Chapter 8, Problem 12GR

Solution

To solve:The equation $\frac{1}{4}{x}^2+\frac{7}{2}x=-12$ .

  $x=-6$ and $x=-8$ .

Given Information:

  $\frac{1}{4}{x}^2+\frac{7}{2}x=-12$

Concept Used:

Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where $a=$ the coefficient in front of or the number beside $x^2$

  $b=$ the coefficient in front of or the number beside $x$

  $c=$ the constant

Calculation:

  $\begin{array}{c}\frac{1}{4}{x}^2+\frac{7}{2}x=-12\\ \frac{1}{4}{x}^2\cdot 4+\frac{7}{2}x\cdot 4=-12\cdot 4\left[\text{multiply both side by 4}\right]\\ x^2+14x=-48\\ x^2+14x+48=-48+48\left[\text{add 48 to both side}\right]\\ x^2+14x+48=0\\ x=\frac{-14\pm \sqrt{{14}^2-4\cdot 1\cdot 48}}{2\cdot 1}\left[\text{using quadratic formula}\right]\\ x=\frac{-14\pm 2}{2\cdot 1}\\ x=\frac{-14+2}{2},x=\frac{-14-2}{2}\\ x=\frac{-12}{2},x=\frac{-16}{2}\\ x=-6,x=-8\end{array}$

Thus,the solutions are $x=-6$ or $x=-8$ .