Solve x^2+left(80-xright)^2=200

$${ x }^{ 2 } + { \left(80-x \right) }^{ 2 } =200$$

Solve for x (complex solution)

$$x=40+10\sqrt{15}i\approx 40+38.729833462i$$

$$x=-10\sqrt{15}i+40\approx 40-38.729833462i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(80-x\right)^{2}.

$$x^{2}+6400-160x+x^{2}=200$$

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

$$2x^{2}+6400-160x=200$$

Subtract 200 from both sides.

$$2x^{2}+6400-160x-200=0$$

Subtract 200 from 6400 to get 6200.

$$2x^{2}+6200-160x=0$$

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.

$$2x^{2}-160x+6200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -160 for b, and 6200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-160\right)±\sqrt{\left(-160\right)^{2}-4\times 2\times 6200}}{2\times 2}$$

Square -160.

$$x=\frac{-\left(-160\right)±\sqrt{25600-4\times 2\times 6200}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-160\right)±\sqrt{25600-8\times 6200}}{2\times 2}$$

Multiply -8 times 6200.

$$x=\frac{-\left(-160\right)±\sqrt{25600-49600}}{2\times 2}$$

Add 25600 to -49600.

$$x=\frac{-\left(-160\right)±\sqrt{-24000}}{2\times 2}$$

Take the square root of -24000.

$$x=\frac{-\left(-160\right)±40\sqrt{15}i}{2\times 2}$$

The opposite of -160 is 160.

$$x=\frac{160±40\sqrt{15}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{160±40\sqrt{15}i}{4}$$

Now solve the equation x=\frac{160±40\sqrt{15}i}{4} when ± is plus. Add 160 to 40i\sqrt{15}.

$$x=\frac{160+40\sqrt{15}i}{4}$$

Divide 160+40i\sqrt{15} by 4.

$$x=40+10\sqrt{15}i$$

Now solve the equation x=\frac{160±40\sqrt{15}i}{4} when ± is minus. Subtract 40i\sqrt{15} from 160.

$$x=\frac{-40\sqrt{15}i+160}{4}$$

Divide 160-40i\sqrt{15} by 4.

$$x=-10\sqrt{15}i+40$$

The equation is now solved.

$$x=40+10\sqrt{15}i x=-10\sqrt{15}i+40$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(80-x\right)^{2}.

$$x^{2}+6400-160x+x^{2}=200$$

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

$$2x^{2}+6400-160x=200$$

Subtract 6400 from both sides.

$$2x^{2}-160x=200-6400$$

Subtract 6400 from 200 to get -6200.

$$2x^{2}-160x=-6200$$

Divide both sides by 2.

$$\frac{2x^{2}-160x}{2}=-\frac{6200}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{160}{2}\right)x=-\frac{6200}{2}$$

Divide -160 by 2.

$$x^{2}-80x=-\frac{6200}{2}$$

Divide -6200 by 2.

$$x^{2}-80x=-3100$$

Divide -80, the coefficient of the x term, by 2 to get -40. Then add the square of -40 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-80x+\left(-40\right)^{2}=-3100+\left(-40\right)^{2}$$

Square -40.

$$x^{2}-80x+1600=-3100+1600$$

Add -3100 to 1600.

$$x^{2}-80x+1600=-1500$$

Factor x^{2}-80x+1600. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-40\right)^{2}=-1500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-40\right)^{2}}=\sqrt{-1500}$$

Simplify.

$$x-40=10\sqrt{15}i x-40=-10\sqrt{15}i$$

Add 40 to both sides of the equation.

$$x=40+10\sqrt{15}i x=-10\sqrt{15}i+40$$