Chapter 8.1, Problem 6MRE

To determine

To solve the equation over real numbers.

Answer to Problem 6MRE

The solution of the equation is $\left\{-4,\frac{1}{2}\right\}$ .

Explanation of Solution

Given information:

The given equation is:

  $\frac{w^2}{2}+\frac{7w}{4}=1$

Formula used:

The following formula is used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation :

The solution can be obtained by:

  $\begin{array}{l}\frac{w^2}{2}+\frac{7w}{4}=1\\ \frac{2w^2+7w}{4}=1\left[\text{Solve by using LCD}\right]\\ 2w^2+7w=4\\ 2w^2+7w-4=0\\ w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ w=\frac{-\left(7\right)\pm \sqrt{{\left(7\right)}^2-4\left(2\right)\left(-4\right)}}{2\left(2\right)}\\ w=\frac{-7\pm \sqrt{49+32}}{4}\\ w=\frac{-7\pm \sqrt{81}}{4}\\ w=\frac{-7+9}{4}\text{ or }\frac{-7-9}{4}\\ w=\frac{2}{4}\text{ or }\frac{-16}{4}\\ w=\frac{1}{2}\text{ or }-4\end{array}$

Hence, the solution of the equation is $\left\{-4,\frac{1}{2}\right\}$ .