Solve 1+x^2=2

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

Solve for x

$$x=1$$

$$x=-1$$

Steps Using Difference of Squares

Subtract 2 from both sides.

$$1+x^{2}-2=0$$

Subtract 2 from 1 to get -1.

$$-1+x^{2}=0$$

Consider -1+x^{2}. Rewrite -1+x^{2} as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

Steps by Finding Square Root

Subtract 1 from both sides.

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

Subtract 1 from 2 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

$$1+x^{2}-2=0$$

Subtract 2 from 1 to get -1.

$$-1+x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{0±2}{2}$$

Now solve the equation x=\frac{0±2}{2} when ± is plus. Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{0±2}{2} when ± is minus. Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$