## Intermediate Algebra (12th Edition)

$x=2$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{x^2-3x+3}=x-1 ,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( \sqrt{x^2-3x+3} \right)^2=(x-1)^2 \\\\ x^2-3x+3=(x-1)^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^2-3x+3=(x)^2-2(x)(1)+(1)^2 \\\\ x^2-3x+3=x^2-2x+1 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} (x^2-x^2)+(-3x+2x)=1-3 \\\\ -x=-2 \\\\ x=2 .\end{array} Upon checking, $x=2$ satisfies the original equation.