# Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 15

$$(-1,-\frac{3}{4})$$ 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: $$(-1,-\frac{3}{4})$$ 