SOLUTION: Okay, so I tried doing this problem numerous times and didn't know what to do. -Find All Roots. x^3-13x^2+56x-78 Please And Thank you for your help (:

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Question 193425: Okay, so I tried doing this problem numerous times and didn't know what to do.
-Find All Roots.
x^3-13x^2+56x-78

Please And Thank you for your help (:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
-Find All Roots.
x^3-13x^2+56x-78
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The only real root is 3.
Divide by (x-3) -->
x^2 - 10x + 26 = 0
Use the quadratic eqn to find the other 2 zeroes.
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-10x%2B26+=+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=%28-10%29%5E2-4%2A1%2A26=-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--10%2B-i%2Asqrt%28+-4+%29%29%2F2%5C1+=++%28--10%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-10%2Ax%2B26+%29

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x = 5+i
x = 5-i