Solve x+sqrt{x^2+64}=2

$$x + \sqrt { x ^ { 2 } + 64 } = 2$$

Solve for x

$$x=-15$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{x^{2}+64}=2-x$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+64}\right)^{2}=\left(2-x\right)^{2}$$

Calculate \sqrt{x^{2}+64} to the power of 2 and get x^{2}+64.

$$x^{2}+64=\left(2-x\right)^{2}$$

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

$$x^{2}+64=4-4x+x^{2}$$

Add 4x to both sides.

$$x^{2}+64+4x=4+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+64+4x-x^{2}=4$$

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

$$64+4x=4$$

Subtract 64 from both sides.

$$4x=4-64$$

Subtract 64 from 4 to get -60.

$$4x=-60$$

Divide both sides by 4.

$$x=\frac{-60}{4}$$

Divide -60 by 4 to get -15.

$$x=-15$$

Substitute -15 for x in the equation x+\sqrt{x^{2}+64}=2.

$$-15+\sqrt{\left(-15\right)^{2}+64}=2$$

Simplify. The value x=-15 satisfies the equation.

$$2=2$$

Equation \sqrt{x^{2}+64}=2-x has a unique solution.

$$x=-15$$