## Precalculus (6th Edition)

The solution is $x=9$
$\sqrt{4x}-x+3=0$ Leave the square root alone on the left side of the equation: $\sqrt{4x}=x-3$ Square both sides: $(\sqrt{4x})^{2}=(x-3)^{2}$ $4x=x^{2}-6x+9$ Take $4x$ to the right side and simplify: $0=x^{2}-6x+9-4x$ $0=x^{2}-10x+9$ Rearrange: $x^{2}-10x+9=0$ Solve by factoring: $(x-1)(x-9)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-1=0$ $x=1$ $x-9=0$ $x=9$ Check the solutions found by substituting them into the original equation: $x=1$ $\sqrt{4(1)}-1+3=0$ $2-1+3=0$ $4\ne0$ False $x=9$ $\sqrt{4(9)}-9+3=0$ $6-9+3=0$ $0=0$ True The solution is $x=9$