Solve sqrt{9-3x}=3-x

$$\sqrt{ 9-3x } =3-x$$

Solve for x

$$x=0$$

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{9-3x}\right)^{2}=\left(3-x\right)^{2}$$

Calculate \sqrt{9-3x} to the power of 2 and get 9-3x.

$$9-3x=\left(3-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$9-3x=9-6x+x^{2}$$

Subtract 9 from both sides.

$$9-3x-9=-6x+x^{2}$$

Subtract 9 from 9 to get 0.

$$-3x=-6x+x^{2}$$

Add 6x to both sides.

$$-3x+6x=x^{2}$$

Combine -3x and 6x to get 3x.

$$3x=x^{2}$$

Subtract x^{2} from both sides.

$$3x-x^{2}=0$$

Factor out x.

$$x\left(3-x\right)=0$$

To find equation solutions, solve x=0 and 3-x=0.

$$x=0 x=3$$

Substitute 0 for x in the equation \sqrt{9-3x}=3-x.

$$\sqrt{9-3\times 0}=3-0$$

Simplify. The value x=0 satisfies the equation.

$$3=3$$

Substitute 3 for x in the equation \sqrt{9-3x}=3-x.

$$\sqrt{9-3\times 3}=3-3$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{9-3x}=3-x.

$$x=0 x=3$$