Solve x=8/x+1/3

$$x = \frac { 8 } { x } + \frac { 1 } { 3 }$$

Solve for x

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

$$x=3$$

Steps Using Factoring By Grouping

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 3 is 3x. Multiply \frac{8}{x} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{x}{x}.

$$x=\frac{8\times 3}{3x}+\frac{x}{3x}$$

Since \frac{8\times 3}{3x} and \frac{x}{3x} have the same denominator, add them by adding their numerators.

$$x=\frac{8\times 3+x}{3x}$$

Do the multiplications in 8\times 3+x.

$$x=\frac{24+x}{3x}$$

Subtract \frac{24+x}{3x} from both sides.

$$x-\frac{24+x}{3x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3x}{3x}.

$$\frac{x\times 3x}{3x}-\frac{24+x}{3x}=0$$

Since \frac{x\times 3x}{3x} and \frac{24+x}{3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 3x-\left(24+x\right)}{3x}=0$$

Do the multiplications in x\times 3x-\left(24+x\right).

$$\frac{3x^{2}-24-x}{3x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x.

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

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

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

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.

$$a+b=-1 ab=3\left(-24\right)=-72$$

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 -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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 3 is 3x. Multiply \frac{8}{x} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{x}{x}.

$$x=\frac{8\times 3}{3x}+\frac{x}{3x}$$

Since \frac{8\times 3}{3x} and \frac{x}{3x} have the same denominator, add them by adding their numerators.

$$x=\frac{8\times 3+x}{3x}$$

Do the multiplications in 8\times 3+x.

$$x=\frac{24+x}{3x}$$

Subtract \frac{24+x}{3x} from both sides.

$$x-\frac{24+x}{3x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3x}{3x}.

$$\frac{x\times 3x}{3x}-\frac{24+x}{3x}=0$$

Since \frac{x\times 3x}{3x} and \frac{24+x}{3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 3x-\left(24+x\right)}{3x}=0$$

Do the multiplications in x\times 3x-\left(24+x\right).

$$\frac{3x^{2}-24-x}{3x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x.

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

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.

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

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}.

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

Multiply -4 times 3.

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

Multiply -12 times -24.

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

Add 1 to 288.

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

Take the square root of 289.

$$x=\frac{-\left(-1\right)±17}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±17}{2\times 3}$$

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.

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

Divide 18 by 6.

$$x=3$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 3 is 3x. Multiply \frac{8}{x} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{x}{x}.

$$x=\frac{8\times 3}{3x}+\frac{x}{3x}$$

Since \frac{8\times 3}{3x} and \frac{x}{3x} have the same denominator, add them by adding their numerators.

$$x=\frac{8\times 3+x}{3x}$$

Do the multiplications in 8\times 3+x.

$$x=\frac{24+x}{3x}$$

Subtract \frac{24+x}{3x} from both sides.

$$x-\frac{24+x}{3x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3x}{3x}.

$$\frac{x\times 3x}{3x}-\frac{24+x}{3x}=0$$

Since \frac{x\times 3x}{3x} and \frac{24+x}{3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 3x-\left(24+x\right)}{3x}=0$$

Do the multiplications in x\times 3x-\left(24+x\right).

$$\frac{3x^{2}-24-x}{3x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x.

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

Add 24 to both sides. Anything plus zero gives itself.

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

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 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.

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