Solve x^2+left(2-xright)^2=6

$${ x }^{ 2 } + { \left(2-x \right) }^{ 2 } =6$$

Solve for x

$$x=\sqrt{2}+1\approx 2.414213562$$

$$x=1-\sqrt{2}\approx -0.414213562$$

Steps Using the Quadratic Formula

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

$$x^{2}+4-4x+x^{2}=6$$

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

$$2x^{2}+4-4x=6$$

Subtract 6 from both sides.

$$2x^{2}+4-4x-6=0$$

Subtract 6 from 4 to get -2.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+16}}{2\times 2}$$

Add 16 to 16.

$$x=\frac{-\left(-4\right)±\sqrt{32}}{2\times 2}$$

Take the square root of 32.

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

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{2}}{4}$$

Now solve the equation x=\frac{4±4\sqrt{2}}{4} when ± is plus. Add 4 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+4}{4}$$

Divide 4+4\sqrt{2} by 4.

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

Now solve the equation x=\frac{4±4\sqrt{2}}{4} when ± is minus. Subtract 4\sqrt{2} from 4.

$$x=\frac{4-4\sqrt{2}}{4}$$

Divide 4-4\sqrt{2} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{2}+4-4x+x^{2}=6$$

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

$$2x^{2}+4-4x=6$$

Subtract 4 from both sides.

$$2x^{2}-4x=6-4$$

Subtract 4 from 6 to get 2.

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

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -4 by 2.

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

Divide 2 by 2.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add 1 to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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