Find the value of $\sqrt{(b-a-4)^2}- \sqrt{(a-b+1)^2}$ if a>0 and b<0

Question

Find the value of $\sqrt{(b-a-4)^2}- \sqrt{(a-b+1)^2}$ if $a>0$ and $b<0$.

How do i find the value? This doesn't make any sense.

Answer 1

$$\because b-a-4<0 \text{ and that }a-b+1>0$$

\begin{align*} \therefore \sqrt{(b-a-4)^2}-\sqrt{(a-b+1)^2}&=-(b-a-4)-(a-b+1)\\ &=-b+a+4-a+b-1\\ &=3 \end{align*}