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

1 Answer
May 23, 2017

Answer:

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]}