Q34. - Algebra 1

Question

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

$y^2-6y+2=0$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are 5.65 and 0.35. 

Step 1. Complete the square.

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

  $$\begin{gathered} y^2-6y+2=0 \\ {y}^2-6y+2-2=0-2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\hspace{0.17em}\hspace{0.17em}subtract\hspace{0.17em}\hspace{0.17em}}2\text{\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}]\hspace{0.17em}\hspace{0.17em}} \\ y^2-6y=-2[\text{\hspace{0.17em}\hspace{0.17em}Since\hspace{0.17em}\hspace{0.17em}}{\left(\frac{-6}{2}\right)}^2=9,\text{\hspace{0.17em}add\hspace{0.17em}\hspace{0.17em}}9\text{\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}}] \\ y^2-6y+9=-2+9 \end{gathered}$$

Step 2. Take the square root of each side.

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

  $$\begin{gathered} y^2-6y+9=7 \\ {\left(y-3\right)}^2=7 \\ \sqrt{{\left(y-3\right)}^2}=\pm \sqrt{7} \\ y-3=\pm \sqrt{7} \end{gathered}$$

 

Step 3. Write the solutions.

Now add 3 to both sides and separate the solutions. 

 $y-3+3=3\pm \sqrt{7}$

            $$\begin{gathered} y=3\pm \sqrt{7} \\ y=3+\sqrt{7}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=3-\sqrt{7} \\ y=5.65\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=0.35 \end{gathered}$$

Therefore, the solutions are 5.65 and 0.35.