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$$\frac{\sqrt{x+4} -2}{x }$$

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.

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    I think the second one is $$\frac{\sqrt{x+h} - \sqrt{h}}{x}$$! – 2017-02-05
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    Hint - use a trick: expansion (you may call it FOIL) on $(a-b)(a+b)$ will get $a^2-b^2.$ Now, in your mind, imagine that $a=\sqrt{x+4}$ and $b=2.$ What do you get when you follow the advice in the Answers? – 2017-02-05
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    @Calculus Help Plz It seems that you really need to read http://meta.math.stackexchange.com/questions/3399/why-should-we-accept-answers – 2017-02-05
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    Hello. It seems you do not know how to accept answers. At the left side of every answer there is a check mark for you to click it. Doing so, blank check mark becomes green and that indicates that an answer is accepted. Thank you. – 2017-02-06

3 Answers 3

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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 $\sqrt 2$.

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

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

$$= \frac{x+h-h}{x(\sqrt{x+h}+\sqrt h)}$$

$$= \frac{x}{x(\sqrt{x+h}+\sqrt h)}$$

$$= \frac{1}{\sqrt{x+h}+\sqrt h}$$

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    And I think you have mistake in typing second question. Its $\sqrt h$ instead of $\sqrt2$ – 2017-02-05
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$$\dfrac{\sqrt{x+4}-2}{x}=\\\dfrac{\sqrt{x+4}-2}{x}\times \dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\\ \dfrac{\sqrt{x+4}^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac{(x+4)-4}{x(\sqrt{x+4}+2)}=\\ \dfrac{x}{x(\sqrt{x+4}+2)}=\dfrac{1}{(\sqrt{x+4}+2)}$$ second one is like that ,As I said before ,I think second is this $$\color{red}{\frac{\sqrt{x+h} - \sqrt{h}}{x}}\\\vdots\\\frac{\sqrt{x+h} - \sqrt{h}}{x}\times \frac{\sqrt{x+h} + \sqrt{h}}{\sqrt{x+h} + \sqrt{h}}=\dfrac{x+h-h}{x(\sqrt{x+h} + \sqrt{h})}=\\ \dfrac{x}{x(\sqrt{x+h} + \sqrt{h})}=\dfrac{1}{(\sqrt{x+h} + \sqrt{h})}$$

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$$\begin{align}\frac{\sqrt{x+4}-2}{x}&=\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\\ &=\frac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}\\ &=\frac{x+4-4}{x(\sqrt{x+4}+2)}\\ &=\frac{x}{x(\sqrt{x+4}+2)}\\ &=\frac{1}{\sqrt{x+4}+2}\end{align}$$

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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    Thanks guys it really helped a lot. I messed up because i kept conjugating the sign of the radicals. Now I'm loving math again lol – 2017-02-05