Rewriting $\sqrt{h(h+1)}-h$ into $\frac{h(h+1)-h^2}{\sqrt{h(h+1)}+h}$ I need help with how to rearrange the equation $\sqrt{h(h+1)}-h$. In the solutions booklet the answer to this question is $$\frac{h(h+1)-h^2}{\sqrt{h(h+1)}+h}$$ however I got $$\frac{h(h+1)-h^2}{\sqrt{h(h+1)}-h}$$ instead (difference is the +/- sign.)
Here's how I did it:
$$\sqrt{h(h+1)}-h$$ = $$(h(h+1))^{1/2})^2-h^2$$ = $$(\sqrt{h(h+1)}+h)(\sqrt{h(h+1)}-h)$$ = ${h(h+1)-h^2}$. At this point I divided ${h(h+1)-h^2}$ by the original equation because I raised it to a power 2. So I did
$$\frac{h(h+1)-h^2}{\sqrt{h(h+1)}-h}$$
This solution doesn't fit the one in the textbook. I'm not sure where to get the $\sqrt{h(h+1)}+h$ in the denominator.
 A: You have to rationalize, specifically as follows:
$$\sqrt{h(h+1)}-h=\left(\sqrt{h(h+1)}-h\right)\color{red}{\frac{\sqrt{h(h+1)}+h}{\sqrt{h(h+1)}+h}}$$
Note: I have multiplied and divided by $\sqrt{h(h+1)}+h$, in fact the idea is to eliminate the square root, using the equality: $\color{green}{(a-b)(a+b)=a^2-b^2}$
So now we have:
$$\left(\sqrt{h(h+1)}-h\right)\frac{\sqrt{h(h+1)}+h}{\sqrt{h(h+1)}+h}=\frac{h(h+1)-h^2}{\sqrt{h(h+1)}+h}$$
where to obtain the numerator I have used the equality in green.
A: $\sqrt a -\sqrt b=\frac {(\sqrt a -\sqrt b) (\sqrt a +\sqrt b)} {\sqrt a +\sqrt b}=\frac {a-b} {(\sqrt a +\sqrt b)} $.
A: $$\sqrt{h(h+1)}-h=\frac{\sqrt{h(h+1)}-h}1=\frac{(\sqrt{h(h+1)}-h)(\sqrt{h(h+1)}+h)}{\sqrt{h(h+1)}+h}==\frac{\sqrt{h(h+1)}-h^2}{\sqrt{h(h+1)}+h}$$
A: If you notice carefully, we have a difference of a square root our variable. We can multiply by its conjugate (that's changing the sign between the terms) and simplify by difference of squares.
$$\sqrt{h(h+1)}-h=(\sqrt{h(h+1)}-h)\frac{\sqrt{h(h+1)}+h}{\sqrt{h(h+1)}+h} = \frac{h(h+1)-h^2}{\sqrt{h(h+1)}}$$
which is the answer in the textbook you are wondering about.
