Solve 8+left(4-2xright)^2+x^2=left(4-xright)^2

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

Solve for x (complex solution)

$$x=1+i$$

$$x=1-i$$

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}.

$$8+16-16x+4x^{2}+x^{2}=\left(4-x\right)^{2}$$

Add 8 and 16 to get 24.

$$24-16x+4x^{2}+x^{2}=\left(4-x\right)^{2}$$

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

$$24-16x+5x^{2}=\left(4-x\right)^{2}$$

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

$$24-16x+5x^{2}=16-8x+x^{2}$$

Subtract 16 from both sides.

$$24-16x+5x^{2}-16=-8x+x^{2}$$

Subtract 16 from 24 to get 8.

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

Add 8x to both sides.

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

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

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

Subtract x^{2} from both sides.

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

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

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

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.

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

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

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

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-\left(-8\right)±\sqrt{64-128}}{2\times 4}$$

Add 64 to -128.

$$x=\frac{-\left(-8\right)±\sqrt{-64}}{2\times 4}$$

Take the square root of -64.

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

The opposite of -8 is 8.

$$x=\frac{8±8i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{8±8i}{8}$$

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

$$x=\frac{8+8i}{8}$$

Divide 8+8i by 8.

$$x=1+i$$

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

$$x=\frac{8-8i}{8}$$

Divide 8-8i by 8.

$$x=1-i$$

The equation is now solved.

$$x=1+i x=1-i$$

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}.

$$8+16-16x+4x^{2}+x^{2}=\left(4-x\right)^{2}$$

Add 8 and 16 to get 24.

$$24-16x+4x^{2}+x^{2}=\left(4-x\right)^{2}$$

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

$$24-16x+5x^{2}=\left(4-x\right)^{2}$$

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

$$24-16x+5x^{2}=16-8x+x^{2}$$

Add 8x to both sides.

$$24-16x+5x^{2}+8x=16+x^{2}$$

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

$$24-8x+5x^{2}=16+x^{2}$$

Subtract x^{2} from both sides.

$$24-8x+5x^{2}-x^{2}=16$$

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

$$24-8x+4x^{2}=16$$

Subtract 24 from both sides.

$$-8x+4x^{2}=16-24$$

Subtract 24 from 16 to get -8.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -8 by 4.

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

Divide -8 by 4.

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

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.

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

Add -2 to 1.

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

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.

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

Add 1 to both sides of the equation.

$$x=1+i x=1-i$$