Answer
$-4\leq x\leq 4$
Work Step by Step
We know that the solution to the inequality:
$x^2\lt a$
Is equivalent to:
$-\sqrt{a}\lt x\lt \sqrt{a}$
Therefore:
$x^{2}\leq 16$
$-\sqrt{16}\leq x\leq\sqrt{16}$
$-4\leq x\leq 4$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 1 - Section 1.6 - Equations and Inequalities Involving Absolute Value - 1.6 Assess Your Understanding - Page 134: 89
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