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

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

Solve for x

$$x=6$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 4x to both sides.

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

Combine 2x and 4x to get 6x.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Factor out x.

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

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

$$x=0 x=6$$

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

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

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

$$2=-2$$

Substitute 6 for x in the equation \sqrt{2x+4}=x-2.

$$\sqrt{2\times 6+4}=6-2$$

Simplify. The value x=6 satisfies the equation.

$$4=4$$

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

$$x=6$$