Solve x^2+left(4-2xright)^2=left(4-xright)^2

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

Solve for x

$$x=2$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-2x\right)^{2}.

$$x^{2}+16-16x+4x^{2}=\left(4-x\right)^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+16-16x=\left(4-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$5x^{2}+16-16x=16-8x+x^{2}$$

Subtract 16 from both sides.

$$5x^{2}+16-16x-16=-8x+x^{2}$$

Subtract 16 from 16 to get 0.

$$5x^{2}-16x=-8x+x^{2}$$

Add 8x to both sides.

$$5x^{2}-16x+8x=x^{2}$$

Combine -16x and 8x to get -8x.

$$5x^{2}-8x=x^{2}$$

Subtract x^{2} from both sides.

$$5x^{2}-8x-x^{2}=0$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-8x=0$$

Factor out x.

$$x\left(4x-8\right)=0$$

To find equation solutions, solve x=0 and 4x-8=0.

$$x=0 x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-2x\right)^{2}.

$$x^{2}+16-16x+4x^{2}=\left(4-x\right)^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+16-16x=\left(4-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$5x^{2}+16-16x=16-8x+x^{2}$$

Subtract 16 from both sides.

$$5x^{2}+16-16x-16=-8x+x^{2}$$

Subtract 16 from 16 to get 0.

$$5x^{2}-16x=-8x+x^{2}$$

Add 8x to both sides.

$$5x^{2}-16x+8x=x^{2}$$

Combine -16x and 8x to get -8x.

$$5x^{2}-8x=x^{2}$$

Subtract x^{2} from both sides.

$$5x^{2}-8x-x^{2}=0$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-8\right)^{2}.

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

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{8±8}{8}$$

Now solve the equation x=\frac{8±8}{8} when ± is plus. Add 8 to 8.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{8±8}{8} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=2 x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-2x\right)^{2}.

$$x^{2}+16-16x+4x^{2}=\left(4-x\right)^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+16-16x=\left(4-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$5x^{2}+16-16x=16-8x+x^{2}$$

Add 8x to both sides.

$$5x^{2}+16-16x+8x=16+x^{2}$$

Combine -16x and 8x to get -8x.

$$5x^{2}+16-8x=16+x^{2}$$

Subtract x^{2} from both sides.

$$5x^{2}+16-8x-x^{2}=16$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+16-8x=16$$

Subtract 16 from both sides.

$$4x^{2}-8x=16-16$$

Subtract 16 from 16 to get 0.

$$4x^{2}-8x=0$$

Divide both sides by 4.

$$\frac{4x^{2}-8x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{8}{4}\right)x=\frac{0}{4}$$

Divide -8 by 4.

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

Divide 0 by 4.

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

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.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=1 x-1=-1$$

Add 1 to both sides of the equation.

$$x=2 x=0$$