SOLUTION: find all zeros: f(x)=x^2+6x+11

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Question 192443: find all zeros: f(x)=x^2+6x+11
Found 2 solutions by Mathtut, edjones:
Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!
set f(x)=0
:

:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -8 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 -8 is + or - .

The solution is , or
Here's your graph:

Answer by edjones(8007)   (Show Source): You can put this solution on YOUR website!
x^2+6x+11=0
x^2+6x =-11
x^2+6x+9=+9-11 completing the square by (6/2)^2=9
(x+3)^2=-2
x+3=+-sqrt(-2) take sqrt of each side
x=-3+sqrt(2)i, x=-3-sqrt(2)i There are no real number zeroes.
.
Ed

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