Question

In the following exercises, solve.

$$\begin{gathered} \left(a\right)\sqrt{x}+10=\sqrt{x+2} \\ \left(b\right)\sqrt{y-2}+2=\sqrt{2y+4} \end{gathered}$$

Short answer

Part (a) : $\frac{2401}{100}$

Part (b) : 6

Step-by-step solution

Step 1. Given Information

The given expressions are

$$\begin{gathered} \left(a\right)\sqrt{x}+10=\sqrt{x+2} \\ \left(b\right)\sqrt{y-2}+2=\sqrt{2y+4} \end{gathered}$$

Step 2. Squaring on both sides 

$$\begin{gathered} \left(a\right)\sqrt{x}+10=\sqrt{x+2} \\ {\left(\sqrt{x}+10\right)}^2={\left(\sqrt{x+2}\right)}^2 \\ 98+20\sqrt{x}=0 \\ 20\sqrt{x}=-98 \\ Squaringonbothsides \\ {\left(20\sqrt{x}\right)}^2=9604 \\ x=\frac{2401}{100} \end{gathered}$$

The solution is $\frac{2401}{100}$

$$\begin{gathered} \left(b\right)\sqrt{y-2}+2=\sqrt{2y+4} \\ {\left(\sqrt{y-2}+2\right)}^2={\left(\sqrt{2y+4}\right)}^2 \\ -2+4\sqrt{y-2}=y \\ 4\sqrt{y-2}=y+2 \\ squaringonbothsides \\ 16\left(y-2\right)=y^2+4+4y \\ 16y-32=y^2+4+4y \\ {y}^2-12y+36=0 \\ y=6 \end{gathered}$$

The solution is 6

Step 3. The solution

The solution for are$\frac{2401}{100}and6$