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

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 4x to both sides.

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

Combine x and 4x to get 5x.

$$5x+4-x^{2}=4$$

Subtract 4 from both sides.

$$5x+4-x^{2}-4=0$$

Subtract 4 from 4 to get 0.

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

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{x+4}=x-2.

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

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

$$2=-2$$

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

$$\sqrt{5+4}=5-2$$

Simplify. The value x=5 satisfies the equation.

$$3=3$$

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

$$x=5$$