SOLUTION: If you are looking at a graph of a quadratic equation, how do you determine where the solutions are?

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Question 167757: If you are looking at a graph of a quadratic equation, how do
you determine where the solutions are?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If you are looking at a graph of a quadratic equation, how do
you determine where the solutions are?
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If it's the usual y = a quadratic in x, it's where it crosses the x-axis.
For example:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-12+=+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=%28-1%29%5E2-4%2A1%2A-12=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+49+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+4
x%5B2%5D+=+%28-%28-1%29-sqrt%28+49+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-1x%2B-12 can be factored:
1x%5E2%2B-1x%2B-12+=+%28x-4%29%2A%28x--3%29
Again, the answer is: 4, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-12+%29

If there are no real solutions, it won't cross the x-axis.
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B9+=+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=%280%29%5E2-4%2A1%2A9=-36.

The discriminant -36 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 -36 is + or - sqrt%28+36%29+=+6.

The solution is x%5B12%5D+=+%28-0%2B-i%2Asqrt%28+-36+%29%29%2F2%5C1+=++%28-0%2B-i%2A6%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%2B0%2Ax%2B9+%29