Q14. - Algebra 1

Question

Solve each equation by completing the square.

$2x^2-36=-6x$

Step-by-Step Solution

Answer

The solutions of the quadratic equation are $-6$ and 3.

Step 1. Complete the square.

Complete the square to solve the quadratic equations by isolating $x^2$ and bx-terms.

$$\begin{gathered} 2x^2-36=-6x \\ 2x^2+6x-36=0 \left[\text{\hspace{0.17em}\hspace{0.17em}Standard\hspace{0.17em}\hspace{0.17em}Form\hspace{0.17em}\hspace{0.17em}}\right] \\ \frac{2x^2+6x-66}{2}=\frac{0}{2} \left[\text{\hspace{0.17em}\hspace{0.17em}Divide\hspace{0.17em}\hspace{0.17em}by\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}}\right] \\ x^2+3x-18=0 \\ x^2+3x-18+\frac{9}{4}=0+\frac{9}{4} \left[\text{\hspace{0.17em}Since\hspace{0.17em}\hspace{0.17em}}{\left(\frac{3}{2}\right)}^2=\frac{9}{4}\text{\hspace{0.17em}\hspace{0.17em}add\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}}\right] \\ {x}^2+3x+\frac{9}{4}-18+18=\frac{9}{4}+18 \left[\text{\hspace{0.17em}Add\hspace{0.17em}\hspace{0.17em}6\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}}\right] \\ x^2+3x+\frac{9}{4}=\frac{81}{4} \end{gathered}$$

Step 2. Take the square root of each side.

First factor $x^2+3x+\frac{9}{4}$, then take square root of each side.

$$\begin{gathered} {\left(x+\frac{3}{2}\right)}^2=\frac{81}{4} \\ \sqrt{{\left(x+\frac{3}{2}\right)}^2}=\pm \sqrt{\frac{81}{4}} \\ x+\frac{3}{2}=\pm \frac{9}{2} \end{gathered}$$

Step 3. Write the solutions.

Subtract $\frac{3}{2}$ from both sides and separate the solutions.

$$\begin{gathered} x+\frac{3}{2}-\frac{3}{2}=-\frac{3}{2}\pm \frac{9}{2} \\ x=-\frac{3}{2}-\frac{9}{2}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-\frac{3}{2}+\frac{9}{2} \\ x=-6\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or \hspace{0.17em}\hspace{0.17em}}x=3 \end{gathered}$$

Therefore, the solutions of the equation are $-6$ and 3.