Answer
$$(-∞, -\frac{2}{3}] ∪ [\frac{1}{3},+∞)$$
Work Step by Step
Let $9x^2 + 3x - 2 \geq 0$ be a function $f$.
Find the $x$-intercepts by solving $9x^2 + 3x - 2=0$
Factor: $9x^2 + 3x - 2 = (3x-1)(3x+2)$
$3x-1=0$ or $3x+2=0$
$x = \frac{1}{3}$ or $x = -\frac{2}{3}$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, -\frac{2}{3})(-\frac{2}{3},\frac{1}{3})(\frac{1}{3}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -\frac{2}{3})$
Test Value = $-1$
$f=(3x-1)(3x+2)$
$f=(3(-1)-1)(3(-1)+2)$
$f=4$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -\frac{2}{3})$.
$(-\frac{2}{3}, \frac{1}{3})$
Test Value = $0$
$f=(3x-1)(3x+2)$
$f=(3(0)-1)(3(0)+2)$
$f=-2$
Conclusion: $f (x) < 0$ for all $x$ in $(-\frac{2}{3}, \frac{1}{3})$.
$(\frac{1}{3},+∞)$
Test Value = $1$
$f=(3x-1)(3x+2)$
$f=(3(1)-1)(3(1)+2)$
$f=10$
Conclusion: $f (x) > 0$ for all $x$ in $(\frac{1}{3},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \geq 0$, where $f (x) = 9x^2+3x-2$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(-∞,-\frac{2}{3})$ or $(\frac{1}{3}, +∞)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $9x^2+3x-2 = 0$, namely $\frac{1}{3}$ and $-\frac{2}{3}$, in the solution set.
Thus, the solution set of the given inequality is:
$$(-∞, -\frac{2}{3}] ∪ [\frac{1}{3},+∞)$$
