Q32. - Algebra 1

Question

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

$-a^2-10a+25=25$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are 0 and $-10$. 

Step 1. Complete the square.

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

$$\begin{gathered} -a^2-10a+25=25 \\ \frac{-a^2-10a+25}{-1}=\frac{25}{-1} \\ a^2+10a-25=-25 \\ {a}^2+10a-25+25=-25+25\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\hspace{0.17em}\hspace{0.17em}add 25\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}]\hspace{0.17em}\hspace{0.17em}} \\ a^2+10a+25=25[\text{\hspace{0.17em}\hspace{0.17em}Since\hspace{0.17em}\hspace{0.17em}}{\left(\frac{10}{2}\right)}^2=25\text{\hspace{0.17em}\hspace{0.17em}add\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}}] \end{gathered}$$

Step 2. Take the square root of each side.

The first factor $a^2+10a+25$, then take the square root of each side.

$$\begin{gathered} a^2+10a+25=25 \\ {\left(a+5\right)}^2=25 \\ \sqrt{{\left(a+5\right)}^2}=\pm \sqrt{25} \\ a+5=\pm 5 \end{gathered}$$

Step 3. Write the solutions.

Now subtract 5 from both sides and separate the solutions. 

$-5+a+5=-5\pm 5$

               $$\begin{gathered} a=-5\pm 5 \\ a=-5+5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}a=-5-5 \\ a=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}a=-10 \end{gathered}$$