## Algebra and Trigonometry 10th Edition

f(x)=g(x) (Solve for x) $\sqrt x -4 = 2-x$ $\sqrt x=6-x$ $x = x^2 -12x + 36$ $x^2-13x+36=0$ (x-9)(x-4)=0 x = 9 or 4 Plug in the solutions in the original equation to verify the solution for x = 9 LHS = $\sqrt 9 -4 = 3-4=-1$ $RHS = 2-9 = -7$ not correct for x = 4 LHS = $\sqrt 4 -4 = 2-4=-2$ RHS = 2-4 = -2 LHS=RHS hence x = 4 is a valid solution