## College Algebra 7th Edition

$-1, 0, \frac{2}{3}, \frac{5}{6}, 1$
Subtract 2 to both sides of the inequality to obtain: $x^2+2-2\lt 4-2 \\x^2 \lt 2$ Thus, the numbers that will satisfy the given inequality are the ones whose squares are less than $2$. Note that among the elements of $S$, the following have squares that are less than 2: $(-1)^2=1 \\0^2=0 \\(\frac{2}{3})^2=\frac{4}{9} \\(\frac{5}{6})^2=\frac{25}{36} \\1^2=1$ Therefore, the elements of $S$ that satisfy the given inequality are: $-1, 0, \frac{2}{3}, \frac{5}{6}, 1$