Solve (2000-200x)^2+(100x)^2=1000^2

$$( 2000 - 200 x ) ^ { 2 } + ( 100 x ) ^ { 2 } = 1000 ^ { 2 }$$

Solve for x

$$x=6$$

$$x=10$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2000-200x\right)^{2}.

$$4000000-800000x+40000x^{2}+\left(100x\right)^{2}=1000^{2}$$

Expand \left(100x\right)^{2}.

$$4000000-800000x+40000x^{2}+100^{2}x^{2}=1000^{2}$$

Calculate 100 to the power of 2 and get 10000.

$$4000000-800000x+40000x^{2}+10000x^{2}=1000^{2}$$

Combine 40000x^{2} and 10000x^{2} to get 50000x^{2}.

$$4000000-800000x+50000x^{2}=1000^{2}$$

Calculate 1000 to the power of 2 and get 1000000.

$$4000000-800000x+50000x^{2}=1000000$$

Subtract 1000000 from both sides.

$$4000000-800000x+50000x^{2}-1000000=0$$

Subtract 1000000 from 4000000 to get 3000000.

$$3000000-800000x+50000x^{2}=0$$

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.

$$50000x^{2}-800000x+3000000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 50000 for a, -800000 for b, and 3000000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-800000\right)±\sqrt{\left(-800000\right)^{2}-4\times 50000\times 3000000}}{2\times 50000}$$

Square -800000.

$$x=\frac{-\left(-800000\right)±\sqrt{640000000000-4\times 50000\times 3000000}}{2\times 50000}$$

Multiply -4 times 50000.

$$x=\frac{-\left(-800000\right)±\sqrt{640000000000-200000\times 3000000}}{2\times 50000}$$

Multiply -200000 times 3000000.

$$x=\frac{-\left(-800000\right)±\sqrt{640000000000-600000000000}}{2\times 50000}$$

Add 640000000000 to -600000000000.

$$x=\frac{-\left(-800000\right)±\sqrt{40000000000}}{2\times 50000}$$

Take the square root of 40000000000.

$$x=\frac{-\left(-800000\right)±200000}{2\times 50000}$$

The opposite of -800000 is 800000.

$$x=\frac{800000±200000}{2\times 50000}$$

Multiply 2 times 50000.

$$x=\frac{800000±200000}{100000}$$

Now solve the equation x=\frac{800000±200000}{100000} when ± is plus. Add 800000 to 200000.

$$x=\frac{1000000}{100000}$$

Divide 1000000 by 100000.

$$x=10$$

Now solve the equation x=\frac{800000±200000}{100000} when ± is minus. Subtract 200000 from 800000.

$$x=\frac{600000}{100000}$$

Divide 600000 by 100000.

$$x=6$$

The equation is now solved.

$$x=10 x=6$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2000-200x\right)^{2}.

$$4000000-800000x+40000x^{2}+\left(100x\right)^{2}=1000^{2}$$

Expand \left(100x\right)^{2}.

$$4000000-800000x+40000x^{2}+100^{2}x^{2}=1000^{2}$$

Calculate 100 to the power of 2 and get 10000.

$$4000000-800000x+40000x^{2}+10000x^{2}=1000^{2}$$

Combine 40000x^{2} and 10000x^{2} to get 50000x^{2}.

$$4000000-800000x+50000x^{2}=1000^{2}$$

Calculate 1000 to the power of 2 and get 1000000.

$$4000000-800000x+50000x^{2}=1000000$$

Subtract 4000000 from both sides.

$$-800000x+50000x^{2}=1000000-4000000$$

Subtract 4000000 from 1000000 to get -3000000.

$$-800000x+50000x^{2}=-3000000$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$50000x^{2}-800000x=-3000000$$

Divide both sides by 50000.

$$\frac{50000x^{2}-800000x}{50000}=-\frac{3000000}{50000}$$

Dividing by 50000 undoes the multiplication by 50000.

$$x^{2}+\left(-\frac{800000}{50000}\right)x=-\frac{3000000}{50000}$$

Divide -800000 by 50000.

$$x^{2}-16x=-\frac{3000000}{50000}$$

Divide -3000000 by 50000.

$$x^{2}-16x=-60$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-60+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-60+64$$

Add -60 to 64.

$$x^{2}-16x+64=4$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-8=2 x-8=-2$$

Add 8 to both sides of the equation.

$$x=10 x=6$$