Solve 9a^2-9a+2

$$9 a ^ { 2 } - 9 a + 2$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9a^{2}+pa+qa+2. To find p and q, set up a system to be solved.

$$p+q=-9 pq=9\times 2=18$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 18.

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

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$p=-6 q=-3$$

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

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

Factor out 3a in the first and -1 in the second group.

$$3a\left(3a-2\right)-\left(3a-2\right)$$

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

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$9a^{2}-9a+2=0$$

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.

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

Square -9.

$$a=\frac{-\left(-9\right)±\sqrt{81-4\times 9\times 2}}{2\times 9}$$

Multiply -4 times 9.

$$a=\frac{-\left(-9\right)±\sqrt{81-36\times 2}}{2\times 9}$$

Multiply -36 times 2.

$$a=\frac{-\left(-9\right)±\sqrt{81-72}}{2\times 9}$$

Add 81 to -72.

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

Take the square root of 9.

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

The opposite of -9 is 9.

$$a=\frac{9±3}{2\times 9}$$

Multiply 2 times 9.

$$a=\frac{9±3}{18}$$

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

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

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

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

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

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

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

$$a=\frac{1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and \frac{1}{3} for x_{2}.

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

Subtract \frac{2}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Subtract \frac{1}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9a^{2}-9a+2=9\times \frac{3a-2}{3}\times \frac{3a-1}{3}$$

Multiply \frac{3a-2}{3} times \frac{3a-1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Multiply 3 times 3.

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

Cancel out 9, the greatest common factor in 9 and 9.

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 9

$$x ^ 2 -1x +\frac{2}{9} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 1 rs = \frac{2}{9}$$

Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{9}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = \frac{2}{9}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1}{4} - u^2 = \frac{2}{9}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = \frac{2}{9}-\frac{1}{4} = -\frac{1}{36}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{1}{2} - \frac{1}{6} = 0.333 s = \frac{1}{2} + \frac{1}{6} = 0.667$$

Evaluate

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