SOLUTION: solve the equation in the complex number system 117x^2+1=18x

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Question 305172: solve the equation in the complex number system
117x^2+1=18x
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
117x%5E2+-+18x+%2B+1+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 117x%5E2%2B-18x%2B1+=+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=%28-18%29%5E2-4%2A117%2A1=-144.

The discriminant -144 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 -144 is + or - sqrt%28+144%29+=+12.

The solution is x%5B12%5D+=+%28--18%2B-i%2Asqrt%28+-144+%29%29%2F2%5C117+=++%28--18%2B-i%2A12%29%2F2%5C117+, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+117%2Ax%5E2%2B-18%2Ax%2B1+%29

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= (27 ± 2i)/39