Solve x^2+(x+10)^2=50^2

$$x ^ { 2 } + ( x + 10 ) ^ { 2 } = 50 ^ { 2 }$$

Solve for x

$$x=-40$$

$$x=30$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+10\right)^{2}.

$$x^{2}+x^{2}+20x+100=50^{2}$$

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

$$2x^{2}+20x+100=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$2x^{2}+20x+100=2500$$

Subtract 2500 from both sides.

$$2x^{2}+20x+100-2500=0$$

Subtract 2500 from 100 to get -2400.

$$2x^{2}+20x-2400=0$$

Divide both sides by 2.

$$x^{2}+10x-1200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-1200. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-1200\right)=-1200$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1200.

$$-1,1200 -2,600 -3,400 -4,300 -5,240 -6,200 -8,150 -10,120 -12,100 -15,80 -16,75 -20,60 -24,50 -25,48 -30,40$$

Calculate the sum for each pair.

$$-1+1200=1199 -2+600=598 -3+400=397 -4+300=296 -5+240=235 -6+200=194 -8+150=142 -10+120=110 -12+100=88 -15+80=65 -16+75=59 -20+60=40 -24+50=26 -25+48=23 -30+40=10$$

The solution is the pair that gives sum 10.

$$a=-30 b=40$$

Rewrite x^{2}+10x-1200 as \left(x^{2}-30x\right)+\left(40x-1200\right).

$$\left(x^{2}-30x\right)+\left(40x-1200\right)$$

Factor out x in the first and 40 in the second group.

$$x\left(x-30\right)+40\left(x-30\right)$$

Factor out common term x-30 by using distributive property.

$$\left(x-30\right)\left(x+40\right)$$

To find equation solutions, solve x-30=0 and x+40=0.

$$x=30 x=-40$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+10\right)^{2}.

$$x^{2}+x^{2}+20x+100=50^{2}$$

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

$$2x^{2}+20x+100=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$2x^{2}+20x+100=2500$$

Subtract 2500 from both sides.

$$2x^{2}+20x+100-2500=0$$

Subtract 2500 from 100 to get -2400.

$$2x^{2}+20x-2400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 20 for b, and -2400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\left(-2400\right)}}{2\times 2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 2\left(-2400\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-20±\sqrt{400-8\left(-2400\right)}}{2\times 2}$$

Multiply -8 times -2400.

$$x=\frac{-20±\sqrt{400+19200}}{2\times 2}$$

Add 400 to 19200.

$$x=\frac{-20±\sqrt{19600}}{2\times 2}$$

Take the square root of 19600.

$$x=\frac{-20±140}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-20±140}{4}$$

Now solve the equation x=\frac{-20±140}{4} when ± is plus. Add -20 to 140.

$$x=\frac{120}{4}$$

Divide 120 by 4.

$$x=30$$

Now solve the equation x=\frac{-20±140}{4} when ± is minus. Subtract 140 from -20.

$$x=-\frac{160}{4}$$

Divide -160 by 4.

$$x=-40$$

The equation is now solved.

$$x=30 x=-40$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+10\right)^{2}.

$$x^{2}+x^{2}+20x+100=50^{2}$$

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

$$2x^{2}+20x+100=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$2x^{2}+20x+100=2500$$

Subtract 100 from both sides.

$$2x^{2}+20x=2500-100$$

Subtract 100 from 2500 to get 2400.

$$2x^{2}+20x=2400$$

Divide both sides by 2.

$$\frac{2x^{2}+20x}{2}=\frac{2400}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{20}{2}x=\frac{2400}{2}$$

Divide 20 by 2.

$$x^{2}+10x=\frac{2400}{2}$$

Divide 2400 by 2.

$$x^{2}+10x=1200$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=1200+5^{2}$$

Square 5.

$$x^{2}+10x+25=1200+25$$

Add 1200 to 25.

$$x^{2}+10x+25=1225$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=1225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{1225}$$

Simplify.

$$x+5=35 x+5=-35$$

Subtract 5 from both sides of the equation.

$$x=30 x=-40$$