Chapter 4.7, Problem 38WE

To determine

To solve the equation.

Answer to Problem 38WE

The values of equationsare $\text{2 and }-\frac{1}{2}$

Explanation of Solution

Given information:

  $\frac{1}{3}{x}^2-\frac{1}{2}x-\frac{1}{3}=0$

Calculation:

The given equation already has 0 as one side,and the other side is already factored. Simply set each factored equal to 0 and solve.

Thedouble rootscan be calculated as

  $\begin{array}{l}\frac{1}{3}{x}^2-\frac{1}{2}x-\frac{1}{3}=0\\ \frac{6}{3}{x}^2-\frac{6}{2}x-\frac{6}{3}=0\left[\text{multiply the whole equation by 6}\right]\\ 2x^2-3x-2=0\left[\text{divide numerator by denominator}\right]\\ 2x^2-\left(4x-x\right)-2=0\left[\text{write 3}x=\left(4x-x\right)\right]\\ 2x^2-4x+x-2=0\\ x\left(2x+1\right)-2\left(2x+1\right)=0\left[\begin{array}{l}\text{the first and third term has a common factor,}x\text{ and}\\ \text{the second and fourth has a common factor,2}\end{array}\right]\\ \left(2x+1\right)\left(x-2\right)=0\end{array}$

Therefore,

  $\begin{array}{l}\left(2x+1\right)\text{=0 or }\left(x-2\right)=0\left[\text{use the zero-product property}\right]\\ x=-\frac{1}{2}\text{ or }x\text{=2 }\end{array}$

Hence,

The values of equations are $\text{2 and }-\frac{1}{2}$