## College Algebra (10th Edition)

Published by Pearson

# Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 117: 27

#### Answer

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

#### Work Step by Step

Add 4 on both sides of the equation to obtain: $\sqrt{3(x+10)}=x+4$ Square both sides to obtain: $3(x+10)=(x+4)^2 \\3(x)+3(10)=(x+4)^2 \\3x+30=(x+4)^2$ Use the rule $(a+b)^2 = a^2+2ab+b^2$ to obtain: $3x+30=x^2+2(x)(4)+4^2 \\3x+30=x^2+8x+16$ Subtract $3x$ and $30$ on both sides of the equation, and then combine like terms to obtain: $\begin{array}{ccc} &3x+30-3x-30&= &x^2+8x+16-3x-30 \\&0 &= &x^2+5x-14 \end{array}$ Factor the trinomial to obtain: $0=(x+7)(x-2)$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+7=0 &\text{ or } &x-2=0 \\&x=-7 &\text{ or } &x=2 \end{array}$ $-7$ is an extraneous solution since it does not satisfy the original equation. Thus, the solution set is $\left\{2\right\}$.

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