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: 19


The solution set is $\left\{3\right\}$.

Work Step by Step

Square both sides to obtain: $15-2x=x^2$ Subtract $15$ and add $2x$ on both sides of the equation to obtain: $15-2x-15+2x=x^2+2x-15 \\0=x^2+2x-15$ Factor the trinomial to obtain: $(x+5)(x-3) = 0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+5=0 &\text{ or } &x-3=0 \\&x=-5 &\text{ or } &x=3 \end{array}$ Note that $-5$ is an extraneous solution since the principal square root is never negative. Thus, the solution set is $\left\{3\right\}$.
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