Solve sqrt{4-x}-x=2

$$\sqrt { 4 - x } - x = 2$$

Solve for x

$$x=0$$

Solution Steps

Subtract -x from both sides of the equation.

$$\sqrt{4-x}=2+x$$

Square both sides of the equation.

$$\left(\sqrt{4-x}\right)^{2}=\left(2+x\right)^{2}$$

Calculate \sqrt{4-x} to the power of 2 and get 4-x.

$$4-x=\left(2+x\right)^{2}$$

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

$$4-x=4+4x+x^{2}$$

Subtract 4 from both sides.

$$4-x-4=4x+x^{2}$$

Subtract 4 from 4 to get 0.

$$-x=4x+x^{2}$$

Subtract 4x from both sides.

$$-x-4x=x^{2}$$

Combine -x and -4x to get -5x.

$$-5x=x^{2}$$

Subtract x^{2} from both sides.

$$-5x-x^{2}=0$$

Factor out x.

$$x\left(-5-x\right)=0$$

To find equation solutions, solve x=0 and -5-x=0.

$$x=0 x=-5$$

Substitute 0 for x in the equation \sqrt{4-x}-x=2.

$$\sqrt{4-0}-0=2$$

Simplify. The value x=0 satisfies the equation.

$$2=2$$

Substitute -5 for x in the equation \sqrt{4-x}-x=2.

$$\sqrt{4-\left(-5\right)}-\left(-5\right)=2$$

Simplify. The value x=-5 does not satisfy the equation.

$$8=2$$

Equation \sqrt{4-x}=x+2 has a unique solution.

$$x=0$$