Chapter 12.3, Problem 23WE

To determine

To solve: the equation $2c^2+\frac{1}{2}c+\frac{2}{3}=0$

Answer to Problem 23WE

The required answer is: solution set is 0.44i, - 0.69i

Explanation of Solution

Given information:

The given equation: $2c^2+\frac{1}{2}c+\frac{2}{3}=0$

Calculation:

  $\begin{array}{l}2c^2+\frac{1}{2}c+\frac{2}{3}=0\\ 2c^2+0.5c+0.67=0\end{array}$

(Convert fractions to decimals)

  $c=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

(Quadratic formula)

Where $a=2,b=0.5andc=0.67$

  $c=\frac{-(0.5)\pm \sqrt{{(0.5)}^2-4(2)(0.67)}}{2(2)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{-0.5\pm \sqrt{0.25-5.36}}{4}\\ =\frac{-0.5\pm \sqrt{-5.11}}{4}\end{array}$

(Simplify)

Since $\sqrt{-5.11}\approx 2.26i$

(The square root of -1 is called an imaginary number and is represented by i)

  $c\approx \frac{-0.5+2.26i}{4}=\frac{1.76i}{4}\approx 0.44i$

Or   $c\approx \frac{-0.5-2.26i}{4}=\frac{-2.76i}{4}\approx -0.69i$

Therefore the solution set is $\frac{-0.5+\sqrt{-5.11}}{4},\frac{-0.5-\sqrt{-5.11}}{4}$ or 0.44i, - 0.69i