Solve x/x^2-2x-5/3x^2-12=2/x

$$\frac { x } { x ^ { 2 } - 2 x } - \frac { 5 } { 3 x ^ { 2 } - 12 } = \frac { 2 } { x }$$

Solve for x

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-2x,3x^{2}-12,x.

$$\left(3x+6\right)x-x\times 5=\left(3x^{2}-12\right)\times 2$$

Use the distributive property to multiply 3x+6 by x.

$$3x^{2}+6x-x\times 5=\left(3x^{2}-12\right)\times 2$$

Use the distributive property to multiply 3x^{2}-12 by 2.

$$3x^{2}+6x-x\times 5=6x^{2}-24$$

Subtract 6x^{2} from both sides.

$$3x^{2}+6x-x\times 5-6x^{2}=-24$$

Combine 3x^{2} and -6x^{2} to get -3x^{2}.

$$-3x^{2}+6x-x\times 5=-24$$

Add 24 to both sides.

$$-3x^{2}+6x-x\times 5+24=0$$

Multiply -1 and 5 to get -5.

$$-3x^{2}+6x-5x+24=0$$

Combine 6x and -5x to get x.

$$-3x^{2}+x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=1 ab=-3\times 24=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$a=9 b=-8$$

Rewrite -3x^{2}+x+24 as \left(-3x^{2}+9x\right)+\left(-8x+24\right).

$$\left(-3x^{2}+9x\right)+\left(-8x+24\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(-x+3\right)+8\left(-x+3\right)$$

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

$$\left(-x+3\right)\left(3x+8\right)$$

To find equation solutions, solve -x+3=0 and 3x+8=0.

$$x=3 x=-\frac{8}{3}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-2x,3x^{2}-12,x.

$$\left(3x+6\right)x-x\times 5=\left(3x^{2}-12\right)\times 2$$

Use the distributive property to multiply 3x+6 by x.

$$3x^{2}+6x-x\times 5=\left(3x^{2}-12\right)\times 2$$

Use the distributive property to multiply 3x^{2}-12 by 2.

$$3x^{2}+6x-x\times 5=6x^{2}-24$$

Subtract 6x^{2} from both sides.

$$3x^{2}+6x-x\times 5-6x^{2}=-24$$

Combine 3x^{2} and -6x^{2} to get -3x^{2}.

$$-3x^{2}+6x-x\times 5=-24$$

Add 24 to both sides.

$$-3x^{2}+6x-x\times 5+24=0$$

Multiply -1 and 5 to get -5.

$$-3x^{2}+6x-5x+24=0$$

Combine 6x and -5x to get x.

$$-3x^{2}+x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 1 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)\times 24}}{2\left(-3\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-3\right)\times 24}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-1±\sqrt{1+12\times 24}}{2\left(-3\right)}$$

Multiply 12 times 24.

$$x=\frac{-1±\sqrt{1+288}}{2\left(-3\right)}$$

Add 1 to 288.

$$x=\frac{-1±\sqrt{289}}{2\left(-3\right)}$$

Take the square root of 289.

$$x=\frac{-1±17}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±17}{-6}$$

Now solve the equation x=\frac{-1±17}{-6} when ± is plus. Add -1 to 17.

$$x=\frac{16}{-6}$$

Reduce the fraction \frac{16}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{8}{3}$$

Now solve the equation x=\frac{-1±17}{-6} when ± is minus. Subtract 17 from -1.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=-\frac{8}{3} x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-2x,3x^{2}-12,x.

$$\left(3x+6\right)x-x\times 5=\left(3x^{2}-12\right)\times 2$$

Use the distributive property to multiply 3x+6 by x.

$$3x^{2}+6x-x\times 5=\left(3x^{2}-12\right)\times 2$$

Use the distributive property to multiply 3x^{2}-12 by 2.

$$3x^{2}+6x-x\times 5=6x^{2}-24$$

Subtract 6x^{2} from both sides.

$$3x^{2}+6x-x\times 5-6x^{2}=-24$$

Combine 3x^{2} and -6x^{2} to get -3x^{2}.

$$-3x^{2}+6x-x\times 5=-24$$

Multiply -1 and 5 to get -5.

$$-3x^{2}+6x-5x=-24$$

Combine 6x and -5x to get x.

$$-3x^{2}+x=-24$$

Divide both sides by -3.

$$\frac{-3x^{2}+x}{-3}=-\frac{24}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{1}{-3}x=-\frac{24}{-3}$$

Divide 1 by -3.

$$x^{2}-\frac{1}{3}x=-\frac{24}{-3}$$

Divide -24 by -3.

$$x^{2}-\frac{1}{3}x=8$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=8+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=8+\frac{1}{36}$$

Add 8 to \frac{1}{36}.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{289}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{289}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{289}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{17}{6} x-\frac{1}{6}=-\frac{17}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=3 x=-\frac{8}{3}$$