Q33. - Algebra 1

Question

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

$y^2-8y+16=36$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are 10 and $-2$. 

Step 1. Complete the square.

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

  $$\begin{gathered} y^2-8y+16=36 \\ {y}^2-8y+16-16=36-16\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\hspace{0.17em}\hspace{0.17em}subtract\hspace{0.17em}\hspace{0.17em}}16\text{\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}]\hspace{0.17em}\hspace{0.17em}} \\ y^2-8y+16=20+16[\text{\hspace{0.17em}\hspace{0.17em}Since\hspace{0.17em}\hspace{0.17em}}{\left(\frac{8}{2}\right)}^2=16,\text{\hspace{0.17em}add\hspace{0.17em}\hspace{0.17em}}16\text{\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}}] \\ y^2-8y+16=36 \end{gathered}$$

Step 2. Take the square root of each side.

The first factor $y^2-8y+16$, then take the square root of each side.

$$\begin{gathered} y^2-8y+16=36 \\ {\left(y-4\right)}^2=36 \\ \sqrt{{\left(y-4\right)}^2}=\pm \sqrt{36} \\ y-4=\pm 6 \end{gathered}$$

 

Step 3. Write the solutions.

Now add 4 to both sides and separate the solutions. 

$y-4+4=4\pm 6$

            $$\begin{gathered} y=4\pm 6 \\ y=4+6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=4-6 \\ y=10\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=-2 \end{gathered}$$

Therefore, the solutions are 10 and -2.