Answer
$$(-3,\frac{5}{2})$$
Work Step by Step
Let $2x^2 + x \lt 15 $ be a function $f$.
Express the inequality in the form $f(x) \lt 0$. Begin by the rewriting the inequality so that $0$ is on the right side.
$2x^2 + x \lt 15 $
$2x^2 + x -15 \lt 15-15 $
$2x^2 + x -15 \lt 0 $
Find the $x$-intercepts by solving $2x^2 + x -15=0$
Factor: $2x^2 + x -15 = (2x-5)(x+3)$
$2x-5=0$ or $x+3=0$
$x = \frac{5}{2}$ or $x = -3$
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:
$(-∞, -3)(-3,\frac{5}{2})(\frac{5}{2}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -3)$
Test Value = $-4$
$f=(2x-5)(x+3)$
$f=(2(-4)-5)(-4+3)$
$f=13$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -3)$.
$(-3, \frac{5}{2})$
Test Value = $0$
$f=(2x-5)(x+3)$
$f=(2(0)-5)(0+3)$
$f=-15$
Conclusion: $f (x) < 0$ for all $x$ in $(-3, \frac{5}{2})$.
$(\frac{5}{2},+∞)$
Test Value = $3$
$f=(2x-5)(x+3)$
$f=(2(3)-5)(3+3)$
$f=6$
Conclusion: $f (x) > 0$ for all $x$ in $(\frac{5}{2},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \lt 0$, where $f (x) = 2x^2+x-15$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(-3,\frac{5}{2})$.
Thus, the solution set of the given inequality is:
$$(-3,\frac{5}{2})$$
