Answer
Solution set: $(-\displaystyle \infty, -\frac{2}{3}]\cup[\frac{1}{3},\infty)$
Work Step by Step
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$f(x)= 9x^{2}+3x-2\geq 0$
factor the trinomial...
find factors of $9(-2)=-18$ that add to $3:$
($6$ and $-3$ )
$9x^{2}+3x-2=9x^{2}+6x-3x-2= \quad$ ... in pairs ...
$=3x(3x+2)-(3x+2)=(3x-1)(3x+2)$
$f(x)=(3x-1)(3x+2)\geq 0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(3x-1)(3x+2)=0$
$x=\displaystyle \frac{1}{3}$ or $x=\displaystyle \frac{-2}{3}$
3. Locate these boundary points on a number line, thereby dividing the number line into intervals.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & (-\infty, \frac{-2}{3}) & (-\frac{2}{3},\frac{1}{3}) & (\frac{1}{3},\infty)\\
a=test.val. & -1 & 0 & 1\\
f(a) & (-1)(-1) & (-1)(2) & (2)(5)\\
f(a) \geq 0 ? & T & F & T
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\displaystyle \infty, -\frac{2}{3}]\cup[\frac{1}{3},\infty)$
