SOLUTION: Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer. x2 + x + 4 = 0

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer. x2 + x + 4 = 0      Log On


   

Question 193763: Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer.
x2 + x + 4 = 0
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
D=b%5E2-4ac=162-4%281%29%284%29=1-16=-15
:
so since D<0 we will have 2 complex solutions
:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B4+=+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%2A1%2A4=-15.

The discriminant -15 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -15 is + or - sqrt%28+15%29+=+3.87298334620742.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B4+%29