## 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: 24

#### Answer

The solution set is $\left\{\right\}$ or $\emptyset$.

#### Work Step by Step

Square both sides to obtain: $3-x+x^2=(x-2)^2$ Use the rule $(a-b)^2 = a^2-2ab+b^2$ to obtain: $3-x+x^2=x^2-2(x)(2)+2^2 \\3-x+x^2=x^2-4x+4$ Put all terms on the left side of the equation, and then combine like terms. Note that when a term is transferred to the other side of an equation, its sign/operation changes to its opposite. $3-x+x^2-x^2+4x-4=0 \\(x^2-x^2)+(-x+4x) + (3-4) = 0 \\3x-1=0$ Add $1$ on both sides of the equation to obtain: $3x=1$ Divide 3 on both sides of the equation to obtain: $x=\dfrac{1}{3}$ However, note that when $\frac{1}{3}$ is substituted to $x$ in the equation, the right side becomes $\frac{1}{3} - 2 = -\frac{5}{3}$ A principal square root is never negative. This means that $\frac{1}{3}$ is an extraneous solution. Thus, the solution set is $\left\{\right\}$ or $\emptyset$.

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