Question

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

Answer 1

There are 2 real number solutions: `x_1=(1+sqrt(97))/6, x_2=(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=(-b+-sqrt(Delta))/(2a)`
We are going to focus on this term here:
`+-sqrt(Delta)`

It's quite obvious that if `Delta>0` then `+-sqrt(Delta)inRR`

It should also be a bit obvious that if `Delta<0` then `+-sqrt(Delta)inCC`
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 `ainCC`

Regarding the number of roots, we can see that the term `+-sqrt(Delta)` has a `+-` 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:
`ax^2+bx+c=(x-b/(2a))^2` and the other interesting thing is that when `Delta=0`, the term `-b/(2a)` 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]}