Answer
$$(-2, 0)$$
Work Step by Step
Let $x^2 + 2x \lt 0$ be a function $f$.
Find the $x$-intercepts by solving $x^2 + 2x=0$
Factor: $x^2 + 2x = x(x+2)$
$x=0$ or $x+2=0$
$x = 0$ or $x = -2$
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:
$(-∞, -2)(-2,0)(0, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -2)$
Test Value = $-3$
$f=x(x+2)$
$f=(-3)(-3+2)$
$f=3$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -2)$.
$(-2, 0)$
Test Value = $-1$
$f=x(x+2)$
$f=(-1)(-1+2)$
$f=-1$
Conclusion: $f (x) < 0$ for all $x$ in $(-2,0)$.
$(0,+∞)$
Test Value = $1$
$f=x(x+2)$
$f=(1)(1+2)$
$f=3$
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 $f (x) \lt 0$, where $f (x) = x^2+2x$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(-2,0)$.
Thus, the solution set of the given inequality is:
$$(-2, 0)$$
