Solve 9x^2-8x+x=2

$$9 x ^ { 2 } - 8 x + x = 2$$

Solve for x

$$x=-\frac{2}{9}\approx -0.222222222$$

$$x=1$$

Steps Using Factoring By Grouping

Combine -8x and x to get -7x.

$$9x^{2}-7x=2$$

Subtract 2 from both sides.

$$9x^{2}-7x-2=0$$

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.

$$a+b=-7 ab=9\left(-2\right)=-18$$

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 -7.

$$a=-9 b=2$$

Rewrite 9x^{2}-7x-2 as \left(9x^{2}-9x\right)+\left(2x-2\right).

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

Factor out 9x in the first and 2 in the second group.

$$9x\left(x-1\right)+2\left(x-1\right)$$

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

$$\left(x-1\right)\left(9x+2\right)$$

To find equation solutions, solve x-1=0 and 9x+2=0.

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

Steps Using the Quadratic Formula

Combine -8x and x to get -7x.

$$9x^{2}-7x=2$$

Subtract 2 from both sides.

$$9x^{2}-7x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -7 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 9\left(-2\right)}}{2\times 9}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 9\left(-2\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-7\right)±\sqrt{49-36\left(-2\right)}}{2\times 9}$$

Multiply -36 times -2.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\times 9}$$

Add 49 to 72.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\times 9}$$

Take the square root of 121.

$$x=\frac{-\left(-7\right)±11}{2\times 9}$$

The opposite of -7 is 7.

$$x=\frac{7±11}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{7±11}{18}$$

Now solve the equation x=\frac{7±11}{18} when ± is plus. Add 7 to 11.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{7±11}{18} when ± is minus. Subtract 11 from 7.

$$x=-\frac{4}{18}$$

Reduce the fraction \frac{-4}{18} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Combine -8x and x to get -7x.

$$9x^{2}-7x=2$$

Divide both sides by 9.

$$\frac{9x^{2}-7x}{9}=\frac{2}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{7}{9}x=\frac{2}{9}$$

Divide -\frac{7}{9}, the coefficient of the x term, by 2 to get -\frac{7}{18}. Then add the square of -\frac{7}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{9}x+\left(-\frac{7}{18}\right)^{2}=\frac{2}{9}+\left(-\frac{7}{18}\right)^{2}$$

Square -\frac{7}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{9}x+\frac{49}{324}=\frac{2}{9}+\frac{49}{324}$$

Add \frac{2}{9} to \frac{49}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{9}x+\frac{49}{324}=\frac{121}{324}$$

Factor x^{2}-\frac{7}{9}x+\frac{49}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{18}\right)^{2}=\frac{121}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{18}\right)^{2}}=\sqrt{\frac{121}{324}}$$

Simplify.

$$x-\frac{7}{18}=\frac{11}{18} x-\frac{7}{18}=-\frac{11}{18}$$

Add \frac{7}{18} to both sides of the equation.

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