Chapter 8.6, Problem 3BCYP

To determine

To find: The value of each equation.

Answer to Problem 3BCYP

The value is $y=\frac{2}{3}$ .

Explanation of Solution

Given information:

The given information is $y^2-\frac{4}{3}y+\frac{4}{9}=0$ .

Calculation:

Consider the equation:

  $y^2-\frac{4}{3}y+\frac{4}{9}=0$

Solve the equation.

  $\begin{array}{c}{y}^2-\frac{4}{3}y+\frac{4}{9}=0\\ \frac{9y^2-12y+4}{9}=0\\ 9y^2-12y+4=0\end{array}$

Using the formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ :

  $\begin{array}{c}y=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\left(9\right)\left(4\right)}}{2\left(9\right)}\\ y=\frac{-\left(-12\right)\pm \sqrt{0}}{2\left(9\right)}\\ y=\frac{12}{2\left(9\right)}\end{array}$

  $\begin{array}{l}y=\frac{6}{9}\\ y=\frac{2}{3}\end{array}$

Therefore, the value is $y=\frac{2}{3}$ .