Q. 8.128 - Intermediate Algebra

Question

Solve: $\sqrt{x}+2=\sqrt{x+16}$

Step-by-Step Solution

Answer

The solution is $x=9$.

Step 1. Square both sides

Consider the equation $\sqrt{x}+2=\sqrt{x+16}$.

The radical on the right is isolated. Now we square both sides and simplify. 

$\begin{array}{rcl}{(\sqrt{x}+2)}^2& =& {\left(\sqrt{x+16}\right)}^2 \\ {\left(\sqrt{x}\right)}^2+2·\sqrt{x}·2+2^2& =& x+16\\ x+4\sqrt{x}+4& =& x+16\end{array}$

Step 2. Isolate the radical and solve for x

Subtract $x+4$ from both sides

$\begin{array}{rcl}x+4\sqrt{x}+4-(x+4)& =& x+16-(x+4)\\ x+4\sqrt{x}+4-x-4& =& x+16-x-4\\ 4\sqrt{x}& =& 12\end{array}$

Divide both side by $4$.

$\begin{array}{rcl}\frac{4\sqrt{x}}{4}& =& \frac{12}{4}\\ \sqrt{x}& =& 3\end{array}$

Now square both sides and solve the equation.

$\begin{array}{rcl}{\left(\sqrt{x}\right)}^2& =& 3^2\\ x& =& 9\end{array}$

Step 3. Check the solution

Substitute $9$ for $x$ in the original equation.

$\begin{array}{rcl}\sqrt{x}+2& =& \sqrt{x+16}\\ \sqrt{9}+2& =& \sqrt{9+16}\\ 3+2& =& \sqrt{25}\\ 5& =& 5\end{array}$

As the statement is true, so the found solution is correct.