Solve -2a^2+9a-9

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

Factor

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

Solution Steps

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

$$p+q=9 pq=-2\left(-9\right)=18$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. 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 -2a^{2}+9a-9 as \left(-2a^{2}+6a\right)+\left(3a-9\right).

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

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

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

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

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

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.

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

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.

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

Square 9.

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

Multiply -4 times -2.

$$a=\frac{-9±\sqrt{81+8\left(-9\right)}}{2\left(-2\right)}$$

Multiply 8 times -9.

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

Add 81 to -72.

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

Take the square root of 9.

$$a=\frac{-9±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$a=\frac{-9±3}{-4}$$

Now solve the equation a=\frac{-9±3}{-4} when ± is plus. Add -9 to 3.

$$a=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation a=\frac{-9±3}{-4} when ± is minus. Subtract 3 from -9.

$$a=-\frac{12}{-4}$$

Divide -12 by -4.

$$a=3$$

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

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

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

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

Cancel out 2, the greatest common factor in -2 and 2.

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

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.

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

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

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

Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{4}. 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>

$$r = \frac{9}{4} - u s = \frac{9}{4} + u$$

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

$$(\frac{9}{4} - u) (\frac{9}{4} + u) = \frac{9}{2}$$

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

$$\frac{81}{16} - u^2 = \frac{9}{2}$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = \frac{9}{2}-\frac{81}{16} = -\frac{9}{16}$$

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

$$u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}$$

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

$$r =\frac{9}{4} - \frac{3}{4} = 1.500 s = \frac{9}{4} + \frac{3}{4} = 3$$

Evaluate

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