SOLUTION: Find the x values at which the following parabolas cross the x-axis. a) y= x^2 + 5x + 4 b) y= x^2 + 4x + 5

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find the x values at which the following parabolas cross the x-axis. a) y= x^2 + 5x + 4 b) y= x^2 + 4x + 5      Log On


   

Question 191250: Find the x values at which the following parabolas cross the x-axis.
a) y= x^2 + 5x + 4
b) y= x^2 + 4x + 5
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
set y=0
:
a)x%5E2%2B5x%2B4=0
:
%28x%2B4%29%28x%2B1%29=0 x=-4 and -1 (-4,0), (-1,0)
:
b)x%5E2%2B4x%2B5=0
:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B5+=+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=%284%29%5E2-4%2A1%2A5=-4.

The discriminant -4 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 -4 is + or - sqrt%28+4%29+=+2.

The solution is x%5B12%5D+=+%28-4%2B-i%2Asqrt%28+-4+%29%29%2F2%5C1+=++%28-4%2B-i%2A2%29%2F2%5C1+, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B5+%29