Solve x^2+left(x-4right)^2+left(x-6right)^2=195

$${ x }^{ 2 } + { \left(x-4 \right) }^{ 2 } + { \left(x-6 \right) }^{ 2 } =195$$

Solve for x

$$x = -\frac{13}{3} = -4\frac{1}{3} \approx -4.333333333$$

$$x=11$$

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}.

$$x^{2}+x^{2}-8x+16+\left(x-6\right)^{2}=195$$

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

$$2x^{2}-8x+16+\left(x-6\right)^{2}=195$$

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

$$2x^{2}-8x+16+x^{2}-12x+36=195$$

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

$$3x^{2}-8x+16-12x+36=195$$

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

$$3x^{2}-20x+16+36=195$$

Add 16 and 36 to get 52.

$$3x^{2}-20x+52=195$$

Subtract 195 from both sides.

$$3x^{2}-20x+52-195=0$$

Subtract 195 from 52 to get -143.

$$3x^{2}-20x-143=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-143. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\left(-143\right)=-429$$

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 -429.

$$1,-429 3,-143 11,-39 13,-33$$

Calculate the sum for each pair.

$$1-429=-428 3-143=-140 11-39=-28 13-33=-20$$

The solution is the pair that gives sum -20.

$$a=-33 b=13$$

Rewrite 3x^{2}-20x-143 as \left(3x^{2}-33x\right)+\left(13x-143\right).

$$\left(3x^{2}-33x\right)+\left(13x-143\right)$$

Factor out 3x in the first and 13 in the second group.

$$3x\left(x-11\right)+13\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(3x+13\right)$$

To find equation solutions, solve x-11=0 and 3x+13=0.

$$x=11 x=-\frac{13}{3}$$

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}.

$$x^{2}+x^{2}-8x+16+\left(x-6\right)^{2}=195$$

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

$$2x^{2}-8x+16+\left(x-6\right)^{2}=195$$

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

$$2x^{2}-8x+16+x^{2}-12x+36=195$$

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

$$3x^{2}-8x+16-12x+36=195$$

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

$$3x^{2}-20x+16+36=195$$

Add 16 and 36 to get 52.

$$3x^{2}-20x+52=195$$

Subtract 195 from both sides.

$$3x^{2}-20x+52-195=0$$

Subtract 195 from 52 to get -143.

$$3x^{2}-20x-143=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -20 for b, and -143 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-143\right)}}{2\times 3}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\left(-143\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\left(-143\right)}}{2\times 3}$$

Multiply -12 times -143.

$$x=\frac{-\left(-20\right)±\sqrt{400+1716}}{2\times 3}$$

Add 400 to 1716.

$$x=\frac{-\left(-20\right)±\sqrt{2116}}{2\times 3}$$

Take the square root of 2116.

$$x=\frac{-\left(-20\right)±46}{2\times 3}$$

The opposite of -20 is 20.

$$x=\frac{20±46}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{20±46}{6}$$

Now solve the equation x=\frac{20±46}{6} when ± is plus. Add 20 to 46.

$$x=\frac{66}{6}$$

Divide 66 by 6.

$$x=11$$

Now solve the equation x=\frac{20±46}{6} when ± is minus. Subtract 46 from 20.

$$x=-\frac{26}{6}$$

Reduce the fraction \frac{-26}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{13}{3}$$

The equation is now solved.

$$x=11 x=-\frac{13}{3}$$

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}.

$$x^{2}+x^{2}-8x+16+\left(x-6\right)^{2}=195$$

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

$$2x^{2}-8x+16+\left(x-6\right)^{2}=195$$

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

$$2x^{2}-8x+16+x^{2}-12x+36=195$$

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

$$3x^{2}-8x+16-12x+36=195$$

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

$$3x^{2}-20x+16+36=195$$

Add 16 and 36 to get 52.

$$3x^{2}-20x+52=195$$

Subtract 52 from both sides.

$$3x^{2}-20x=195-52$$

Subtract 52 from 195 to get 143.

$$3x^{2}-20x=143$$

Divide both sides by 3.

$$\frac{3x^{2}-20x}{3}=\frac{143}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{20}{3}x=\frac{143}{3}$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=\frac{143}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{143}{3}+\frac{100}{9}$$

Add \frac{143}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{529}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{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}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{529}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{529}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{23}{3} x-\frac{10}{3}=-\frac{23}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=11 x=-\frac{13}{3}$$