# How do I simplify $\frac{\sqrt{4+h}-2}{h}$? [duplicate]

I know this looks like a dumb question, but how do I simplify this? Does it uses some square root property or factorization? The wolfram alpha has no step-by-step solution for this, so it may use some crazy technique, I guess.

$$\frac{\sqrt{4+h}-2}{h} = \frac{1}{\sqrt{4+h}-2}$$

## marked as duplicate by user63181, user127096, user61527, Dan, Hans EnglerMar 30 '14 at 0:27

• It is simple already, but very useful is to multiply by $\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}$. We get $\frac{1}{\sqrt{4+h}+2}$. – André Nicolas Mar 29 '14 at 21:37
$$\dfrac{\sqrt{4+h}-2}{h}$$ $$=\dfrac{\sqrt{4+h}-2}{h}\times\dfrac{\sqrt{4+h}+2}{\sqrt{4+h}+2}$$ $$=\dfrac{(4+h)-4}{h(\sqrt{4+h}+2)}$$ $$=\dfrac{h}{h(\sqrt{4+h}+2)}$$ $$=\dfrac{1}{\sqrt{4+h}+2}$$
Hint: Multiply with $$\sqrt{4+h}+2$$ both the numerator and the denominator. The RHS shoud be $$\frac{1}{\sqrt{4+h}+2}$$ and not with $-2$ as you have it.
$$\frac{\sqrt{4+h}-2}{h}=\frac{(\sqrt{4+h}-2)\cdot(\sqrt{4+h}+2)}{h\cdot(\sqrt{4+h}+2)}=\frac{(\sqrt{4+h})^2-4}{h(\sqrt{4+h}+2)} = \frac{1}{\sqrt{4+h}+2}$$ Obviously we suppose $h\ne0$.