Q. 8.123 - Intermediate Algebra

Question

Solve: $2\sqrt{4a+4}-16=16$

Step-by-Step Solution

Answer

The solution is $a=63$.

Step 1. Isolate the radical

In the equation, isolate the radical by first adding $16$ both sides.

$\begin{array}{rcl}2\sqrt{4a+4}-16& =& 16\\ 2\sqrt{4a+4}-16+16& =& 16+16\\ 2\sqrt{4a+4}& =& 32\end{array}$

Now divide both sides by $2$

$\begin{array}{rcl}\frac{2\sqrt{4a+4}}{2}& =& \frac{32}{2}\\ \sqrt{4a+4}& =& 16\end{array}$

Step 2. Square both sides and solve for a

Remove the radical by squaring both sides.

$\begin{array}{rcl}{\left(\sqrt{4a+4}\right)}^2& =& {16}^2\\ 4a+4& =& 256\end{array}$

Subtract $4$ from both sides

$\begin{array}{rcl}4a+4-4& =& 256-4\\ 4a& =& 252\end{array}$

Now divide both sides by $4$ and solve for $a$.

$\begin{array}{rcl}\frac{4a}{4}& =& \frac{252}{4}\\ a& =& 63\end{array}$

Step 3. Check the solution

Substitute $63$ for $a$ in the given equation.

$\begin{array}{rcl}2\sqrt{4a+4}-16& =& 16\\ 2\sqrt{4·63+4}-16& =& 16\\ 2\sqrt{252+4}-16& =& 16\\ 2\sqrt{256}-16& =& 16\\ 2·16-16& =& 16\\ 32-16& =& 16\\ 16& =& 16\end{array}$

So, we get a true statement. Thus the solution found is correct.