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If $\alpha$ and $\beta$ are the root of equation $ax^2+bx+c=0$.
Prove that the equation whose root is $\alpha^n$ and $\beta^n$ is $$a(x^\frac 1n)^2+b(x^\frac 1n)+c=0$$

I had already found the equation whose root is whose root is $\alpha^n$ and $\beta^n$ by using $$x^2-(\text{sum of roots})x+\text{product of roots}\implies x^2-(\alpha^n+\beta^n)x+\alpha^n\beta^n=0$$
Is this equation is same as what is given to prove?

Asaf Karagila
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  • You are right about the last part but you have to express $\alpha^n ,\beta^n$ in terms of $a,b,c$. – Vineet Jun 12 '19 at 14:41
  • I can represent $\alpha^n \beta^n$ in terms of $a,b,c$ but, It is quite difficult to represent $\alpha^n+\beta^n$ in terms of it. –  Jun 12 '19 at 14:44
  • So far as I remember these kinds of problem have to do with recurrence relations and vieta formula. – Vineet Jun 12 '19 at 16:15
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    You can't delete question which contain answer. People give effort on your question. Then why should you delete? @deletedaccount – emonHR Aug 17 '21 at 18:30
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    Do not vandalize your question. – Noah Schweber Aug 25 '21 at 02:55
  • @emonHR When this happens you should try and upvote the answer(s): questions with no positive-score answers or accepted answer are automatically deleted in certain circumstances. See here. – user1729 Aug 26 '21 at 07:54
  • I’m voting to close this question because it has been repeatedly vandalised. – Joe Aug 27 '21 at 14:21

1 Answers1

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You are trying to find the new quadratic equation whose roots are $\alpha^n$ and $\beta^n$ your second equation is OK and it does the required. But the method of transformation of equation will not work, because your first equation when rationalized for an equation with rational powers either does not get rationalized or it becomes an equation of much higher degree which is definitely not a quadratic. So your first method would work only if new roots are $\alpha^2$ and $\beta^2$. You may check that you cannot construct a new quadratic equation whose roots are $\alpha^3$ and $\beta^3$.

Z Ahmed
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