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: 25


The solution set is $\left\{8\right\}$.

Work Step by Step

Subtract 3 on both sides of the equation to obtain: $\sqrt{3x+1}=x-3$ Square both sides to obtain: $3x+1=(x-3)^2$ Use the rule $(a-b)^2 = a^2-2ab+b^2$ to obtain: $3x+1=x^2-2(x)(3) + 3^2 \\3x+1=x^2-6x+9$ Subtract $3x$ and $1$ on both sides of the equation to obtain: $\begin{array}{ccc} &3x+1-3x-1 &= &x^2-6x+9-3x-1 \\&0 &= &x^2-9x+8 \end{array}$ Factor the trinomial to obtain: $0=(x-8)(x-1)$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x-8=0 &\text{ or } &x-1=0 \\&x=8 &\text{ or } &x=1 \end{array}$ $1$ is an extraneous solution since it does not satisfy the original equation. Thus, the solution set is $\left\{8\right\}$.
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