Solve 4=(x+1)^2+(x-5)^2

$$4 = ( x + 1 ) ^ { 2 } + ( x - 5 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=2+\sqrt{7}i\approx 2+2.645751311i$$

$$x=-\sqrt{7}i+2\approx 2-2.645751311i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$4=x^{2}+2x+1+\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$4=x^{2}+2x+1+x^{2}-10x+25$$

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

$$4=2x^{2}+2x+1-10x+25$$

Combine 2x and -10x to get -8x.

$$4=2x^{2}-8x+1+25$$

Add 1 and 25 to get 26.

$$4=2x^{2}-8x+26$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-8x+26=4$$

Subtract 4 from both sides.

$$2x^{2}-8x+26-4=0$$

Subtract 4 from 26 to get 22.

$$2x^{2}-8x+22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 22}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 22}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 22}}{2\times 2}$$

Multiply -8 times 22.

$$x=\frac{-\left(-8\right)±\sqrt{64-176}}{2\times 2}$$

Add 64 to -176.

$$x=\frac{-\left(-8\right)±\sqrt{-112}}{2\times 2}$$

Take the square root of -112.

$$x=\frac{-\left(-8\right)±4\sqrt{7}i}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{7}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±4\sqrt{7}i}{4}$$

Now solve the equation x=\frac{8±4\sqrt{7}i}{4} when ± is plus. Add 8 to 4i\sqrt{7}.

$$x=\frac{8+4\sqrt{7}i}{4}$$

Divide 8+4i\sqrt{7} by 4.

$$x=2+\sqrt{7}i$$

Now solve the equation x=\frac{8±4\sqrt{7}i}{4} when ± is minus. Subtract 4i\sqrt{7} from 8.

$$x=\frac{-4\sqrt{7}i+8}{4}$$

Divide 8-4i\sqrt{7} by 4.

$$x=-\sqrt{7}i+2$$

The equation is now solved.

$$x=2+\sqrt{7}i x=-\sqrt{7}i+2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$4=x^{2}+2x+1+\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$4=x^{2}+2x+1+x^{2}-10x+25$$

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

$$4=2x^{2}+2x+1-10x+25$$

Combine 2x and -10x to get -8x.

$$4=2x^{2}-8x+1+25$$

Add 1 and 25 to get 26.

$$4=2x^{2}-8x+26$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-8x+26=4$$

Subtract 26 from both sides.

$$2x^{2}-8x=4-26$$

Subtract 26 from 4 to get -22.

$$2x^{2}-8x=-22$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=-\frac{22}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=-\frac{22}{2}$$

Divide -8 by 2.

$$x^{2}-4x=-\frac{22}{2}$$

Divide -22 by 2.

$$x^{2}-4x=-11$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-11+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-11+4$$

Add -11 to 4.

$$x^{2}-4x+4=-7$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=-7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-7}$$

Simplify.

$$x-2=\sqrt{7}i x-2=-\sqrt{7}i$$

Add 2 to both sides of the equation.

$$x=2+\sqrt{7}i x=-\sqrt{7}i+2$$