Answer
$$(-1,-\frac{3}{4})$$
Work Step by Step
Let $4x^2 + 7x \lt -3 $ 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.
$4x^2 + 7x \lt -3 $
$4x^2 + 7x +3\lt -3+3 $
$4x^2 + 7x +3\lt 0$
Find the $x$-intercepts by solving $4x^2 + 7x +3=0$
Factor: $4x^2 + 7x +3 = (4x+3)(x+1)$
$4x+3=0$ or $x+1=0$
$x = -\frac{3}{4}$ or $x = -1$
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:
$(-∞, -1)(-1,-\frac{3}{4})(-\frac{3}{4}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -1)$
Test Value = $-2$
$f=(4x+3)(x+1)$
$f=(4(-2)+3)(-2+1)$
$f=5$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -1)$.
$(-1, -\frac{3}{4})$
Test Value = $-\frac{7}{8}$
$f=(4x+3)(x+1)$
$f=(4(-\frac{7}{8})+3)(-\frac{7}{8}+1)$
$f=-\frac{1}{16}$
Conclusion: $f (x) < 0$ for all $x$ in $(-1, -\frac{3}{4})$.
$(-\frac{3}{4},+∞)$
Test Value = $0$
$f=(4x+3)(x+1)$
$f=(4(0)+3)(0+1)$
$f=3$
Conclusion: $f (x) > 0$ for all $x$ in $(-\frac{3}{4},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \lt 0$, where $f (x) = 4x^2+7x+3$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(-1,-\frac{3}{4})$.
Thus, the solution set of the given inequality is:
$$(-1,-\frac{3}{4})$$
