Solve sqrt{x^2+2}+x=1

$$\sqrt { x ^ { 2 } + 2 } + x = 1$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{x^{2}+2}=1-x$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+2}\right)^{2}=\left(1-x\right)^{2}$$

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

$$x^{2}+2=\left(1-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$x^{2}+2=1-2x+x^{2}$$

Add 2x to both sides.

$$x^{2}+2+2x=1+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+2+2x-x^{2}=1$$

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

$$2+2x=1$$

Subtract 2 from both sides.

$$2x=1-2$$

Subtract 2 from 1 to get -1.

$$2x=-1$$

Divide both sides by 2.

$$x=\frac{-1}{2}$$

Fraction \frac{-1}{2} can be rewritten as -\frac{1}{2} by extracting the negative sign.

$$x=-\frac{1}{2}$$

Substitute -\frac{1}{2} for x in the equation \sqrt{x^{2}+2}+x=1.

$$\sqrt{\left(-\frac{1}{2}\right)^{2}+2}-\frac{1}{2}=1$$

Simplify. The value x=-\frac{1}{2} satisfies the equation.

$$1=1$$

Equation \sqrt{x^{2}+2}=1-x has a unique solution.

$$x=-\frac{1}{2}$$