Solve sqrt{4+2x-x^2}=x-2

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Combine -x^{2} and -x^{2} to get -2x^{2}.

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

Add 4x to both sides.

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

Combine 2x and 4x to get 6x.

$$4+6x-2x^{2}=4$$

Subtract 4 from both sides.

$$4+6x-2x^{2}-4=0$$

Subtract 4 from 4 to get 0.

$$6x-2x^{2}=0$$

Factor out x.

$$x\left(6-2x\right)=0$$

To find equation solutions, solve x=0 and 6-2x=0.

$$x=0 x=3$$

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

$$\sqrt{4+2\times 0-0^{2}}=0-2$$

Simplify. The value x=0 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Substitute 3 for x in the equation \sqrt{4+2x-x^{2}}=x-2.

$$\sqrt{4+2\times 3-3^{2}}=3-2$$

Simplify. The value x=3 satisfies the equation.

$$1=1$$

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

$$x=3$$