}{x} × \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$$
$$= \frac{(\sqrt{x+4})^2-(2)^2}{x(\sqrt{x+4}+2)}$$
$$= \frac{x+4-4}{x(\sqrt{x+4}+2)}$$
$$= \frac{x}{x(\sqrt{x+4}+2)}$$
$$= \frac{1}{\sqrt{x+4}+2}$$
Similarly other question, but I think you have typo in that its $\sqrt h$ instead of $\sqrt 2$.
$$\frac{\sqrt{x+h}-\sqrt h}{x} × \frac{\sqrt{x+h}+\sqrt h}{\sqrt{x+h}+\sqrt h}$$
$$= \frac{(\sqrt{x+h})^2-(\sqrt h)^2}{x (\sqrt{x+h}+\sqrt h)}$$
$$= \frac{x+h-h}{x(\sqrt{x+h}+\sqrt h)}$$
$$= \frac{x}{x(\sqrt{x+h}+\sqrt h)}$$
$$= \frac{1}{\sqrt{x+h}+\sqrt h}$$
$\endgroup$ 1 $\begingroup$$$\dfrac{\sqrt{x+4}-2}{x}=\\\dfrac{\sqrt{x+4}-2}{x}\times \dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\\ \dfrac{\sqrt{x+4}^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac{(x+4)-4}{x(\sqrt{x+4}+2)}=\\ \dfrac{x}{x(\sqrt{x+4}+2)}=\dfrac{1}{(\sqrt{x+4}+2)}$$ second one is like that ,As I said before ,I think second is this $$\color{red}{\frac{\sqrt{x+h} - \sqrt{h}}{x}}\\\vdots\\\frac{\sqrt{x+h} - \sqrt{h}}{x}\times \frac{\sqrt{x+h} + \sqrt{h}}{\sqrt{x+h} + \sqrt{h}}=\dfrac{x+h-h}{x(\sqrt{x+h} + \sqrt{h})}=\\ \dfrac{x}{x(\sqrt{x+h} + \sqrt{h})}=\dfrac{1}{(\sqrt{x+h} + \sqrt{h})}$$
$\endgroup$ $\begingroup$$$\begin{align}\frac{\sqrt{x+4}-2}{x}&=\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\\ &=\frac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}\\ &=\frac{x+4-4}{x(\sqrt{x+4}+2)}\\ &=\frac{x}{x(\sqrt{x+4}+2)}\\ &=\frac{1}{\sqrt{x+4}+2}\end{align}$$
Your 2nd question does not match with your answer. But if your question is how rationalize the numerator of $\frac{\sqrt{x+h}-\sqrt{x}}{h}$ then your answer $\frac{1}{\sqrt{x+h}+\sqrt{x}}$ is correct. The solution is similar to the above solution.
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