# Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 26

$$(-∞,\frac{5-\sqrt{10}}{3})∪(\frac{5+\sqrt{10}}{3},+∞)$$ Let $3x^2 \gt 10x -5$ be a function $f$. Express the inequality in the form $f(x) \gt 0$. Begin by the rewriting the inequality so that $0$ is on the right side. $3x^2 \gt 10x -5$ $3x^2 -10x +5\gt 10x -5 -10x +5$ $3x^2 -10x +5\gt 0$ Find the $x$-intercepts by solving $3x^2 -10x +5$ Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ where $a=3$, $b=-10$, $c=5$ $x = \frac{-(-10)±\sqrt{-10^2-4⋅3⋅5}}{2⋅3}$ $x = \frac{5+\sqrt{10}}{3}$ or $\frac{5-\sqrt{10}}{3}$ 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: $(-∞, \frac{5-\sqrt{10}}{3})(\frac{5-\sqrt{10}}{3},\frac{5+\sqrt{10}}{3})(\frac{5-\sqrt{10}}{3}, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, \frac{5-\sqrt{10}}{3})$ Test Value = $0$ $f=3x^2-10x+5$ $f=3(0)^2-10(0)-5$ $f=5$ Conclusion: $f (x) > 0$ for all $x$ in $(-∞,\frac{5-\sqrt{10}}{3})$. $(\frac{5-\sqrt{10}}{3}, \frac{5+\sqrt{10}}{3})$ Test Value = $1$ $f=3x^2-10x+5$ $f=3(1)^2-10(1)+5$ $f=-2$ Conclusion: $f (x) < 0$ for all $x$ in $(\frac{5-\sqrt{10}}{3}, \frac{5+\sqrt{10}}{3})$. $(\frac{5+\sqrt{10}}{3},+∞)$ Test Value = $3$ $f=3x^2-10x+5$ $f=3(3)^2-10(3)+5$ $f=2$ Conclusion: $f (x) > 0$ for all $x$ in $(\frac{5+\sqrt{10}}{3},+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \gt 0$, where $f (x) = 3x^2-10x-5$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(-∞,\frac{5-\sqrt{10}}{3})$ or $(\frac{5+\sqrt{10}}{3},+∞)$. Thus, the solution set of the given inequality is: $$(-∞,\frac{5-\sqrt{10}}{3})∪(\frac{5+\sqrt{10}}{3},+∞)$$ 