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

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

Factor

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

Solution Steps

Factor out \frac{1}{2}.

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

Consider 2x^{2}-3x-9. 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.

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

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

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

Factor out 2x in the first and 3 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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