Solve (x-3)^2+(x-1)^2=0

$$( x - 3 ) ^ { 2 } + ( x - 1 ) ^ { 2 } = 0$$

Solve for x (complex solution)

$$x=2+i$$

$$x=2-i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+\left(x-1\right)^{2}=0$$

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

$$x^{2}-6x+9+x^{2}-2x+1=0$$

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

$$2x^{2}-6x+9-2x+1=0$$

Combine -6x and -2x to get -8x.

$$2x^{2}-8x+9+1=0$$

Add 9 and 1 to get 10.

$$2x^{2}-8x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 10}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-\left(-8\right)±\sqrt{64-80}}{2\times 2}$$

Add 64 to -80.

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

Take the square root of -16.

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

The opposite of -8 is 8.

$$x=\frac{8±4i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±4i}{4}$$

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

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

Divide 8+4i by 4.

$$x=2+i$$

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

$$x=\frac{8-4i}{4}$$

Divide 8-4i by 4.

$$x=2-i$$

The equation is now solved.

$$x=2+i x=2-i$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+\left(x-1\right)^{2}=0$$

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

$$x^{2}-6x+9+x^{2}-2x+1=0$$

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

$$2x^{2}-6x+9-2x+1=0$$

Combine -6x and -2x to get -8x.

$$2x^{2}-8x+9+1=0$$

Add 9 and 1 to get 10.

$$2x^{2}-8x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-8x=-10$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=-\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=-\frac{10}{2}$$

Divide -8 by 2.

$$x^{2}-4x=-\frac{10}{2}$$

Divide -10 by 2.

$$x^{2}-4x=-5$$

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.

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

Square -2.

$$x^{2}-4x+4=-5+4$$

Add -5 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=i x-2=-i$$

Add 2 to both sides of the equation.

$$x=2+i x=2-i$$