Answer
$$(-∞, -\frac{3}{2}] ∪ [0,+∞)$$
Work Step by Step
Let $2x^2 +3x \gt 0$ be a function $f$.
Find the $x$-intercepts by solving $2x^2 +3x=0$
Factor: $2x^2 +3x = x(2x+3)$
$x=0$ or $2x+3=0$
$x = 0$ or $x = -\frac{3}{2}$
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:
$(-∞, -\frac{3}{2})(-\frac{3}{2},0)(0, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -\frac{3}{2})$
Test Value = $-2$
$f=x(2x+3)$
$f=(-2)(2(-2)+3)$
$f=2$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -\frac{3}{2})$.
$(-\frac{3}{2}, 0)$
Test Value = $-1$
$f=x(2x+3)$
$f=(-1)(2(-1)+3)$
$f=-1$
Conclusion: $f (x) < 0$ for all $x$ in $(-\frac{3}{2}, 0)$.
$(0,+∞)$
Test Value = $1$
$f=x(2x+3)$
$f=(1)(2(1)+3)$
$f=5$
Conclusion: $f (x) > 0$ for all $x$ in $(0,+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \gt 0$, where $f (x) = 2x^2+3x$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(-∞,-\frac{3}{2})$ or $(0, +∞)$.
Thus, the solution set of the given inequality is:
$$(-∞, -\frac{3}{2}] ∪ [0,+∞)$$
