Which values of $x$ satisfy $x^{2}<1$
(a) $x<1$
(b) $x>1$
(c) $-2<x<2$
(d) $x<-2$
(e) $x<0$

## Solutions to List 8
Let's observe that if a number $a$ satisfies the inequality, then $-a$ also satisfies the inequality, so the values that satisfy the inequality form a symmetric set. Therefore, it suffices to consider the case where $x$ is positive. However, $(2-x)(1+x)=x+3-x^{2}>0$ is positive if $2-x$ is positive, so $x<2$. Since the solution is symmetric, we have that $-2<x<2$ is the solution to the initial equation. The correct option is (c).