Question

In the following exercises, solve.

(a) $\sqrt{v-10}+10=v$

(b)role="math" localid="1647988729108" $\sqrt{y+4}-y+2=0$

Short answer

Part (a) : $v=10,11$

Part (b) : $y=0,5$

Step-by-step solution

Part (a) . Step 1. Given information.

An equation is given as$\sqrt{v-10}+10=v$.

Part (a) Step 2. Solution.

Isolate radical to one side, take square and simplify.

$$\begin{gathered} \sqrt{v-10}+10=v \\ \sqrt{v-10}+10-10=v-10 \\ \sqrt{v-10}=v-10 \\ {\left(\sqrt{v-10}\right)}^2={(v-10)}^2 \\ v-10=v^2-20v+100 \\ 0=v^2-20v+100-v+10 \\ 0=v^2-21v+110 \\ {v}^2-21v+110=0 \\ {v}^2-11v-10v+110=0 \\ v(v-11)-10(v-11)=0 \\ (v-10)(v-11)=0 \\ v=10orv=11 \end{gathered}$$

Part (b) Step 1. Solution.

To solve $\sqrt{y+4}-y+2=0$.

Isolate radical to one side, take square and simplify.

localid="1657332045697" $$\begin{gathered} \sqrt{y+4}-y+2=0 \\ \sqrt{y+4}-y+2+y-2=0+y-2 \\ \sqrt{y+4}=y-2 \\ {\left(\sqrt{y+4}\right)}^2={(y-2)}^2 \\ y+4=y^2-4y+4 \\ 0=y^2-4y-y+4-4 \\ 0=y^2-5y \\ {y}^2-5y=0 \\ y(y-5)=0 \\ y=0ory=5 \end{gathered}$$