Solve 1/3x^2-1/6x=1

$$\frac { 1 } { 3 x ^ { 2 } } - \frac { 1 } { 6 x } = 1$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=\frac{1}{2}=0.5$$

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 6x^{2}, the least common multiple of 3x^{2},6x.

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

Subtract 6x^{2} from both sides.

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

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

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

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

$$a+b=-1 ab=-6\times 2=-12$$

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.

$$a=3 b=-4$$

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

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

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

$$-3x\left(2x-1\right)-2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-3x-2\right)$$

To find equation solutions, solve 2x-1=0 and -3x-2=0.

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

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 6x^{2}, the least common multiple of 3x^{2},6x.

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

Subtract 6x^{2} from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -1 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)\times 2}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24\times 2}}{2\left(-6\right)}$$

Multiply 24 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-6\right)}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-6\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\left(-6\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{1±7}{-12}$$

Now solve the equation x=\frac{1±7}{-12} when ± is plus. Add 1 to 7.

$$x=\frac{8}{-12}$$

Reduce the fraction \frac{8}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{1±7}{-12} when ± is minus. Subtract 7 from 1.

$$x=-\frac{6}{-12}$$

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

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

The equation is now solved.

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

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 6x^{2}, the least common multiple of 3x^{2},6x.

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

Subtract 6x^{2} from both sides.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -6.

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

Dividing by -6 undoes the multiplication by -6.

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

Divide -1 by -6.

$$x^{2}+\frac{1}{6}x=-\frac{2}{-6}$$

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

$$x^{2}+\frac{1}{6}x=\frac{1}{3}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{6}x+\left(\frac{1}{12}\right)^{2}=\frac{1}{3}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{1}{3}+\frac{1}{144}$$

Add \frac{1}{3} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{49}{144}$$

Factor x^{2}+\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{12}\right)^{2}=\frac{49}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{12}\right)^{2}}=\sqrt{\frac{49}{144}}$$

Simplify.

$$x+\frac{1}{12}=\frac{7}{12} x+\frac{1}{12}=-\frac{7}{12}$$

Subtract \frac{1}{12} from both sides of the equation.

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