Solve (x-4)^2+(x-6)^2=100

$$( x - 4 ) ^ { 2 } + ( x - 6 ) ^ { 2 } = 100$$

Solve for x

$$x=-2$$

$$x=12$$

Steps Using Factoring By Grouping

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

$$x^{2}-8x+16+\left(x-6\right)^{2}=100$$

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

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

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

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

Combine -8x and -12x to get -20x.

$$2x^{2}-20x+16+36=100$$

Add 16 and 36 to get 52.

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

Subtract 100 from both sides.

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

Subtract 100 from 52 to get -48.

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

Divide both sides by 2.

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-12 b=2$$

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

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

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

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

Factor out common term x-12 by using distributive property.

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

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

$$x=12 x=-2$$

Steps Using the Quadratic Formula

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

$$x^{2}-8x+16+\left(x-6\right)^{2}=100$$

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

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

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

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

Combine -8x and -12x to get -20x.

$$2x^{2}-20x+16+36=100$$

Add 16 and 36 to get 52.

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

Subtract 100 from both sides.

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

Subtract 100 from 52 to get -48.

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

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}.

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

Square -20.

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

Multiply -4 times 2.

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

Multiply -8 times -48.

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

Add 400 to 384.

$$x=\frac{-\left(-20\right)±\sqrt{784}}{2\times 2}$$

Take the square root of 784.

$$x=\frac{-\left(-20\right)±28}{2\times 2}$$

The opposite of -20 is 20.

$$x=\frac{20±28}{2\times 2}$$

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.

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

Divide 48 by 4.

$$x=12$$

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

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

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=12 x=-2$$

Steps for Completing the Square

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

$$x^{2}-8x+16+\left(x-6\right)^{2}=100$$

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

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

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

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

Combine -8x and -12x to get -20x.

$$2x^{2}-20x+16+36=100$$

Add 16 and 36 to get 52.

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

Subtract 52 from both sides.

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

Subtract 52 from 100 to get 48.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{20}{2}\right)x=\frac{48}{2}$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=24+\left(-5\right)^{2}$$

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.

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

Add 5 to both sides of the equation.

$$x=12 x=-2$$