Solve (100)^2+(x+100)^2=(2x+100)^2

$$( 100 ) ^ { 2 } + ( x + 100 ) ^ { 2 } = ( 2 x + 100 ) ^ { 2 }$$

Solve for x

$$x = \frac{100}{3} = 33\frac{1}{3} \approx 33.333333333$$

$$x=-100$$

Steps Using Factoring By Grouping

Calculate 100 to the power of 2 and get 10000.

$$10000+\left(x+100\right)^{2}=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+100\right)^{2}.

$$10000+x^{2}+200x+10000=\left(2x+100\right)^{2}$$

Add 10000 and 10000 to get 20000.

$$20000+x^{2}+200x=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+100\right)^{2}.

$$20000+x^{2}+200x=4x^{2}+400x+10000$$

Subtract 4x^{2} from both sides.

$$20000+x^{2}+200x-4x^{2}=400x+10000$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$20000-3x^{2}+200x=400x+10000$$

Subtract 400x from both sides.

$$20000-3x^{2}+200x-400x=10000$$

Combine 200x and -400x to get -200x.

$$20000-3x^{2}-200x=10000$$

Subtract 10000 from both sides.

$$20000-3x^{2}-200x-10000=0$$

Subtract 10000 from 20000 to get 10000.

$$10000-3x^{2}-200x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-200x+10000=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+10000. To find a and b, set up a system to be solved.

$$a+b=-200 ab=-3\times 10000=-30000$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -30000.

$$1,-30000 2,-15000 3,-10000 4,-7500 5,-6000 6,-5000 8,-3750 10,-3000 12,-2500 15,-2000 16,-1875 20,-1500 24,-1250 25,-1200 30,-1000 40,-750 48,-625 50,-600 60,-500 75,-400 80,-375 100,-300 120,-250 125,-240 150,-200$$

Calculate the sum for each pair.

$$1-30000=-29999 2-15000=-14998 3-10000=-9997 4-7500=-7496 5-6000=-5995 6-5000=-4994 8-3750=-3742 10-3000=-2990 12-2500=-2488 15-2000=-1985 16-1875=-1859 20-1500=-1480 24-1250=-1226 25-1200=-1175 30-1000=-970 40-750=-710 48-625=-577 50-600=-550 60-500=-440 75-400=-325 80-375=-295 100-300=-200 120-250=-130 125-240=-115 150-200=-50$$

The solution is the pair that gives sum -200.

$$a=100 b=-300$$

Rewrite -3x^{2}-200x+10000 as \left(-3x^{2}+100x\right)+\left(-300x+10000\right).

$$\left(-3x^{2}+100x\right)+\left(-300x+10000\right)$$

Factor out -x in the first and -100 in the second group.

$$-x\left(3x-100\right)-100\left(3x-100\right)$$

Factor out common term 3x-100 by using distributive property.

$$\left(3x-100\right)\left(-x-100\right)$$

To find equation solutions, solve 3x-100=0 and -x-100=0.

$$x=\frac{100}{3} x=-100$$

Steps Using the Quadratic Formula

Calculate 100 to the power of 2 and get 10000.

$$10000+\left(x+100\right)^{2}=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+100\right)^{2}.

$$10000+x^{2}+200x+10000=\left(2x+100\right)^{2}$$

Add 10000 and 10000 to get 20000.

$$20000+x^{2}+200x=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+100\right)^{2}.

$$20000+x^{2}+200x=4x^{2}+400x+10000$$

Subtract 4x^{2} from both sides.

$$20000+x^{2}+200x-4x^{2}=400x+10000$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$20000-3x^{2}+200x=400x+10000$$

Subtract 400x from both sides.

$$20000-3x^{2}+200x-400x=10000$$

Combine 200x and -400x to get -200x.

$$20000-3x^{2}-200x=10000$$

Subtract 10000 from both sides.

$$20000-3x^{2}-200x-10000=0$$

Subtract 10000 from 20000 to get 10000.

