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) About Me  (Show Source):
You can put this solution on YOUR website!
set f(x)=0
:
x%5E2%2B6x%2B11=0
:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B6x%2B11+=+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=%286%29%5E2-4%2A1%2A11=-8.

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 - sqrt%28+8%29+=+2.82842712474619.

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%2B6%2Ax%2B11+%29

Answer by edjones(8007) About Me  (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