SOLUTION: solve by graphing 4x^2-x+8=y or 4x^2-x+8=0 vertex= ( 1/8 , 127/16 )

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve by graphing 4x^2-x+8=y or 4x^2-x+8=0 vertex= ( 1/8 , 127/16 )       Log On


   

Question 249707: solve by graphing 4x^2-x+8=y or 4x^2-x+8=0 vertex= ( 1/8 , 127/16 )
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2-x+8=0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B-1x%2B8+=+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=%28-1%29%5E2-4%2A4%2A8=-127.

The discriminant -127 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 -127 is + or - sqrt%28+127%29+=+11.2694276695846.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-1%2Ax%2B8+%29

--------------------------
There are no real zeroes, so the graph doesn't help much.