Solve (x-3)^2+(x-6)^2=x^2

$$( x - 3 ) ^ { 2 } + ( x - 6 ) ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x=3$$

$$x=15$$

Steps Using Factoring

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

$$x^{2}-6x+9+\left(x-6\right)^{2}=x^{2}$$

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

$$x^{2}-6x+9+x^{2}-12x+36=x^{2}$$

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

$$2x^{2}-6x+9-12x+36=x^{2}$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x+9+36=x^{2}$$

Add 9 and 36 to get 45.

$$2x^{2}-18x+45=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-18x+45-x^{2}=0$$

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

$$x^{2}-18x+45=0$$

To solve the equation, factor x^{2}-18x+45 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.

$$a+b=-18 ab=45$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

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

$$\left(x-15\right)\left(x-3\right)$$

To find equation solutions, solve x-15=0 and x-3=0.

$$x=15 x=3$$

Steps Using Factoring By Grouping

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

$$x^{2}-6x+9+\left(x-6\right)^{2}=x^{2}$$

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

$$x^{2}-6x+9+x^{2}-12x+36=x^{2}$$

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

$$2x^{2}-6x+9-12x+36=x^{2}$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x+9+36=x^{2}$$

Add 9 and 36 to get 45.

$$2x^{2}-18x+45=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-18x+45-x^{2}=0$$

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

$$x^{2}-18x+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 45=45$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

Rewrite x^{2}-18x+45 as \left(x^{2}-15x\right)+\left(-3x+45\right).

$$\left(x^{2}-15x\right)+\left(-3x+45\right)$$

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

$$x\left(x-15\right)-3\left(x-15\right)$$

Factor out common term x-15 by using distributive property.

$$\left(x-15\right)\left(x-3\right)$$

To find equation solutions, solve x-15=0 and x-3=0.

$$x=15 x=3$$

Steps Using the Quadratic Formula

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

$$x^{2}-6x+9+\left(x-6\right)^{2}=x^{2}$$

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

$$x^{2}-6x+9+x^{2}-12x+36=x^{2}$$

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

$$2x^{2}-6x+9-12x+36=x^{2}$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x+9+36=x^{2}$$

Add 9 and 36 to get 45.

$$2x^{2}-18x+45=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-18x+45-x^{2}=0$$

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

$$x^{2}-18x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 45}}{2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 45}}{2}$$

Multiply -4 times 45.

$$x=\frac{-\left(-18\right)±\sqrt{324-180}}{2}$$

Add 324 to -180.

$$x=\frac{-\left(-18\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-\left(-18\right)±12}{2}$$

The opposite of -18 is 18.

$$x=\frac{18±12}{2}$$

Now solve the equation x=\frac{18±12}{2} when ± is plus. Add 18 to 12.

$$x=\frac{30}{2}$$

Divide 30 by 2.

$$x=15$$

Now solve the equation x=\frac{18±12}{2} when ± is minus. Subtract 12 from 18.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=15 x=3$$

Steps for Completing the Square

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

$$x^{2}-6x+9+\left(x-6\right)^{2}=x^{2}$$

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

$$x^{2}-6x+9+x^{2}-12x+36=x^{2}$$

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

$$2x^{2}-6x+9-12x+36=x^{2}$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x+9+36=x^{2}$$

Add 9 and 36 to get 45.

$$2x^{2}-18x+45=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-18x+45-x^{2}=0$$

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

$$x^{2}-18x+45=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-18x=-45$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-18x+\left(-9\right)^{2}=-45+\left(-9\right)^{2}$$

Square -9.

$$x^{2}-18x+81=-45+81$$

Add -45 to 81.

$$x^{2}-18x+81=36$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-9=6 x-9=-6$$

Add 9 to both sides of the equation.

$$x=15 x=3$$