Q13. - Algebra 1

Question

Solve each equation by completing the square.

$x^2-x-6=0$

Step-by-Step Solution

Answer

The solutions of the quadratic equation are $-2$ and 3.

Step 1. Complete the square

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

$$\begin{gathered} x^2-x-6=0 \\ x^2-x-6+\frac{1}{4}=0+\frac{1}{4} \left[\text{Since\hspace{0.17em}\hspace{0.17em}}{\left(\frac{1}{2}\right)}^2=\frac{1}{4}\text{\hspace{0.17em}\hspace{0.17em}add\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side}\right] \\ {x}^2-x+\frac{1}{4}-6+6=\frac{1}{4}+6 \left[\text{\hspace{0.17em}Add\hspace{0.17em}\hspace{0.17em}6\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}}\right] \\ x^2-x+\frac{1}{4}=\frac{25}{4} \end{gathered}$$

Step 2. Take the square root of each side

First factor $x^2-x+\frac{1}{4}$, then take square root of each side.

$$\begin{gathered} {\left(x-\frac{1}{2}\right)}^2=\frac{25}{4} \\ \sqrt{{\left(x-\frac{1}{2}\right)}^2}=\pm \sqrt{\frac{25}{4}} \\ x-\frac{1}{2}=\pm \frac{5}{2} \end{gathered}$$

Step 3. Write the solutions

Add $\frac{1}{2}$ to both sides and separate the solutions.

$$\begin{gathered} x-\frac{1}{2}+\frac{1}{2}=\frac{1}{2}\pm \frac{5}{2} \\ x=\frac{1}{2}-\frac{5}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}x=\frac{1}{2}+\frac{5}{2} \\ x=-2\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or \hspace{0.17em}\hspace{0.17em}}x=3 \end{gathered}$$ 

Therefore, the solutions of the equation are $-2$ and 3.