Solve left(6+xright)^2+left(4+xright)^2=100

$${ \left(6+x \right) }^{ 2 } + { \left(4+x \right) }^{ 2 } =100$$

Solve for x

$$x=-12$$

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}+\left(4+x\right)^{2}=100$$

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

$$36+12x+x^{2}+16+8x+x^{2}=100$$

Add 36 and 16 to get 52.

$$52+12x+x^{2}+8x+x^{2}=100$$

Combine 12x and 8x to get 20x.

$$52+20x+x^{2}+x^{2}=100$$

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

$$52+20x+2x^{2}=100$$

Subtract 100 from both sides.

$$52+20x+2x^{2}-100=0$$

Subtract 100 from 52 to get -48.

$$-48+20x+2x^{2}=0$$

Divide both sides by 2.

$$-24+10x+x^{2}=0$$

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

$$x^{2}+10x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-24\right)=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 10.

$$a=-2 b=12$$

Rewrite x^{2}+10x-24 as \left(x^{2}-2x\right)+\left(12x-24\right).

$$\left(x^{2}-2x\right)+\left(12x-24\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x-2\right)+12\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+12\right)$$

To find equation solutions, solve x-2=0 and x+12=0.

$$x=2 x=-12$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}+\left(4+x\right)^{2}=100$$

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

$$36+12x+x^{2}+16+8x+x^{2}=100$$

Add 36 and 16 to get 52.

$$52+12x+x^{2}+8x+x^{2}=100$$

Combine 12x and 8x to get 20x.

$$52+20x+x^{2}+x^{2}=100$$

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

$$52+20x+2x^{2}=100$$

Subtract 100 from both sides.

$$52+20x+2x^{2}-100=0$$

Subtract 100 from 52 to get -48.

$$-48+20x+2x^{2}=0$$

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.

$$2x^{2}+20x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 20 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\left(-48\right)}}{2\times 2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 2\left(-48\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-20±\sqrt{400-8\left(-48\right)}}{2\times 2}$$

Multiply -8 times -48.

$$x=\frac{-20±\sqrt{400+384}}{2\times 2}$$

Add 400 to 384.

$$x=\frac{-20±\sqrt{784}}{2\times 2}$$

Take the square root of 784.

$$x=\frac{-20±28}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-20±28}{4}$$

Now solve the equation x=\frac{-20±28}{4} when ± is plus. Add -20 to 28.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-20±28}{4} when ± is minus. Subtract 28 from -20.

$$x=-\frac{48}{4}$$

Divide -48 by 4.

$$x=-12$$

The equation is now solved.

$$x=2 x=-12$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}+\left(4+x\right)^{2}=100$$

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

$$36+12x+x^{2}+16+8x+x^{2}=100$$

Add 36 and 16 to get 52.

$$52+12x+x^{2}+8x+x^{2}=100$$

Combine 12x and 8x to get 20x.

$$52+20x+x^{2}+x^{2}=100$$

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

$$52+20x+2x^{2}=100$$

Subtract 52 from both sides.

$$20x+2x^{2}=100-52$$

Subtract 52 from 100 to get 48.

$$20x+2x^{2}=48$$

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.

$$2x^{2}+20x=48$$

Divide both sides by 2.

$$\frac{2x^{2}+20x}{2}=\frac{48}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{20}{2}x=\frac{48}{2}$$

Divide 20 by 2.

$$x^{2}+10x=\frac{48}{2}$$

Divide 48 by 2.

$$x^{2}+10x=24$$

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.

$$x^{2}+10x+5^{2}=24+5^{2}$$

Square 5.

$$x^{2}+10x+25=24+25$$

Add 24 to 25.

$$x^{2}+10x+25=49$$

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}.

$$\left(x+5\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+5=7 x+5=-7$$

Subtract 5 from both sides of the equation.

$$x=2 x=-12$$