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 - Page 117: 26


The solution set is $\left\{4\right\}$.

Work Step by Step

Subtract 2 on both sides of the equation to obtain: $\sqrt{12-2x}=x-2$ Square both sides to obtain: $12-2x=(x-2)^2$ Use the rule $(a-b)^2 = a^2-2ab+b^2$ to obtain: $12-2x=x^2-2(x)(2) + 2^2 \\12-2x=x^2-4x+4$ Subtract $12$ and add $2x$ on both sides of the equation, and then combine like terms to obtain: $\begin{array}{ccc} &12-2x-12+2x &= &x^2-4x+4-12+2x \\&0 &= &x^2-2x-8 \end{array}$ Factor the trinomial to obtain: $0=(x-4)(x+2)$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x-4=0 &\text{ or } &x+2=0 \\&x=4 &\text{ or } &x=-2 \end{array}$ $-2$ is an extraneous solution since it does not satisfy the original equation. Thus, the solution set is $\left\{4\right\}$.
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