Solve x^2-x/6-1/3=0

$$x ^ { 2 } - \frac { x } { 6 } - \frac { 1 } { 3 } = 0$$

Solve for x

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

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

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6, the least common multiple of 6,3.

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

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.

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

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

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

Factor out 2x in 6x^{2}-4x.

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

Factor out common term 3x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 6,3.

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

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}.

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

Multiply -4 times 6.

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

Multiply -24 times -2.

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

Add 1 to 48.

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

Take the square root of 49.

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

The opposite of -1 is 1.

$$x=\frac{1±7}{2\times 6}$$

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

Multiply both sides of the equation by 6, the least common multiple of 6,3.

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

Add 2 to both sides. Anything plus zero gives itself.

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

Divide both sides by 6.

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

Dividing by 6 undoes the multiplication by 6.

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

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.

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

Add \frac{1}{12} to both sides of the equation.

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