SOLUTION: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; jus

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; jus      Log On


   

Question 36300: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions.
2x^2 + x - 1 = 0
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B1x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A2%2A-1=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+9+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+9+%29%29%2F2%5C2+=+0.5
x%5B2%5D+=+%28-%281%29-sqrt%28+9+%29%29%2F2%5C2+=+-1

Quadratic expression 2x%5E2%2B1x%2B-1 can be factored:
2x%5E2%2B1x%2B-1+=+2%28x-0.5%29%2A%28x--1%29
Again, the answer is: 0.5, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B1%2Ax%2B-1+%29

....since the discriminant is greater than 0, the answers are either two rational or irrational real numbers....