SOLUTION: in simplest a + bi form, what would be the roots of the equation x^2 + 5 = 4x ?

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Question 389882: in simplest a + bi form, what would be the roots of the equation x^2 + 5 = 4x ?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + 5 = 4x
x%5E2+-+4x+%2B+5+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%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=%28-4%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%2B-4%2Ax%2B5+%29

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x = 2 ± i