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

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

Solve for x

$$x=2$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

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

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

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

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

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

$$3x=6x^{2}-36x+54$$

Subtract 6x^{2} from both sides.

$$3x-6x^{2}=-36x+54$$

Add 36x to both sides.

$$3x-6x^{2}+36x=54$$

Combine 3x and 36x to get 39x.

$$39x-6x^{2}=54$$

Subtract 54 from both sides.

$$39x-6x^{2}-54=0$$

Divide both sides by 3.

$$13x-2x^{2}-18=0$$

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

$$-2x^{2}+13x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=13 ab=-2\left(-18\right)=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=9 b=4$$

Rewrite -2x^{2}+13x-18 as \left(-2x^{2}+9x\right)+\left(4x-18\right).

$$\left(-2x^{2}+9x\right)+\left(4x-18\right)$$

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

$$-x\left(2x-9\right)+2\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(-x+2\right)$$

To find equation solutions, solve 2x-9=0 and -x+2=0.

$$x=\frac{9}{2} x=2$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$3x=6x^{2}-36x+54$$

Subtract 6x^{2} from both sides.

$$3x-6x^{2}=-36x+54$$

Add 36x to both sides.

$$3x-6x^{2}+36x=54$$

Combine 3x and 36x to get 39x.

$$39x-6x^{2}=54$$

Subtract 54 from both sides.

$$39x-6x^{2}-54=0$$

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.

$$-6x^{2}+39x-54=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 39 for b, and -54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-39±\sqrt{39^{2}-4\left(-6\right)\left(-54\right)}}{2\left(-6\right)}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\left(-6\right)\left(-54\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-39±\sqrt{1521+24\left(-54\right)}}{2\left(-6\right)}$$

Multiply 24 times -54.

$$x=\frac{-39±\sqrt{1521-1296}}{2\left(-6\right)}$$

Add 1521 to -1296.

$$x=\frac{-39±\sqrt{225}}{2\left(-6\right)}$$

Take the square root of 225.

$$x=\frac{-39±15}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-39±15}{-12}$$

Now solve the equation x=\frac{-39±15}{-12} when ± is plus. Add -39 to 15.

$$x=-\frac{24}{-12}$$

Divide -24 by -12.

$$x=2$$

Now solve the equation x=\frac{-39±15}{-12} when ± is minus. Subtract 15 from -39.

$$x=-\frac{54}{-12}$$

Reduce the fraction \frac{-54}{-12} to lowest terms by extracting and canceling out 6.

$$x=\frac{9}{2}$$

The equation is now solved.

$$x=2 x=\frac{9}{2}$$

Steps for Completing the Square

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

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

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

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

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

$$3x=6x^{2}-36x+54$$

Subtract 6x^{2} from both sides.

$$3x-6x^{2}=-36x+54$$

Add 36x to both sides.

$$3x-6x^{2}+36x=54$$

Combine 3x and 36x to get 39x.

$$39x-6x^{2}=54$$

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.

$$-6x^{2}+39x=54$$

Divide both sides by -6.

$$\frac{-6x^{2}+39x}{-6}=\frac{54}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{39}{-6}x=\frac{54}{-6}$$

Reduce the fraction \frac{39}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{13}{2}x=\frac{54}{-6}$$

Divide 54 by -6.

$$x^{2}-\frac{13}{2}x=-9$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-9+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-9+\frac{169}{16}$$

Add -9 to \frac{169}{16}.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{5}{4} x-\frac{13}{4}=-\frac{5}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=\frac{9}{2} x=2$$