Q. 8.124 - Intermediate Algebra

Question

Solve: $3\sqrt{2b+3}-25=50$

Step-by-Step Solution

Answer

The solution is $b=311$.

Step 1. Isolate the radical

Consider the equation $3\sqrt{2b+3}-25=50$.

In the equation, isolate the radical by first adding $25$ both sides.

$\begin{array}{rcl}3\sqrt{2b+3}-25+25& =& 50+25\\ 3\sqrt{2b+3}& =& 75\end{array}$

Now divide both sides by $3$.

$\begin{array}{rcl}\frac{3\sqrt{2b+3}}{3}& =& \frac{75}{3}\\ \sqrt{2b+3}& =& 25\end{array}$

Step 2. Square both sides and solve for b

Remove the radical by squaring both sides.

$\begin{array}{rcl}{\left(\sqrt{2b+3}\right)}^2& =& {25}^2\\ 2b+3& =& 625\end{array}$

Subtract $3$ from both sides

$\begin{array}{rcl}2b+3-3& =& 625-3\\ 2b& =& 622\end{array}$

Now divide both sides by $2$ and solve for $b$.

$\begin{array}{rcl}\frac{2b}{2}& =& \frac{622}{2}\\ b& =& 311\end{array}$

Step 3. Check the solution

Substitute $311$ for $b$ in the given equation.

$\begin{array}{rcl}3\sqrt{2b+3}-25& =& 50\\ 3\sqrt{2·311+3}-25& =& 50\\ 3\sqrt{622+3}-25& =& 50\\ 3\sqrt{625}-25& =& 50\\ 3·25-25& =& 50\\ 75-25& =& 50\\ 50& =& 50\end{array}$

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