Chapter 8 Rational Functions - 8.1 Model Inverse and Joint Variation - 8.1 Exercises - Mixed Review - Page 557: 55

$x=12$

Given: $x-6=\sqrt 3x$ Square both sides: $(x-6)^2=(\sqrt 3x)^2\\x^2-12x+36=3x\\x^2-15x+36=0$ Solve for x: $x=12 \lor x=3$ Check: $12-6=6=\sqrt 3(12)=6$ $3-6=-3\ne\sqrt 3(3)=3$ Hence, the solution is $x=12$; there is an extraneous solution.