$$10000-3x^{2}-200x=0$$

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.

$$-3x^{2}-200x+10000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -200 for b, and 10000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-200\right)±\sqrt{\left(-200\right)^{2}-4\left(-3\right)\times 10000}}{2\left(-3\right)}$$

Square -200.

$$x=\frac{-\left(-200\right)±\sqrt{40000-4\left(-3\right)\times 10000}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-200\right)±\sqrt{40000+12\times 10000}}{2\left(-3\right)}$$

Multiply 12 times 10000.

$$x=\frac{-\left(-200\right)±\sqrt{40000+120000}}{2\left(-3\right)}$$

Add 40000 to 120000.

$$x=\frac{-\left(-200\right)±\sqrt{160000}}{2\left(-3\right)}$$

Take the square root of 160000.

$$x=\frac{-\left(-200\right)±400}{2\left(-3\right)}$$

The opposite of -200 is 200.

$$x=\frac{200±400}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{200±400}{-6}$$

Now solve the equation x=\frac{200±400}{-6} when ± is plus. Add 200 to 400.

$$x=\frac{600}{-6}$$

Divide 600 by -6.

$$x=-100$$

Now solve the equation x=\frac{200±400}{-6} when ± is minus. Subtract 400 from 200.

$$x=-\frac{200}{-6}$$

Reduce the fraction \frac{-200}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{100}{3}$$

The equation is now solved.

$$x=-100 x=\frac{100}{3}$$

Steps for Completing the Square

Calculate 100 to the power of 2 and get 10000.

$$10000+\left(x+100\right)^{2}=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+100\right)^{2}.

$$10000+x^{2}+200x+10000=\left(2x+100\right)^{2}$$

Add 10000 and 10000 to get 20000.

$$20000+x^{2}+200x=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+100\right)^{2}.

$$20000+x^{2}+200x=4x^{2}+400x+10000$$

Subtract 4x^{2} from both sides.

$$20000+x^{2}+200x-4x^{2}=400x+10000$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$20000-3x^{2}+200x=400x+10000$$

Subtract 400x from both sides.

$$20000-3x^{2}+200x-400x=10000$$

Combine 200x and -400x to get -200x.

$$20000-3x^{2}-200x=10000$$

Subtract 20000 from both sides.

$$-3x^{2}-200x=10000-20000$$

Subtract 20000 from 10000 to get -10000.

$$-3x^{2}-200x=-10000$$

Divide both sides by -3.

$$\frac{-3x^{2}-200x}{-3}=-\frac{10000}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{200}{-3}\right)x=-\frac{10000}{-3}$$

Divide -200 by -3.

$$x^{2}+\frac{200}{3}x=-\frac{10000}{-3}$$

Divide -10000 by -3.

$$x^{2}+\frac{200}{3}x=\frac{10000}{3}$$

Divide \frac{200}{3}, the coefficient of the x term, by 2 to get \frac{100}{3}. Then add the square of \frac{100}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{200}{3}x+\left(\frac{100}{3}\right)^{2}=\frac{10000}{3}+\left(\frac{100}{3}\right)^{2}$$

Square \frac{100}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{200}{3}x+\frac{10000}{9}=\frac{10000}{3}+\frac{10000}{9}$$

Add \frac{10000}{3} to \frac{10000}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{200}{3}x+\frac{10000}{9}=\frac{40000}{9}$$

Factor x^{2}+\frac{200}{3}x+\frac{10000}{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}.

$$\left(x+\frac{100}{3}\right)^{2}=\frac{40000}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{100}{3}\right)^{2}}=\sqrt{\frac{40000}{9}}$$

Simplify.

$$x+\frac{100}{3}=\frac{200}{3} x+\frac{100}{3}=-\frac{200}{3}$$

Subtract \frac{100}{3} from both sides of the equation.

$$x=\frac{100}{3} x=-100$$