## Intermediate Algebra (6th Edition)

$x=6$
Using the properties of equality, the given equation, $x-\sqrt[]{x-2}=4 ,$ is equivalent to \begin{array}{l}\require{cancel} x-4=\sqrt[]{x-2} .\end{array} Raising both sides of the equation above to the second power, then the solution/s is/are \begin{array}{l}\require{cancel} (x-4)^2=x-2 \\\\ (x)^2+2(x)(-4)+(-4)^2=x-2 \\\\ x^2-8x+16=x-2 \\\\ x^2+(-8x-x)+(16+2)=0 \\\\ x^2-9x+18=0 \\\\ (x-3)(x-6)=0 \\\\ x=\{ 3,6 \} .\end{array} Upon checking, only $x=6$ satisfies the original equation.