anonymous
  • anonymous
polynomial problem
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Given polynomial p(x) = x^3 + ax^2 + bx + c with distinct roots. a,b,c are constants. If there is exactly one value of y that satisfies p(y) = y, determine c in terms of a and b
primeralph
  • primeralph
|dw:1371276676727:dw|
anonymous
  • anonymous
how do we eliminate the y then?

Looking for something else?

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

More answers

anonymous
  • anonymous
I've spent days on this problem, I am sure the problem is right though, because it's an math olympiad problem
primeralph
  • primeralph
Yeah, I can solve it. Hold on.
anonymous
  • anonymous
let y be 0. c = 0.
anonymous
  • anonymous
then equate to y. factorize and get quadratic value of y i think.
anonymous
  • anonymous
yeah. y^2 + ay + b = 1. solve quadratic and i think you will get value of y. sub back into former equation that has c as subject and get answer.
anonymous
  • anonymous
there is exactly one value of y
anonymous
  • anonymous
so if you let y = 0, there won't be any other value that satisfies p(y) = y
anonymous
  • anonymous
yes. but that is an arbitrary value to find how y relates to other unknowns.
primeralph
  • primeralph
@exraven can you give me till tomorrow? I'm a little sleepy now.
anonymous
  • anonymous
yeah don't worry, I'll open this question till solved
anonymous
  • anonymous
Given polynomial p(x) = x^3 + ax^2 + bx + c with distinct roots. a,b,c are constants. If there is exactly one value of y that satisfies p(y) = y, determine c in terms of a and b $$p(x)=x^3+ax^2+bx+c$$. Observe that \(p(y)=y\) has one solution is equivalent to \(p(x)-x\) having a single root with multiplicty three, let's call it \(\delta\):$$p(x)-x=0\\x^3+ax^2+(b-1)x+c=0\\x^3+3\delta x^2+3\delta^2x+\delta^3=0$$Equating coefficients we find:$$3\delta=a\\3\delta^2=b-1\\\delta^3=c$$Clearly \(3\delta\cdot3\delta^2=9\delta^3=9c\) but also \(3\delta\cdot3\delta^2=a(b-1)\) hence we find:$$9c=a(b-1)\\c=\frac19a(b-1)$$
anonymous
  • anonymous
Expand \(p(x)-x=(x-\delta)^3\):$$p(x)-x=(x-\delta)^3\\x^3+ax^2+(b-1)x+c=x^3-3\delta x^2+3\delta^2x-\delta^3\\a=-3\delta,b-1=3\delta^2,c=-\delta^3\\-9\delta^3=a(b-1)\\9c=a(b-1)\\c=\frac19a(b-1)$$
anonymous
  • anonymous
@oldrin.bataku nice, thanks a lot

Looking for something else?

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