# Inequalities

Announcements
#1
Struggling to approach part ii, I understand that for any cubic there will always be 2 turning points, hence why g^2>4h is not necessary.

To go about showing it isn't I thought I would consider the cases of the discriminate of the quadratic polynomial for the other two cases and try subbing into the equation that we're trying to prove but it doesn't seem to work
Last edited by Student 999; 1 month ago
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#2
Is this the approach, I can't seem to think of any way to reason that the last line is valid unless somehow proving that gk is always >0
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1 month ago
#3
(Original post by Student 999)
Is this the approach, I can't seem to think of any way to reason that the last line is valid unless somehow proving that gk is always >0
For me, it looks like they want you to use the part a) result and note that g^2 > 4h is just a rewritten discrimant inequality > 0.
Last edited by mqb2766; 1 month ago
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#4
(Original post by mqb2766)
For me, it looks like they want you to use the part a) result and note that g^2 > 4h is just a rewritten discrimant inequality > 0.
I see so just proof of the same inequality and a reason why's its sufficient not necessary is enough
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1 month ago
#5
(Original post by Student 999)
I see so just proof of the same inequality and a reason why's its sufficient not necessary is enough
Im not sure what youve done, but that could be ok.
Id probably interpret it using the previous result, rather than proving it again.
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