SOLUTION: help! Find zeros for the function: g(x) = 6 - x^2 -5x h(x) = x^2 + 6x + 12

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Question 172630: help!
Find zeros for the function:
g(x) = 6 - x^2 -5x
h(x) = x^2 + 6x + 12

Answer by KnightOwlTutor(293) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -1x%5E2%2B-5x%2B-6+=+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=%28-5%29%5E2-4%2A-1%2A-6=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+1+%29%29%2F2%5C-1+=+-3
x%5B2%5D+=+%28-%28-5%29-sqrt%28+1+%29%29%2F2%5C-1+=+-2

Quadratic expression -1x%5E2%2B-5x%2B-6 can be factored:
-1x%5E2%2B-5x%2B-6+=+-1%28x--3%29%2A%28x--2%29
Again, the answer is: -3, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-1%2Ax%5E2%2B-5%2Ax%2B-6+%29

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B6x%2B12+=+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%2A12=-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

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B6%2Ax%2B12+%29