Question

Solve by completing the square.

$2x^2+2x=4$

Short answer

The Solution is$x=-2$and$x=1$.

Step-by-step solution

Step 1. Given information.

The given equation is$2x^2+2x=4$.

Step 2. Simplify the given equation.

$\begin{array}{rcl}2x^2+2x& =& 4\\ \frac{2x^2}{2}+\frac{2x}{2}& =& \frac{4}{2}\\ x^2+x& =& 2\\ x^2+2·x·\frac{1}{2}& =& 2\\ x^2+2·x·\frac{1}{2}+{\left(\frac{1}{2}\right)}^2& =& 2+{\left(\frac{1}{2}\right)}^2\\ {\left(x+\frac{1}{2}\right)}^2& =& 2+\frac{1}{4}\\ {\left(x+\frac{1}{2}\right)}^2& =& \frac{9}{4}\\ & & \\ & & \end{array}$

Step 3. Continue the above calculation.

$\begin{array}{rcl}{\left(x+\frac{1}{2}\right)}^2& =& \frac{9}{4}\\ x+\frac{1}{2}& =& \pm \sqrt{\frac{9}{4}}\\ x+\frac{1}{2}& =& \pm \frac{3}{2}\end{array}$

Step 4. Solve the equation for x.

$\begin{array}{rcl}x+\frac{1}{2}& =& \frac{3}{2}\\ x& =& \frac{3}{2}-\frac{1}{2}\\ x& =& \frac{3-1}{2}\\ x& =& \frac{2}{2}\\ x& =& 1\end{array}$ and$\begin{array}{rcl}x+\frac{1}{2}& =& -\frac{3}{2}\\ x& =& -\frac{3}{2}-\frac{1}{2}\\ x& =& \frac{-3-1}{2}\\ x& =& \frac{-4}{2}\\ x& =& -2\end{array}$

Step 5. Simplified answer.

Hence, the required solution is$x=1$and$x=-2$.