Solve x^2-x=2/9-2x/3

$$x ^ { 2 } - x = \frac { 2 } { 9 } - \frac { 2 x } { 3 }$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

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

Steps Using Factoring By Grouping

Multiply both sides of the equation by 9, the least common multiple of 9,3.

$$9x^{2}-9x=2-3\times 2x$$

Multiply -3 and 2 to get -6.

$$9x^{2}-9x=2-6x$$

Subtract 2 from both sides.

$$9x^{2}-9x-2=-6x$$

Add 6x to both sides.

$$9x^{2}-9x-2+6x=0$$

Combine -9x and 6x to get -3x.

$$9x^{2}-3x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=9\left(-2\right)=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=-6 b=3$$

Rewrite 9x^{2}-3x-2 as \left(9x^{2}-6x\right)+\left(3x-2\right).

$$\left(9x^{2}-6x\right)+\left(3x-2\right)$$

Factor out 3x in 9x^{2}-6x.

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

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

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

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

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 9, the least common multiple of 9,3.

$$9x^{2}-9x=2-3\times 2x$$

Multiply -3 and 2 to get -6.

$$9x^{2}-9x=2-6x$$

Subtract 2 from both sides.

$$9x^{2}-9x-2=-6x$$

Add 6x to both sides.

$$9x^{2}-9x-2+6x=0$$

Combine -9x and 6x to get -3x.

$$9x^{2}-3x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 9\left(-2\right)}}{2\times 9}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 9\left(-2\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-3\right)±\sqrt{9-36\left(-2\right)}}{2\times 9}$$

Multiply -36 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 9}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\times 9}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2\times 9}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{3±9}{18}$$

Now solve the equation x=\frac{3±9}{18} when ± is plus. Add 3 to 9.

$$x=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{3±9}{18} when ± is minus. Subtract 9 from 3.

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

Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 9, the least common multiple of 9,3.

$$9x^{2}-9x=2-3\times 2x$$

Multiply -3 and 2 to get -6.

$$9x^{2}-9x=2-6x$$

Add 6x to both sides.

$$9x^{2}-9x+6x=2$$

Combine -9x and 6x to get -3x.

$$9x^{2}-3x=2$$

Divide both sides by 9.

$$\frac{9x^{2}-3x}{9}=\frac{2}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{3}{9}\right)x=\frac{2}{9}$$

Reduce the fraction \frac{-3}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{3}x=\frac{2}{9}$$

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.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{2}{9}+\left(-\frac{1}{6}\right)^{2}$$

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

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{2}{9}+\frac{1}{36}$$

Add \frac{2}{9} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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