Solve (x+2)^2+(x+4)^2=8x+2

$$( x + 2 ) ^ { 2 } + ( x + 4 ) ^ { 2 } = 8 x + 2$$

Solve for x (complex solution)

$$x=-1+2\sqrt{2}i\approx -1+2.828427125i$$

$$x=-2\sqrt{2}i-1\approx -1-2.828427125i$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Combine 4x and 8x to get 12x.

$$2x^{2}+12x+4+16=8x+2$$

Add 4 and 16 to get 20.

$$2x^{2}+12x+20=8x+2$$

Subtract 8x from both sides.

$$2x^{2}+12x+20-8x=2$$

Combine 12x and -8x to get 4x.

$$2x^{2}+4x+20=2$$

Subtract 2 from both sides.

$$2x^{2}+4x+20-2=0$$

Subtract 2 from 20 to get 18.

$$2x^{2}+4x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\times 18}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\times 18}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\times 18}}{2\times 2}$$

Multiply -8 times 18.

$$x=\frac{-4±\sqrt{16-144}}{2\times 2}$$

Add 16 to -144.

$$x=\frac{-4±\sqrt{-128}}{2\times 2}$$

Take the square root of -128.

$$x=\frac{-4±8\sqrt{2}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±8\sqrt{2}i}{4}$$

Now solve the equation x=\frac{-4±8\sqrt{2}i}{4} when ± is plus. Add -4 to 8i\sqrt{2}.

$$x=\frac{-4+2\times 2^{\frac{5}{2}}i}{4}$$

Divide -4+2i\times 2^{\frac{5}{2}} by 4.

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

Now solve the equation x=\frac{-4±8\sqrt{2}i}{4} when ± is minus. Subtract 8i\sqrt{2} from -4.

$$x=\frac{-2\times 2^{\frac{5}{2}}i-4}{4}$$

Divide -4-2i\times 2^{\frac{5}{2}} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

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

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

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

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

Combine 4x and 8x to get 12x.

$$2x^{2}+12x+4+16=8x+2$$

Add 4 and 16 to get 20.

$$2x^{2}+12x+20=8x+2$$

Subtract 8x from both sides.

$$2x^{2}+12x+20-8x=2$$

Combine 12x and -8x to get 4x.

$$2x^{2}+4x+20=2$$

Subtract 20 from both sides.

$$2x^{2}+4x=2-20$$

Subtract 20 from 2 to get -18.

$$2x^{2}+4x=-18$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=-\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=-\frac{18}{2}$$

Divide 4 by 2.

$$x^{2}+2x=-\frac{18}{2}$$

Divide -18 by 2.

$$x^{2}+2x=-9$$

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.

$$x^{2}+2x+1^{2}=-9+1^{2}$$

Square 1.

$$x^{2}+2x+1=-9+1$$

Add -9 to 1.

$$x^{2}+2x+1=-8$$

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}.

$$\left(x+1\right)^{2}=-8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{-8}$$

Simplify.

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

Subtract 1 from both sides of the equation.

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