Chapter 12.3, Problem 25WE

To determine

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

Answer to Problem 25WE

The required answer is: solution set is 0.54i, - 0.77i

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{l}\frac{3}{2}{x}^2+\frac{1}{3}x+\frac{2}{3}=0\\ 1.5x^2+0.34x+0.67=0\end{array}$

(Convert fractions to decimals)

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

(Quadratic formula)

Where $a=1.5,b=0.34andc=0.67$

  $x=\frac{-(0.34)\pm \sqrt{{(0.34)}^2-4(1.5)(0.67)}}{2(1.5)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{-0.34\pm \sqrt{0.12-4.02}}{3}\\ =\frac{-0.34\pm \sqrt{-3.9}}{3}\end{array}$

(Simplify)

Since $\sqrt{-3.9}\approx 1.97i$

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

  $x\approx \frac{-0.34+1.97i}{3}=\frac{1.63i}{3}\approx 0.54i$

Or   $x\approx \frac{-0.34-1.97i}{3}=\frac{-2.31i}{3}\approx -0.77i$

Therefore the solution set is $\frac{-0.34+\sqrt{-3.9}}{3},\frac{-0.34-\sqrt{-3.9}}{3}$ or 0.54i, - 0.77i