Chapter 12.3, Problem 22WE

To determine

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

Answer to Problem 22WE

The required answer is: solution set is - 0.42i, - 1.08i

Explanation of Solution

Given information:

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

Calculation:

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

(Convert fractions to decimals)

  $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

(Quadratic formula)

Where $a=1,b=1.5andc=0.67$

  $t=\frac{-(1.5)\pm \sqrt{{(1.5)}^2-4(1)(0.67)}}{2(1)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{-1.5\pm \sqrt{2.25-2.68}}{2}\\ =\frac{-1.5\pm \sqrt{-0.43}}{2}\end{array}$

(Simplify)

Since $\sqrt{-0.43}\approx 0.66i$

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

  $t\approx \frac{-1.5+0.66i}{2}=\frac{-0.84i}{2}\approx -0.42i$

Or   $t\approx \frac{-1.5-0.66i}{2}=\frac{-2.16i}{2}\approx -1.08i$

Therefore the solution set is $\frac{-1.5+\sqrt{-0.43}}{2},\frac{-1.5-\sqrt{-0.43}}{2}$ or - 0.42i, - 1.08i