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

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

Solve for a

$$a=-2$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract a^{2} from both sides.

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

Combine a^{2} and -a^{2} to get 0.

$$-4=4a+4$$

Swap sides so that all variable terms are on the left hand side.

$$4a+4=-4$$

Subtract 4 from both sides.

$$4a=-4-4$$

Subtract 4 from -4 to get -8.

$$4a=-8$$

Divide both sides by 4.

$$a=\frac{-8}{4}$$

Divide -8 by 4 to get -2.

$$a=-2$$

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

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

Simplify. The value a=-2 satisfies the equation.

$$0=0$$

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

$$a=-2$$