Chapter SH, Problem 9.4.14EP

To determine

The value of c that makes each trinomial a perfect square.

Answer to Problem 9.4.14EP

The solution set is $\{-2.3,-2.8\}.$

Explanation of Solution

Given:

The given equation is:

  $c^2+6=-5c$

Calculation:

The solution of the given equation will be:

  $\begin{array}{l}{\text{Step 1: Isolate c}}^2\text{ and c terms}\text{.}\\ c^2+6=-5c\\ c^2+5c+6-6=-5c+5c-6\\ c^2+5c=-6\\ \text{Step 2: Complete the square and solve}\text{.}\\ \text{As }{\left(\frac{\text{-5}}{\text{2}}\right)}^{\text{2}}\text{=}\frac{25}{4}\text{, add }\frac{25}{4}\text{ to each side;}\\ c^2+5c+\frac{25}{4}=-6+\frac{25}{4}\\ {\text{Factorize c}}^2+5c+\frac{25}{4}:\\ {\left(c+\frac{5}{2}\right)}^2=\frac{1}{4}\\ c+\frac{5}{2}=\pm \frac{1}{4}\\ c+\frac{5}{2}-\frac{5}{2}=\pm \frac{1}{4}-\frac{5}{2}\\ c=-\frac{5}{2}\pm \frac{1}{4}\\ \text{Calculate each value of c:}\\ c=-\frac{5}{2}+\frac{1}{4}\text{ or }c=-\frac{5}{2}-\frac{1}{4}\\ c=-2.3\text{ or c = -2}\text{.8}\end{array}$

The solution set is $\{-2.3,-2.8\}.$