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Old 2019-09-25, 17:15   #1
wildrabbitt
 
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Default roots of cubics

Hi,


say a cubic equation intersects the y-axis at a point where it's derivative is 0 and also intersects the y-axis at one other point (so it touches the y-axis twice), wouldn't it have to have one root that isn't real?


Will
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Old 2019-09-25, 17:26   #2
R.D. Silverman
 
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Quote:
Originally Posted by wildrabbitt View Post
Hi,


say a cubic equation intersects the y-axis at a point where it's derivative is 0 and also intersects the y-axis at one other point (so it touches the y-axis twice), wouldn't it have to have one root that isn't real?


Will
If you mean: y = f(x) where f is a cubic polynomial:

IMPOSSIBLE. It is an elementary exercise to see why. There are two different reasons.
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Old 2019-09-25, 17:28   #3
henryzz
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Quote:
Originally Posted by wildrabbitt View Post
Hi,


say a cubic equation intersects the y-axis at a point where it's derivative is 0 and also intersects the y-axis at one other point (so it touches the y-axis twice), wouldn't it have to have one root that isn't real?


Will
It is possible to have repeated roots of cubic equations.
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Old 2019-09-25, 17:38   #4
fivemack
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Quote:
Originally Posted by wildrabbitt View Post
Hi,


say a cubic equation intersects the y-axis at a point where it's derivative is 0 and also intersects the y-axis at one other point (so it touches the y-axis twice), wouldn't it have to have one root that isn't real?


Will
No: if any polynomial has f(t)=0 and f'(t)=0 for the same t, that t is a multiple root of the polynomial (that is, the polynomial is divisible by (x-t)^2 )
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Old 2019-09-25, 18:25   #5
R.D. Silverman
 
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No: if any polynomial has f(t)=0 and f'(t)=0 for the same t, that t is a multiple root of the polynomial (that is, the polynomial is divisible by (x-t)^2 )

Sigh..... Bob runs screaming from the classroom...…...Has everyone forgotten basic
algebra????

Reread what the OP wrote!! He said that the curve itself hits the y-axis twice......Once
where its derivative is 0. i.e. he wants f(0) to have TWO DIFFERENT VALUES.
This is not a function!!! [y = cubic polynomial in x]

And, of course, a cubic can NEVER have a single imaginary root......Imaginary roots
come in pairs!
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Old 2019-09-25, 18:54   #6
VBCurtis
 
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Bob-
It's pretty clear the OP meant x-axis, rather than y-axis. Your answer about imaginary (complex) roots always coming in pairs helps whether he typo'ed y-axis or not, though.
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Old 2019-09-25, 19:16   #7
R.D. Silverman
 
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Bob-
It's pretty clear the OP meant x-axis, rather than y-axis. Your answer about imaginary (complex) roots always coming in pairs helps whether he typo'ed y-axis or not, though.
I assume that people mean what they write. I assume that they proofread before posting.
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Old 2019-09-25, 19:18   #8
R.D. Silverman
 
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I assume that people mean what they write. I assume that they proofread before posting.
Note that he wrote 'y-axis' three different times.....
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Old 2019-09-25, 19:20   #9
fivemack
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Quote:
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Note that he wrote 'y-axis' three different times.....
This makes it even clearer that he meant 'the y=0 axis' (IE the X axis)
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Old 2019-09-25, 19:24   #10
R.D. Silverman
 
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Quote:
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This makes it even clearer that he meant 'the y=0 axis' (IE the X axis)
I disagree. The content of the post speaks for itself. I assume that people mean what they
write, especially when discussing a subject (such as mathematics) where it is possible
to always use precise language.
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Old 2019-09-25, 20:01   #11
VBCurtis
 
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Quote:
Originally Posted by R.D. Silverman View Post
I assume that people mean what they write. I assume that they proofread before posting.
Two bad assumptions, even around here.
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