Question 182673: 5a.
(i) How many solutions exist for a quadratic equation? Explain.
(ii) How do we determine whether the solutions are real or complex?
I have no idea how to approach either of these. thanks for the help.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! This is well covered by the onsite solver.
As an example 
| 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 :  .
Discriminant d=49 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3, -4.
Here's your graph:
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Example 2
| 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 -15 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 -15 is + or -  .
The solution is  , or
Here's your graph:
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