Solve x^2+left(x+1right)^2+left(x+2right)^2=left(x+3right)^2+{left(x+4right)}^2

$${ x }^{ 2 } + { \left(x+1 \right) }^{ 2 } + { \left(x+2 \right) }^{ 2 } = { \left(x+3 \right) }^{ 2 } + { \left(x+4 \right) }^{ 2 }$$

Solve for x

$$x=-2$$

$$x=10$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+x^{2}+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$3x^{2}+2x+1+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Combine 2x and 4x to get 6x.

$$3x^{2}+6x+1+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Add 1 and 4 to get 5.

$$3x^{2}+6x+5=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+x^{2}+8x+16$$

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

$$3x^{2}+6x+5=2x^{2}+6x+9+8x+16$$

Combine 6x and 8x to get 14x.

$$3x^{2}+6x+5=2x^{2}+14x+9+16$$

Add 9 and 16 to get 25.

$$3x^{2}+6x+5=2x^{2}+14x+25$$

Subtract 2x^{2} from both sides.

$$3x^{2}+6x+5-2x^{2}=14x+25$$

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

$$x^{2}+6x+5=14x+25$$

Subtract 14x from both sides.

$$x^{2}+6x+5-14x=25$$

Combine 6x and -14x to get -8x.

$$x^{2}-8x+5=25$$

Subtract 25 from both sides.

$$x^{2}-8x+5-25=0$$

Subtract 25 from 5 to get -20.

$$x^{2}-8x-20=0$$

To solve the equation, factor x^{2}-8x-20 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x+2\right)$$

To find equation solutions, solve x-10=0 and x+2=0.

$$x=10 x=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+x^{2}+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$3x^{2}+2x+1+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Combine 2x and 4x to get 6x.

$$3x^{2}+6x+1+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Add 1 and 4 to get 5.

$$3x^{2}+6x+5=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+x^{2}+8x+16$$

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

$$3x^{2}+6x+5=2x^{2}+6x+9+8x+16$$

Combine 6x and 8x to get 14x.

$$3x^{2}+6x+5=2x^{2}+14x+9+16$$

Add 9 and 16 to get 25.

$$3x^{2}+6x+5=2x^{2}+14x+25$$

Subtract 2x^{2} from both sides.

$$3x^{2}+6x+5-2x^{2}=14x+25$$

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

$$x^{2}+6x+5=14x+25$$

Subtract 14x from both sides.

$$x^{2}+6x+5-14x=25$$

Combine 6x and -14x to get -8x.

$$x^{2}-8x+5=25$$

Subtract 25 from both sides.

$$x^{2}-8x+5-25=0$$

Subtract 25 from 5 to get -20.

$$x^{2}-8x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-20\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

Rewrite x^{2}-8x-20 as \left(x^{2}-10x\right)+\left(2x-20\right).

$$\left(x^{2}-10x\right)+\left(2x-20\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-10\right)+2\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+2\right)$$

To find equation solutions, solve x-10=0 and x+2=0.

$$x=10 x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+x^{2}+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$3x^{2}+2x+1+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Combine 2x and 4x to get 6x.

$$3x^{2}+6x+1+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Add 1 and 4 to get 5.

$$3x^{2}+6x+5=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+x^{2}+8x+16$$

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

$$3x^{2}+6x+5=2x^{2}+6x+9+8x+16$$

Combine 6x and 8x to get 14x.

$$3x^{2}+6x+5=2x^{2}+14x+9+16$$

Add 9 and 16 to get 25.

$$3x^{2}+6x+5=2x^{2}+14x+25$$

Subtract 2x^{2} from both sides.

$$3x^{2}+6x+5-2x^{2}=14x+25$$

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

$$x^{2}+6x+5=14x+25$$

Subtract 14x from both sides.

$$x^{2}+6x+5-14x=25$$

Combine 6x and -14x to get -8x.

$$x^{2}-8x+5=25$$

Subtract 25 from both sides.

$$x^{2}-8x+5-25=0$$

Subtract 25 from 5 to get -20.

$$x^{2}-8x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-20\right)}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2}$$

Add 64 to 80.

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

Take the square root of 144.

$$x=\frac{-\left(-8\right)±12}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±12}{2}$$

Now solve the equation x=\frac{8±12}{2} when ± is plus. Add 8 to 12.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{8±12}{2} when ± is minus. Subtract 12 from 8.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=10 x=-2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+\left(x+2\right)^{2}=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$2x^{2}+2x+1+x^{2}+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

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

$$3x^{2}+2x+1+4x+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Combine 2x and 4x to get 6x.

$$3x^{2}+6x+1+4=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Add 1 and 4 to get 5.

$$3x^{2}+6x+5=\left(x+3\right)^{2}+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3x^{2}+6x+5=x^{2}+6x+9+x^{2}+8x+16$$

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

$$3x^{2}+6x+5=2x^{2}+6x+9+8x+16$$

Combine 6x and 8x to get 14x.

$$3x^{2}+6x+5=2x^{2}+14x+9+16$$

Add 9 and 16 to get 25.

$$3x^{2}+6x+5=2x^{2}+14x+25$$

Subtract 2x^{2} from both sides.

$$3x^{2}+6x+5-2x^{2}=14x+25$$

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

$$x^{2}+6x+5=14x+25$$

Subtract 14x from both sides.

$$x^{2}+6x+5-14x=25$$

Combine 6x and -14x to get -8x.

$$x^{2}-8x+5=25$$

Subtract 5 from both sides.

$$x^{2}-8x=25-5$$

Subtract 5 from 25 to get 20.

$$x^{2}-8x=20$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=20+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=20+16$$

Add 20 to 16.

$$x^{2}-8x+16=36$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-4=6 x-4=-6$$

Add 4 to both sides of the equation.

$$x=10 x=-2$$