Chapter 7, Problem 32CLR

To determine

Tofind: the roots of the equation.

Answer to Problem 32CLR

The roots of the equation is $y=\frac{-2+i}{3}\text{,}\frac{-2-i}{3}$ .

Explanation of Solution

Given information:

The equation is $9y^2+12y+5=0$ .

Formula used: the quadratic formula is,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation: to find the roots of the equation, first simplify and then use the formula of quadratic equation and then substitute and then simplify,

$\begin{array}{l}9y^2+12y+5=0\\ a=9,b=12,c=5\end{array}$

Now, use the quadratic formula,

$\begin{array}{l}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ y=\frac{-\left(12\right)\pm \sqrt{{\left(12\right)}^2-4\left(9\right)\left(5\right)}}{2\left(9\right)}\text{ [substitute]}\\ y=\frac{-12\pm \sqrt{144-180}}{18}\text{ [simplify]}\\ y=\frac{-12+6i}{18},\frac{-12-6i}{18}\\ y=\frac{-2+i}{3}\text{,}\frac{-2-i}{3}\text{ [simplify] }\end{array}$

Thus, Theroots of the equation is $y=\frac{-2+i}{3}\text{,}\frac{-2-i}{3}$ .