Solve -2x^2-3x+9

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

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-2\times 9=-18$$

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.

$$a=3 b=-6$$

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

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

Factor out -x in the first and -3 in the second group.

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

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

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

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.

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

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.

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

Square -3.

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

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\times 9}}{2\left(-2\right)}$$

Multiply 8 times 9.

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

Add 9 to 72.

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

Take the square root of 81.

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

The opposite of -3 is 3.

$$x=\frac{3±9}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±9}{-4}$$

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

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

Divide 12 by -4.

$$x=-3$$

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

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

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

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

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

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

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

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

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

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

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.

$$x ^ 2 +\frac{3}{2}x -\frac{9}{2} = 0$$

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

$$r + s = -\frac{3}{2} rs = -\frac{9}{2}$$

Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{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>

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

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

$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = -\frac{9}{2}$$

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

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

Simplify the expression by subtracting \frac{9}{16} on both sides

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

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

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

Evaluate

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