SOLUTION: find the root of equation for X^3=8=0 trying ans f(1)=9 f(-2)=0 so, x=-2 is the root x+2 is factor of f(x) (x+2)^3=0

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: find the root of equation for X^3=8=0 trying ans f(1)=9 f(-2)=0 so, x=-2 is the root x+2 is factor of f(x) (x+2)^3=0      Log On


   

Question 1141576: find the root of equation for X^3=8=0
trying ans
f(1)=9
f(-2)=0
so, x=-2 is the root
x+2 is factor of f(x)
(x+2)^3=0
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
find the root of equation for X^3=8=0
====================
Too many equal signs.
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From your attempts, I can see that you meant
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x^3 + 8 = 0
Factor it.
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(x+2)*(x^2 - 2x + 4) = 0
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x = -2
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(x^2 - 2x + 4) = 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-2x%2B4+=+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=%28-2%29%5E2-4%2A1%2A4=-12.

The discriminant -12 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 -12 is + or - sqrt%28+12%29+=+3.46410161513775.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B4+%29



---> roots are -2 and
1 + i*sqrt(3) and
1 - i*sqrt(3)