# How do you find the discriminant for 3x^2-x=8 and determine the number and type of solutions?

May 23, 2017

There are 2 real number solutions: ${x}_{1} = \frac{1 + \sqrt{97}}{6} , {x}_{2} = \frac{1 - \sqrt{97}}{6}$

#### Explanation:

Using the discriminant, we can evaluate the type and number of roots to a quadratic using these rules (explanation comes after):

1. if $\Delta = 0$ then there is 1 root
2. if $\Delta > 0$ then there are 2 real number roots
3. if $\Delta < 0$ then there are 2 complex roots

Note that $\Delta$ here is the discriminant.

But why?

Well, let's take a look at the quadratic formula:
$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$
We are going to focus on this term here:
$\pm \sqrt{\Delta}$

It's quite obvious that if $\Delta > 0$ then $\pm \sqrt{\Delta} \in \mathbb{R}$

It should also be a bit obvious that if $\Delta < 0$ then $\pm \sqrt{\Delta} \in \mathbb{C}$
as there is no real number such that that number squared gives a negative, using the wonderful language of Math:
If ${a}^{2} < 0$ then $a \in \mathbb{C}$

Regarding the number of roots, we can see that the term $\pm \sqrt{\Delta}$ has a $\pm$ sign, which means it has two values, and therefore there will be two values to $x$ **except...#

When $\Delta = 0$, $\sqrt{\Delta} = 0$, so then $\sqrt{\Delta} = - \sqrt{\Delta}$ and we converge on one value, $0$. An interesting thing is that when the discriminant is $0$, then the equation is a perfect square and can be factored like this:
$a {x}^{2} + b x + c = {\left(x - \frac{b}{2 a}\right)}^{2}$ and the other interesting thing is that when $\Delta = 0$, the term $- \frac{b}{2 a}$ is actually the vertex of the quadratic!
Wow! The two roots and the vertex is one point!! Let me leave you off with this pretty example of the discriminant equal to $0$.
graph{x^2-4x+4 [-10, 10, -5, 5]}