## Intermediate Algebra (12th Edition)

$x=1$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{5-x}=x+1 ,$ 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{5-x} \right)^2=(x+1)^2 \\\\ 5-x=(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} 5-x=(x)^2+2(x)(1)+(1)^2 \\\\ 5-x=x^2+2x+1 .\end{array} Using the propertis of equality, in the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -x^2+(-x-2x)+(5-1)=0 \\\\ -x^2-3x+4=0 \\\\ -1(-x^2-3x+4)=(0)(-1) \\\\ x^2+3x-4=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+4)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property) results to \begin{array}{l}\require{cancel} x+4=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+4=0 \\\\ x=-4 \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking, only $x=1$ satisfies the original equation.