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.

Solution 1:

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.