# Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 24

$$[1-\sqrt{3},1+\sqrt{3}]$$ Let $x^2 \leq 2x +2$ 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. $x^2 \leq 2x +2$ $x^2 -2x -2\leq 2x +2-2x -2$ $x^2 -2x -2\leq 0$ Find the $x$-intercepts by solving $x^2 -2x -2=0$ Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ where $a=1$, $b=-2$, $c=-2$ $x = \frac{-(-2)±\sqrt{-2^2-4⋅1⋅-2}}{2⋅1}$ $x = 1+ \sqrt{3}$ or $1- \sqrt{3}$ 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: $(-∞, 1- \sqrt{3})(1- \sqrt{3},1+ \sqrt{3})(1+ \sqrt{3}, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, 1- \sqrt{3})$ Test Value = $-1$ $f=x^2-2x-2$ $f=-1^2-2(-1)-2$ $f=-1$ Conclusion: $f (x) < 0$ for all $x$ in $(-∞,1-\sqrt{3})$. $(1-\sqrt{3}, 1+sqrt{3})$ Test Value = $0$ $f=x^2-2x-2$ $f=0^2-2(0)-2$ $f=-2$ Conclusion: $f (x) < 0$ for all $x$ in $(1-\sqrt{3}, 1+\sqrt{3})$. $(1+\sqrt{3},+∞)$ Test Value = $3$ $f=x^2-2x-2$ $f=3^2-2(3)-2$ $f=1$ Conclusion: $f (x) > 0$ for all $x$ in $(1+\sqrt{3},+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \leq 0$, where $f (x) = x^2-2x-2$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(1-\sqrt{3},1+\sqrt{3})$. However, because the inequality involves $\leq$ (less than or equal to), we must also include the solutions of $x^2-2x-2$, namely $1-\sqrt{3}$ and $1+\sqrt{3}$, in the solution set. Thus, the solution set of the given inequality is: $$[1-\sqrt{3},1+\sqrt{3}]$$ 