Solve 1/(x-3)^2+4/x(x-3)+2/3-x=0

$$\frac { 1 } { ( x - 3 ) ^ { 2 } } + \frac { 4 } { x ( x - 3 ) } + \frac { 2 } { 3 - x } = 0$$

Solve for x

$$x=4$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of \left(x-3\right)^{2},x\left(x-3\right),3-x.

$$x+\left(x-3\right)\times 4-x\left(-3+x\right)\times 2=0$$

Use the distributive property to multiply x-3 by 4.

$$x+4x-12-x\left(-3+x\right)\times 2=0$$

Combine x and 4x to get 5x.

$$5x-12-x\left(-3+x\right)\times 2=0$$

Multiply -1 and 2 to get -2.

$$5x-12-2x\left(-3+x\right)=0$$

Use the distributive property to multiply -2x by -3+x.

$$5x-12+6x-2x^{2}=0$$

Combine 5x and 6x to get 11x.

$$11x-12-2x^{2}=0$$

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

$$-2x^{2}+11x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=11 ab=-2\left(-12\right)=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

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

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=8 b=3$$

Rewrite -2x^{2}+11x-12 as \left(-2x^{2}+8x\right)+\left(3x-12\right).

$$\left(-2x^{2}+8x\right)+\left(3x-12\right)$$

Factor out 2x in the first and -3 in the second group.

$$2x\left(-x+4\right)-3\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(2x-3\right)$$

To find equation solutions, solve -x+4=0 and 2x-3=0.

$$x=4 x=\frac{3}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of \left(x-3\right)^{2},x\left(x-3\right),3-x.

$$x+\left(x-3\right)\times 4-x\left(-3+x\right)\times 2=0$$

Use the distributive property to multiply x-3 by 4.

$$x+4x-12-x\left(-3+x\right)\times 2=0$$

Combine x and 4x to get 5x.

$$5x-12-x\left(-3+x\right)\times 2=0$$

Multiply -1 and 2 to get -2.

$$5x-12-2x\left(-3+x\right)=0$$

Use the distributive property to multiply -2x by -3+x.

$$5x-12+6x-2x^{2}=0$$

Combine 5x and 6x to get 11x.

$$11x-12-2x^{2}=0$$

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.

$$-2x^{2}+11x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-12\right)}}{2\left(-2\right)}$$

Multiply 8 times -12.

$$x=\frac{-11±\sqrt{121-96}}{2\left(-2\right)}$$

Add 121 to -96.

$$x=\frac{-11±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-11±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±5}{-4}$$

Now solve the equation x=\frac{-11±5}{-4} when ± is plus. Add -11 to 5.

$$x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{-11±5}{-4} when ± is minus. Subtract 5 from -11.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=\frac{3}{2} x=4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of \left(x-3\right)^{2},x\left(x-3\right),3-x.

$$x+\left(x-3\right)\times 4-x\left(-3+x\right)\times 2=0$$

Use the distributive property to multiply x-3 by 4.

$$x+4x-12-x\left(-3+x\right)\times 2=0$$

Combine x and 4x to get 5x.

$$5x-12-x\left(-3+x\right)\times 2=0$$

Multiply -1 and 2 to get -2.

$$5x-12-2x\left(-3+x\right)=0$$

Use the distributive property to multiply -2x by -3+x.

$$5x-12+6x-2x^{2}=0$$

Combine 5x and 6x to get 11x.

$$11x-12-2x^{2}=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$11x-2x^{2}=12$$

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.

$$-2x^{2}+11x=12$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{12}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{12}{-2}$$

Divide 12 by -2.

$$x^{2}-\frac{11}{2}x=-6$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-6+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-6+\frac{121}{16}$$

Add -6 to \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{5}{4} x-\frac{11}{4}=-\frac{5}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=4 x=\frac{3}{2}$$