1. Originally Posted by CaptainBlank
Pointing at a diagram does not constitute a proof. You will have the
Bourbaki police after you at this rate.
You do not need to lecture me on not using diagrams. I do not use them for proofs and I agree we cannot use them for proofs. I was making it easier for the user to understand.

However, you can use the fact that any quadradic has a maximum with the main coefficient negative. That is what I was saying.

2. Originally Posted by ThePerfectHacker
You do not need to lecture me on not using diagrams. I do not use them for proofs and I agree we cannot use them for proofs. I was making it easier for the user to understand.
You fail to detect humor in this case.

However, you can use the fact that any quadradic has a maximum with the main coefficient negative. That is what I was saying.
And what I am saying is that the student cannot assume that in this case
as it is (almost) what they are being asked to proove.

RonL

3. what if i say m and n are both not even because it is the special case of elimination of common factors property.

4. how did u get k<=13 and i got k <= -5

5. Originally Posted by ruprotein
how did u get k<=13 and i got k <= -5
We are looking for real solutions of:

y^2-6y+(k-4)=0

the discriminant of this is:

d=(-6)^2-4(k-4)=52-4k,

and this has to be >=0,m so:

52-4k>=0

which implies thet:

52>=4k,

13>=k,

or k<=13.

If you have something else tell us how you got what you have so we
can see where we/you have gone down different paths.

RonL

RonL

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