College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding: 22

Answer

The solution set is $\left\{-2\right\}$.

Work Step by Step

Square both sides to obtain: $x^2=(2\sqrt{-x-1})^2$ Use the rule $(ab)^m = a^mb^m$ to obtain: $x^2=2^2(\sqrt{-x-1})^2 \\x^2=4(-x-1) \\x^2=-4x-4$ Add $4x$ and $4$ on both sides of the equation to obtain: $x^2+4x+4 = -4x-4+4x+4 \\x^2+4x+4=0$ Factor the trinomial to obtain: $(x+2)(x+2) = 0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+2=0 &\text{ or } &x+2=0 \\&x=-2 &\text{ or } &x=-2 \end{array}$ Thus, the solution set is $\left\{-2\right\}$.
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