## Algebra: A Combined Approach (4th Edition)

$x=5$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\sqrt{2x-1}+2=x ,$ isolate first the radical. Then raise both sides to the exponent equal to the index of the radical. Use concepts of solving quadratic equations to find the values of the variable. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} \sqrt{2x-1}=x-2 .\end{array} Get rid of the radical symbol by raising both sides to the exponent equal to $2 ,$ the same index as the radical. Hence, the equation above becomes \begin{array}{l}\require{cancel} (\sqrt{2x-1})^2=(x-2)^2 \\\\ 2x-1=(x-2)^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} 2x-1=(x)^2-2(x)(2)+(2)^2 \\\\ 2x-1=x^2-4x+4 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -x^2+(2x+4x)+(-1-4)=0 \\\\ -x^2+6x-5=0 \\\\ -1(-x^2+6x-5)=-1(0) \\\\ x^2-6x+5=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} (x-5)(x-1)=0 .\end{array} Equating each factor to $0$ (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} x-5=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-5=0 \\\\ x=5 \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking, only $x=5$ satisfies the original equation.