Solve x^2+(x-6)^2=16

$$x ^ { 2 } + ( x - 6 ) ^ { 2 } = 16$$

Solve for x (complex solution)

$$x=3+i$$

$$x=3-i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}+x^{2}-12x+36=16$$

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

$$2x^{2}-12x+36=16$$

Subtract 16 from both sides.

$$2x^{2}-12x+36-16=0$$

Subtract 16 from 36 to get 20.

$$2x^{2}-12x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 20}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 20}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\times 20}}{2\times 2}$$

Multiply -8 times 20.

$$x=\frac{-\left(-12\right)±\sqrt{144-160}}{2\times 2}$$

Add 144 to -160.

$$x=\frac{-\left(-12\right)±\sqrt{-16}}{2\times 2}$$

Take the square root of -16.

$$x=\frac{-\left(-12\right)±4i}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±4i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±4i}{4}$$

Now solve the equation x=\frac{12±4i}{4} when ± is plus. Add 12 to 4i.

$$x=\frac{12+4i}{4}$$

Divide 12+4i by 4.

$$x=3+i$$

Now solve the equation x=\frac{12±4i}{4} when ± is minus. Subtract 4i from 12.

$$x=\frac{12-4i}{4}$$

Divide 12-4i by 4.

$$x=3-i$$

The equation is now solved.

$$x=3+i x=3-i$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}+x^{2}-12x+36=16$$

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

$$2x^{2}-12x+36=16$$

Subtract 36 from both sides.

$$2x^{2}-12x=16-36$$

Subtract 36 from 16 to get -20.

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

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=-\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=-\frac{20}{2}$$

Divide -12 by 2.

$$x^{2}-6x=-\frac{20}{2}$$

Divide -20 by 2.

$$x^{2}-6x=-10$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-10+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-10+9$$

Add -10 to 9.

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

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=-1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-1}$$

Simplify.

$$x-3=i x-3=-i$$

Add 3 to both sides of the equation.

$$x=3+i x=3-i$$