## College Algebra (10th Edition)

$-1\lt x\lt 1$
We know that the solution to the inequality $x^2\lt a$ is equivalent to: $-\sqrt{a}\lt x\lt \sqrt{a}$ Therefore: $x^{2}\lt 1$ $-\sqrt{1}\lt x\lt \sqrt{1}$ $-1\lt x\lt 1$