Chapter 9.2, Problem 60SR

To determine

Solve the equation and check the solutions.

Answer to Problem 60SR

The solution is: $-4$

Explanation of Solution

Given:

The equation: $2x^2+16x=-32$

Concept Used:

In algebra, a quadratic equation is any equation that can be rearranged in standard form $a{x}^2+bx+c=0$ as, where x represents an unknown, and a, b, and c represent known numbers, where a ? 0.

Calculation:

The equation: $2x^2+16x=-32$

  $\begin{array}{l}2x^2+16x=-32\text{Original}\text{Equation}\\ 2x^2+16x+32=32-32\text{Add}\text{ 32 from both sides}\text{.}\\ 2x^2+16x+32=0\text{Simplify}\\ \text{2(}{x}^2+8x-16)=0\text{Factor}\text{GCF}\text{2}\text{from}\text{each}\text{terms}\\ 2(x+4)(x+4)=0\text{Factor}\end{array}$

Solve for x using the zero product property.

  $\begin{array}{l}\text{ 2}\ne \text{0}\\ x+4=0\text{and}x+4=0\\ x=-4\text{and}x=-4\end{array}$

The solution is: $-4$

Check:

  $\begin{array}{l}2x^2+16x=-32\\ $ 2{(-4)}^2+16·(-4)=-32$2·16-64=-32\\ -32=-32\end{array}$

Thus, the solution is: $-4$