anonymous
  • anonymous
A cubic polynomial F has three "different" real zeros a,b and c. The coefficient of x^3 is positive. Show that F`(a)+F`(b)+F`(c) > 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[F(x)=p(x-a)(x-b)(x-c)\] is probably a good start, where \(p>0\) take the derivative via the product rule and see what you get i didn't actually do it, but is seems like a start to me
anonymous
  • anonymous
Thanks, would that be all I need to do?
anonymous
  • anonymous
i don't know i am writing it now but it seems like the only way to start

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[F'(x)=p(x-b)(x-c)+p(x-a)(x-b)+p(x-a)(x-c)\] is a start then compute \(F'(a)\) etc. add. and see what you get
anonymous
  • anonymous
Not sure how to do that either, I've only just begun attempting this type of problems.
anonymous
  • anonymous
All help is very appreciated
anonymous
  • anonymous
man this is not nearly as easy as i though
anonymous
  • anonymous
lets ignore the \(p\) for a second, as it is positive, lets say it is 1
anonymous
  • anonymous
\[F'(x)=(x-b)(x-c)+(x-a)(x-b)+(x-a)(x-c)\] \[F'(a)=(a-b)(a-c)\] \[F'(b)=(b-a)(b-c)\] \[F'(c)=(c-a)(c-b)\]
anonymous
  • anonymous
add them up and get \[(a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\]
anonymous
  • anonymous
there might be some elementary reason why this is positive, but i don't see it if you do a raft of algebra you can write this as the sum of two squares, and so it has to be positive (non negative actually)
anonymous
  • anonymous
oh actually positive, since it is only zero if \(a=b\) or \(b=c\) and you are told that they are different
anonymous
  • anonymous
@Zarkon do you see right of the bat that this must be positive?
Zarkon
  • Zarkon
look at \[\frac{1}{2}[(a-b)^2+(a-c)^2+(b-c)^2]\]
anonymous
  • anonymous
*off
Zarkon
  • Zarkon
\[f'(a)+f'(b)+f'(c)=\frac{1}{2}[(a-b)^2+(a-c)^2+(b-c)^2]\]
Zarkon
  • Zarkon
my f being your F ;)
Zarkon
  • Zarkon
suppressing the p
anonymous
  • anonymous
i believe you but i have to figure out why
anonymous
  • anonymous
Same here
anonymous
  • anonymous
oh, algebra is why
anonymous
  • anonymous
Heh, could you explain it? I feel as though I am not smart enough to figure it out by myself at the moment :/
anonymous
  • anonymous
it is algebra: multiply out, combine like terms, and see what you get although how @Zarkon recognized it as \[\frac{1}{2}\left((a-b)^2+(a-c)^2+(b-c)^2\right)\] is for him to know
anonymous
  • anonymous
Does that prove that F`(a)+F`(b)+F`(c) > 0 is possible?
anonymous
  • anonymous
a square is always positive unless it is zero
Zarkon
  • Zarkon
it was a guess
anonymous
  • anonymous
really? damn! good guess
Zarkon
  • Zarkon
I thought it might be that ...so I expanded both
anonymous
  • anonymous
@Dahlioz multiply it out and see that it works
anonymous
  • anonymous
i cheated and used wolfram, probably would have been better to do it by hand, which i did afterwards
anonymous
  • anonymous
wolf gives this http://www.wolframalpha.com/input/?i=%28a-b%29%28a-c%29%2B%28b-a%29%28b-c%29%2B%28c-a%29%28c-b%29 although not \[\frac{1}{4}\left(2a-b-c\right)^2+\frac{3}{4}\left(b-c\right)^2\] which also gives it
Zarkon
  • Zarkon
ah...cool
anonymous
  • anonymous
not nearly as cool as the sum of three squares
Zarkon
  • Zarkon
I agree ;)
anonymous
  • anonymous
Thanks guys, you both have been very helpful
anonymous
  • anonymous
Just one question, about what grade/year of school would this be suitable for?
anonymous
  • anonymous
But why can't (2a-b-c) be equal to 0?
anonymous
  • anonymous
As wolfram gives 1/4 (2 a-b-c)^2+3/4 (b^2-2 b c+c^2) rather than 1/4(2a−b−c)^2+3/4(b−c)^2. @Zarkon Sorry for disturbing you guys with this.
Zarkon
  • Zarkon
those two expressions are equal
anonymous
  • anonymous
Ahh, I see

Looking for something else?

Not the answer you are looking for? Search for more explanations.