Solve 24x-72/x=12

$$24 x - \frac { 72 } { x } = 12$$

Solve for x

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

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$24xx-72=12x$$

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

$$24x^{2}-72=12x$$

Subtract 12x from both sides.

$$24x^{2}-72-12x=0$$

Divide both sides by 12.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-6\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite 2x^{2}-x-6 as \left(2x^{2}-4x\right)+\left(3x-6\right).

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$24xx-72=12x$$

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

$$24x^{2}-72=12x$$

Subtract 12x from both sides.

$$24x^{2}-72-12x=0$$

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.

$$24x^{2}-12x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, -12 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 24\left(-72\right)}}{2\times 24}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 24\left(-72\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-12\right)±\sqrt{144-96\left(-72\right)}}{2\times 24}$$

Multiply -96 times -72.

$$x=\frac{-\left(-12\right)±\sqrt{144+6912}}{2\times 24}$$

Add 144 to 6912.

$$x=\frac{-\left(-12\right)±\sqrt{7056}}{2\times 24}$$

Take the square root of 7056.

$$x=\frac{-\left(-12\right)±84}{2\times 24}$$

The opposite of -12 is 12.

$$x=\frac{12±84}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{12±84}{48}$$

Now solve the equation x=\frac{12±84}{48} when ± is plus. Add 12 to 84.

$$x=\frac{96}{48}$$

Divide 96 by 48.

$$x=2$$

Now solve the equation x=\frac{12±84}{48} when ± is minus. Subtract 84 from 12.

$$x=-\frac{72}{48}$$

Reduce the fraction \frac{-72}{48} to lowest terms by extracting and canceling out 24.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$24xx-72=12x$$

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

$$24x^{2}-72=12x$$

Subtract 12x from both sides.

$$24x^{2}-72-12x=0$$

Add 72 to both sides. Anything plus zero gives itself.

$$24x^{2}-12x=72$$

Divide both sides by 24.

$$\frac{24x^{2}-12x}{24}=\frac{72}{24}$$

Dividing by 24 undoes the multiplication by 24.

$$x^{2}+\left(-\frac{12}{24}\right)x=\frac{72}{24}$$

Reduce the fraction \frac{-12}{24} to lowest terms by extracting and canceling out 12.

$$x^{2}-\frac{1}{2}x=\frac{72}{24}$$

Divide 72 by 24.

$$x^{2}-\frac{1}{2}x=3$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=3+\frac{1}{16}$$

Add 3 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{7}{4} x-\frac{1}{4}=-\frac{7}{4}$$

Add \frac{1}{4} to both sides of the equation.

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