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Given that $_{ }$$x^{4}+x^{3}+px^{2}+4x-2=0$ where $p$ is a constant, has roots $x_{1}, x_{2}, x_{3}\,and\,x_{4}$

a) Find the equation whose roots are $\frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}}\,and\,\frac{1}{x_{4}}$

b) Given that $ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{3}^{2}=\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\frac{1}{x_{3}^{2}}+\frac{1}{x_{4}^{2}} $, find the value of $p$.

I know (from Vieta's formulas) that the sum of roots = -1 and product of roots = -2. I feel that it can be solved using Vieta's theorem but I am stuck. Please help with a hint on how to proceed. Thank you for any help you can offer.

Jane T
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1 Answers1

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Hints:  (a)  if $P(x)$ is a polynomial of degree $n$ in $x$ with roots $x_k \ne 0$, then $Q(x)=x^nP\left(\frac{1}{x}\right)$ is a polynomial of degree $n$ with roots $\frac{1}{x_k}$.  (b)  $x_1^2+x_2^2+x_3^2+x_4^2=\left(\sum_k x_k\right)^2 - 2\,\sum_{i \lt j} x_ix_j\,$.

dxiv
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  • Thanks for your prompt response. I am a high school student and I have no idea what you've written above. Are these formulas derived from Vieta's formulas for the quartic? If so, then how have they been derived? – Jane T Oct 01 '18 at 22:55
  • @JaneT (a) $x^{4}+x^{3}+px^{2}+4x-2=0 \iff x^4 \left(1 + \frac{1}{x}+p\left(\frac{1}{x}\right)^2+4\left(\frac{1}{x}\right)^3-2\left(\frac{1}{x}\right)^4\right) = 0,$, so the polynomial with roots $\frac{1}{x_k}$ is $-2x^4+4x^3+px^2+x+1$. (b) $x_1^2+x_2^2+x_3^2+x_4^2$ $=\left(x_1+x_2+x_3+x_4\right)^2$ $- 2,\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\right),$ where both terms on the RHS are known by Vieta's. – dxiv Oct 01 '18 at 23:06
  • Many thanks for your prompt reply, I now understand part (a), however, for part (b) I am unsure on how to proceed with the equation which you have provided (which I fully understand). How does this equation help in calculation of parameter p? – Jane T Oct 01 '18 at 23:31
  • @JaneT Using Vieta's relations, it follows that $,x_1^2+x_2^2+x_3^2+x_4^2 = (-1)^2 - 2p,$. Now that you know the polynomial with roots $,1/x_k,$, do the same to calculate $,1/x_1^2+1/x_2^2+1/x_3^2+1/x_4^2,$. Equate the two, and you get a simple equation to solve for $,p,$. – dxiv Oct 01 '18 at 23:35
  • Ah, thank you so much for your help I really appreciate it. – Jane T Oct 01 '18 at 23:41
  • @JaneT Glad you worked it out. – dxiv Oct 01 '18 at 23:48