Question

In the following exercises, solve.

(a) $\sqrt{r-1}-r=-1$

(b)role="math" localid="1647990004799" $\sqrt{z+100}-z+10=0$

Short answer

Part (a) : $r=1,2$

Part (b) : $z=0,21$

Step-by-step solution

Part (a) Step 1. Given information.

An equation is given as$\sqrt{r-1}-r=-1$.

Part (a) Step 2. Solution.

Isolate radical to one side, take square and simplify.

$$\begin{gathered} \sqrt{r-1}-r=-1 \\ \sqrt{r-1}-r+r=-1+r \\ \sqrt{r-1}=r-1 \\ {\left(\sqrt{r-1}\right)}^2={(r-1)}^{2}r-1=r^2-2r+1 \\ 0=r^2-2r+1-r+1 \\ 0=r^2-3r+2 \\ {r}^2-3r+2=0 \\ {r}^2-r-2r+2=0 \\ r(r-1)-2(r-1)=0 \\ (r-1)(r-2)=0 \\ r=1orr=2 \end{gathered}$$

Part (b) Step 1. Solution.

To solve $\sqrt{z+100}-z+10=0$.

Isolate radical to one side, take square and simplify.

$$\begin{gathered} \sqrt{z+100}-z+10=0 \\ \sqrt{z+100}-z+10+z-10=0+z-10 \\ \sqrt{z+100}=z-10 \\ {\left(\sqrt{z+100}\right)}^2={(z-10)}^{2}z+100=z^2-20z+100 \\ 0=z^2-20z+100-z-100 \\ 0=z^2-21z \\ {z}^2-21z=0 \\ z(z-21)=0 \\ z=0orz=21 \end{gathered}$$