Question 71840
If the discriminant is positive you get two real answers to the quadratic. 

We could do this by saying that we want {{{b^2}}} to be larger than {{{4*a*c}}}.
So let's 
say that {{{b^2 = 36}}} so that {{{b = 6}}}. Now let's assume {{{a = 1}}} 
and 
therefore, as long as 4*c is less than 36 we will have a positive discriminant. So
let's say that {{{4*c = 20}}}.  Solving this we see that {{{c = 5}}}.

So we have a = 1, b = 6, and c = 5. This makes the quadratic equation:

{{{x^2 + 6x + 5 = 0}}}
.
and the discriminant is:
.
{{{(6)^2 - 4(1)(5) = 36 - 20 = +16}}}
.
(Just for your info, if you solve {{{x^2 + 6x + 5 = 0}}} you will find that {{{x= -1}}} or
{{{x = -5}}} are the answers.
.
Now let's find a case where the discriminant equals zero.  For this case, we need to
make {{{b^2}}} equal to {{{4*a*c}}}.  Again, we will assume that a = 1 just to simplify
things.  This reduces the problem to making {{{b^2}}} equal to {{{4*c}}}.
.
Let's assign a value of 8 to b.  This means that {{{b^2 = 64}}}. Therefore, we need to 
make {{{4*c}}} equal to 64.  Solving this we find that {{{c = 16}}}. So our values for
the quadratic equation are a = 1, b = 8, and c = 16.  Plugging these into the standard
form of a quadratic equation we get:
.
{{{x^2 + 8x + 16 = 0}}}
.
If you calculate the discriminant you will find that it equals zero, just as we figured it
would.  And the answer for x in this equation is x= 4.
.
To find an equation that has no real solution, all we need to do is make {{{4*a*c}}}
greater than {{{b^2}}}.  Again, for simplification let a = 1.  Then we just need to have
{{{b^2}}} be less than {{{4*c}}}.  Let's assume b = 2.  Then {{{b^2 = 4}}}.  So all we 
now need to do is make sure that {{{4*c}}} is greater than 4.  So let's make c equal 3.
Then {{{4*c = 12}}} and this is greater than {{{b^2}}}.
.
We have now found the following values:  a = 1, b = 2, and c = 3. So the corresponding
quadratic equation is:
.
{{{x^2 + 2x + 3 = 0}}}
.
If you solve for the discriminant you will find that it has a negative value, and if
you further solve for x using the quadratic formula you will find that the two answers are:
.
{{{x = -1-sqrt(2)*i}}} and {{{x = -1+sqrt(2)*i}}}
.
I hope this helps you to see your way through this problem. Check the math above. I
think it's correct, but it's pretty late at night to work error free.