Solve x^2+(2-x)^2=(2x)^2

$$x ^ { 2 } + ( 2 - x ) ^ { 2 } = ( 2 x ) ^ { 2 }$$

Solve for x (complex solution)

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-\left(\sqrt{3}+1\right)\approx -2.732050808$$

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}.

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

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

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

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.

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

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

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

Square -4.

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

Multiply -4 times -2.

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

Multiply 8 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-2\right)}$$

Add 16 to 32.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-2\right)}$$

Take the square root of 48.

$$x=\frac{-\left(-4\right)±4\sqrt{3}}{2\left(-2\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±4\sqrt{3}}{-4}$$

Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is plus. Add 4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+4}{-4}$$

Divide 4+4\sqrt{3} by -4.

$$x=-\left(\sqrt{3}+1\right)$$

Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is minus. Subtract 4\sqrt{3} from 4.

$$x=\frac{4-4\sqrt{3}}{-4}$$

Divide 4-4\sqrt{3} by -4.

$$x=\sqrt{3}-1$$

The equation is now solved.

$$x=-\left(\sqrt{3}+1\right) x=\sqrt{3}-1$$

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}.

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

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

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

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

Divide -4 by -2.

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

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.

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

Square 1.

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

Add 2 to 1.

$$x^{2}+2x+1=3$$

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}.

$$\left(x+1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{3}$$

Simplify.

$$x+1=\sqrt{3} x+1=-\sqrt{3}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

Solve for x

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-\sqrt{3}-1\approx -2.732050808$$

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}.

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

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

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

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.

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

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

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

Square -4.

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

Multiply -4 times -2.

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

Multiply 8 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-2\right)}$$

Add 16 to 32.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-2\right)}$$

Take the square root of 48.

$$x=\frac{-\left(-4\right)±4\sqrt{3}}{2\left(-2\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±4\sqrt{3}}{-4}$$

Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is plus. Add 4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+4}{-4}$$

Divide 4+4\sqrt{3} by -4.

$$x=-\left(\sqrt{3}+1\right)$$

Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is minus. Subtract 4\sqrt{3} from 4.

$$x=\frac{4-4\sqrt{3}}{-4}$$

Divide 4-4\sqrt{3} by -4.

$$x=\sqrt{3}-1$$

The equation is now solved.

$$x=-\left(\sqrt{3}+1\right) x=\sqrt{3}-1$$

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}.

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

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

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

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

Divide -4 by -2.

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

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.

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

Square 1.

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

Add 2 to 1.

$$x^{2}+2x+1=3$$

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}.

$$\left(x+1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{3}$$

Simplify.

$$x+1=\sqrt{3} x+1=-\sqrt{3}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$