## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=4$
Squaring both sides and then isolating the radical expression, the given equation is equivalent to \begin{array}{l}\require{cancel} 1+\sqrt{x}=\sqrt{3x-3} \\\\ \left( 1+\sqrt{x} \right)^2=\left(\sqrt{3x-3}\right)^2 \\\\ (1)^2+2(1)(\sqrt{x})+(\sqrt{x})^2=3x-3 \\\\ 1+2\sqrt{x}+x=3x-3 \\\\ 2\sqrt{x}=3x-x-3-1 \\\\ 2\sqrt{x}=2x-4 .\end{array} Squaring both sides again, the given equation is equivalent to \begin{array}{l}\require{cancel} 2\sqrt{x}=2x-4 \\\\ \left( 2\sqrt{x} \right)^2=(2x-4)^2 \\\\ 4x=(2x)^2-2(2x)(4)+(-4)^2 \\\\ 4x=4x^2-16x+16 \\\\ 0=4x^2-16x-4x+16 \\\\ 0=4x^2-12x+16 \\\\ \dfrac{0}{4}=\dfrac{4x^2-12x+16}{4} \\\\ x^2-3x+4=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-4)(x+1)=0 .\end{array} Equating each factor to zero and then isolating the variable, the solutions are $x=\{-1,4 \}$ If $x=-1,$ then \begin{array}{l}\require{cancel} 1+\sqrt{x}=\sqrt{3x-3} \\\\ 1+\sqrt{-1}=\sqrt{3(-1)-3} \text{ (not a real number)} .\end{array} Hence, only $x=4$ satisfies the original equation.