How do you solve 2x^2-3x-9=0?

Jul 13, 2017

$x = - \frac{3}{2} \text{ }$ or $\text{ } x = 3$

Explanation:

Given:

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

Note that this quadratic is in the standard form:

$a {x}^{2} + b x + c = 0$

with $a = 2$, $b = - 3$ and $c = - 9$.

Its discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 3}\right)}^{2} - 4 \left(\textcolor{b l u e}{2}\right) \left(\textcolor{b l u e}{- 9}\right) = 9 + 72 = 81 = {9}^{2}$

Since this is positive and a perfect square, we can deduce that this quadratic has rational zeros that we can find by factoring with integer coefficients...

Use an AC method:

Look for a pair of factors of $A C = 2 \cdot 9 = 18$ which differ by $B = 3$.

The pair $6 , 3$ works in that $6 \cdot 3 = 18$ and $6 - 3 = 3$.

Use this pair to split the middle term and factor by grouping:

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

$\textcolor{w h i t e}{0} = \left(2 {x}^{2} - 6 x\right) + \left(3 x - 9\right)$

$\textcolor{w h i t e}{0} = 2 x \left(x - 3\right) + 3 \left(x - 3\right)$

$\textcolor{w h i t e}{0} = \left(2 x + 3\right) \left(x - 3\right)$

Hence zeros:

$x = - \frac{3}{2} \text{ }$ or $\text{ } x = 3$