Q31. - Algebra 1

Question

Solve each equation by completing the square. Round to the nearest tenth if necessary.

$x^2+6x+9=16$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are 1 and $\mathbf{-}\mathbf{7}$. 

Step 1. Complete the square.

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

$$\begin{gathered} x^2+6x+9=16 \\ {x}^2+6x+9-9=16-9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\hspace{0.17em}\hspace{0.17em}subtract\hspace{0.17em}\hspace{0.17em}9\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}]\hspace{0.17em}\hspace{0.17em}} \\ x^2+6x+9=7+9[\text{\hspace{0.17em}\hspace{0.17em}Since\hspace{0.17em}\hspace{0.17em}}{\left(\frac{6}{2}\right)}^2=9\text{\hspace{0.17em}\hspace{0.17em}add\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}}] \end{gathered}$$

Step 2. Take the square root of each side.

The first factor $x^2+6x+9$, then take the square root of each side.

$$\begin{gathered} {\left(x+3\right)}^2=16 \\ \sqrt{{(x+3)}^2}=\pm \sqrt{16} \\ x+3=\pm 4 \end{gathered}$$

Step 3. Write the solutions.

Now subtract 3 from both sides and separate the solutions. 

$$\begin{gathered} x+3-3=-3\pm 4 \\ x=-3+4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-3-4 \\ x=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-7 \end{gathered}$$

Therefore, the solutions are 1 and $-7$.