Solve (3-x)^2/9=x(3-x)/12

$$\frac { ( 3 - x ) ^ { 2 } } { 9 } = \frac { x ( 3 - x ) } { 12 }$$

Solve for x

$$x = \frac{12}{7} = 1\frac{5}{7} \approx 1.714285714$$

$$x=3$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 36, the least common multiple of 9,12.

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

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

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

Use the distributive property to multiply 4 by 9-6x+x^{2}.

$$36-24x+4x^{2}=3x\left(3-x\right)$$

Use the distributive property to multiply 3x by 3-x.

$$36-24x+4x^{2}=9x-3x^{2}$$

Subtract 9x from both sides.

$$36-24x+4x^{2}-9x=-3x^{2}$$

Combine -24x and -9x to get -33x.

$$36-33x+4x^{2}=-3x^{2}$$

Add 3x^{2} to both sides.

$$36-33x+4x^{2}+3x^{2}=0$$

Combine 4x^{2} and 3x^{2} to get 7x^{2}.

$$36-33x+7x^{2}=0$$

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

$$7x^{2}-33x+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=-33 ab=7\times 36=252$$

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 252.

$$-1,-252 -2,-126 -3,-84 -4,-63 -6,-42 -7,-36 -9,-28 -12,-21 -14,-18$$

Calculate the sum for each pair.

$$-1-252=-253 -2-126=-128 -3-84=-87 -4-63=-67 -6-42=-48 -7-36=-43 -9-28=-37 -12-21=-33 -14-18=-32$$

The solution is the pair that gives sum -33.

$$a=-21 b=-12$$

Rewrite 7x^{2}-33x+36 as \left(7x^{2}-21x\right)+\left(-12x+36\right).

$$\left(7x^{2}-21x\right)+\left(-12x+36\right)$$

Factor out 7x in the first and -12 in the second group.

$$7x\left(x-3\right)-12\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(7x-12\right)$$

To find equation solutions, solve x-3=0 and 7x-12=0.

$$x=3 x=\frac{12}{7}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 36, the least common multiple of 9,12.

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

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

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

Use the distributive property to multiply 4 by 9-6x+x^{2}.

$$36-24x+4x^{2}=3x\left(3-x\right)$$

Use the distributive property to multiply 3x by 3-x.

$$36-24x+4x^{2}=9x-3x^{2}$$

Subtract 9x from both sides.

$$36-24x+4x^{2}-9x=-3x^{2}$$

Combine -24x and -9x to get -33x.

$$36-33x+4x^{2}=-3x^{2}$$

Add 3x^{2} to both sides.

$$36-33x+4x^{2}+3x^{2}=0$$

Combine 4x^{2} and 3x^{2} to get 7x^{2}.

$$36-33x+7x^{2}=0$$

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.

$$7x^{2}-33x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -33 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 7\times 36}}{2\times 7}$$

Square -33.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\times 7\times 36}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-33\right)±\sqrt{1089-28\times 36}}{2\times 7}$$

Multiply -28 times 36.

$$x=\frac{-\left(-33\right)±\sqrt{1089-1008}}{2\times 7}$$

Add 1089 to -1008.

$$x=\frac{-\left(-33\right)±\sqrt{81}}{2\times 7}$$

Take the square root of 81.

$$x=\frac{-\left(-33\right)±9}{2\times 7}$$

The opposite of -33 is 33.

$$x=\frac{33±9}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{33±9}{14}$$

Now solve the equation x=\frac{33±9}{14} when ± is plus. Add 33 to 9.

$$x=\frac{42}{14}$$

Divide 42 by 14.

$$x=3$$

Now solve the equation x=\frac{33±9}{14} when ± is minus. Subtract 9 from 33.

$$x=\frac{24}{14}$$

Reduce the fraction \frac{24}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{12}{7}$$

The equation is now solved.

$$x=3 x=\frac{12}{7}$$

Steps for Completing the Square

Multiply both sides of the equation by 36, the least common multiple of 9,12.

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

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

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

Use the distributive property to multiply 4 by 9-6x+x^{2}.

$$36-24x+4x^{2}=3x\left(3-x\right)$$

Use the distributive property to multiply 3x by 3-x.

$$36-24x+4x^{2}=9x-3x^{2}$$

Subtract 9x from both sides.

$$36-24x+4x^{2}-9x=-3x^{2}$$

Combine -24x and -9x to get -33x.

$$36-33x+4x^{2}=-3x^{2}$$

Add 3x^{2} to both sides.

$$36-33x+4x^{2}+3x^{2}=0$$

Combine 4x^{2} and 3x^{2} to get 7x^{2}.

$$36-33x+7x^{2}=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$-33x+7x^{2}=-36$$

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.

$$7x^{2}-33x=-36$$

Divide both sides by 7.

$$\frac{7x^{2}-33x}{7}=-\frac{36}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{33}{7}x=-\frac{36}{7}$$

Divide -\frac{33}{7}, the coefficient of the x term, by 2 to get -\frac{33}{14}. Then add the square of -\frac{33}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{33}{7}x+\left(-\frac{33}{14}\right)^{2}=-\frac{36}{7}+\left(-\frac{33}{14}\right)^{2}$$

Square -\frac{33}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{33}{7}x+\frac{1089}{196}=-\frac{36}{7}+\frac{1089}{196}$$

Add -\frac{36}{7} to \frac{1089}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{33}{7}x+\frac{1089}{196}=\frac{81}{196}$$

Factor x^{2}-\frac{33}{7}x+\frac{1089}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{33}{14}\right)^{2}=\frac{81}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{33}{14}\right)^{2}}=\sqrt{\frac{81}{196}}$$

Simplify.

$$x-\frac{33}{14}=\frac{9}{14} x-\frac{33}{14}=-\frac{9}{14}$$

Add \frac{33}{14} to both sides of the equation.

$$x=3 x=\frac{12}{7}$$