Chapter 12.2, Problem 14WE

a.

To determine

To Solve: The equation $y^2+15y+56=0$ by completing the square.

Answer to Problem 14WE

The required solution set is $\left\{-7,-8\right\}$ .

Explanation of Solution

Given information:

  $y^2+15y+56=0$

Calculation:

  $\begin{array}{l}{y}^2+15y+56=0\\ y^2+15y=-56\end{array}$

(Subtract 56 from each side and rewrite the equation)

  $y^2+15y+{\left(\frac{15}{2}\right)}^2=-56+{\left(\frac{15}{2}\right)}^2$

(Add square of half the coefficient of y to both sides)

  $\begin{array}{l}{y}^2+15y+\left(\frac{225}{4}\right)=-56+\left(\frac{225}{4}\right)\\ {\left(y+\frac{15}{2}\right)}^2=\frac{1}{4}\end{array}$

(Simplify)

  $y+\frac{15}{2}=\pm \frac{1}{2}$

(Perform square root on both sides)

  $y=-\frac{15}{2}\pm \frac{1}{2}$

(Subtract $\frac{15}{2}$ from each side)

Therefore,

  $\begin{array}{l}y=-\frac{15}{2}+\frac{1}{2}\\ y=\frac{-15+1}{2}\\ y=\frac{-14}{2}\\ y=-7\end{array}$

Or

  $\begin{array}{l}y=-\frac{15}{2}-\frac{1}{2}\\ y=\frac{-15-1}{2}\\ y=\frac{-16}{2}\\ y=-8\end{array}$

Therefore, the solution set is $\left\{-7,-8\right\}$ .

b.

To determine

To Solve: The equation $y^2+15y+56=0$ by factoring.

Answer to Problem 14WE

The required solution set is $\left\{-7,-8\right\}$ .

Explanation of Solution

Given information:

  $y^2+15y+56=0$

Calculation:

  $\begin{array}{l}{y}^2+15y+56=0\\ y^2+7y+8y+56=0\\ y\left(y+7\right)+8\left(y+7\right)=0\\ \left(y+7\right)\left(y+8\right)=0\end{array}$

(Splitting the middle term and factoring)

Therefore,

  $\begin{array}{l}y=-7\\ or\\ y=-8\end{array}$

Therefore, the solution set is $\left\{-7,-8\right\}$ .