SOLUTION: Compute the discriminant of the quadratic equation 3x^2+x+2=0 Then write a brief sentence describing the number and type of solutions for this equation. How do I solve this?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Compute the discriminant of the quadratic equation 3x^2+x+2=0 Then write a brief sentence describing the number and type of solutions for this equation. How do I solve this?      Log On


   

Question 574827: Compute the discriminant of the quadratic equation 3x^2+x+2=0
Then write a brief sentence describing the number and type of solutions for this
equation. How do I solve this?
Found 2 solutions by jerryguo41, josmiceli:
Answer by jerryguo41(197) About Me  (Show Source):
You can put this solution on YOUR website!
Discriminant (b^2-4ac)
3x%5E2%2Bx%2B2
b=1 (1x)
a=3 (3x^2)
c=2 (2)
1%5E2-4%2A3%2A2
1-12%2A2
1-24
-23
A negative discriminant means that there are No Real Solutions
or aka Two Imaginary Solutions
A great website about discriminant
http://www.mathwarehouse.com/quadratic/discriminant-in-quadratic-equation.php

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic formula to find the roots of any
quadratic in the form +f%28x%29+=+ax%5E2+%2B+bx+%2B+c+
is x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Notice there is a (+) and (-) in front of the
square root so you get 2 solutions, as you
should with a quadratic.
-----------
The expression inside the square root is called
the discriminant. If the discriminant is a negative
number, you get a (+) and a (-) imaginaries
------------
Here's the possibilities:
+b%5E2+-+4a%2Ac+ = negative > 2 imaginary roots
+b%5E2+-+4a%2Ac+ = positive > 2 real roots
+b%5E2+-+4a%2Ac+ = 0 > 1 real root
------------------------
Here's your problem:
+3x%5E2+%2B+x+%2B+2+=+0+
+a+=+3+
+b+=+1+
+c+=+2+
+b%5E2+-+4a%2Ac+=+1%5E2+-+4%2A3%2A2+
+b%5E2+-+4a%2Ac+=+1+-+24+
+b%5E2+-+4a%2Ac+=+-23+
This is an example of the 1st case, 2 imaginary roots
What this means is: the quadratic never touches the
x-axis, so there are no "real" roots.
Here's a plot to show this:
+graph%28+400%2C+400%2C+-6%2C+6%2C+-6%2C+6%2C+3x%5E2+%2B+x+%2B+2+%29+
Notice that if a or c were negative, you would
get real roots