Polynomials question (1 Viewer)

leehuan · Jul 14, 2016

I only need the second part of part (iii). Everything else I could get out but I get something wrong for the proof.

$P(x)=x^4+Ax^3+Bx^2+Ax+1\\ x=\alpha\text{ is a zero.}$

$\text{(i) Given that }(2+B)^2\neq 4A^2\text{, show that }\alpha \neq 0, \pm 1$

$\text{(ii) Show that }x=\frac{1}{\alpha}\text{ is a root.}$

$\text{(iii) Deduce that if }\alpha\text{ is a multiple root, it has multiplicity 2 }\textbf{and this: }4B=8+A^2$

I had the four roots being alpha, alpha, 1/alpha, 1/alpha

But then I get 4B = A^2, without the eight

InteGrand · Jul 14, 2016

leehuan said:
I only need the second part of part (iii). Everything else I could get out but I get something wrong for the proof.

$P(x)=x^4+Ax^3+Bx^2+Ax+1\\ x=\alpha\text{ is a zero.}$

$\text{(i) Given that }(2+B)^2\neq 4A^2\text{, show that }\alpha \neq 0, \pm 1$

$\text{(ii) Show that }x=\frac{1}{\alpha}\text{ is a root.}$

$\text{(iii) Deduce that if }\alpha\text{ is a multiple root, it has multiplicity 2 }\textbf{and this: }4B=8+A^2$

I had the four roots being alpha, alpha, 1/alpha, 1/alpha

But then I get 4B = A^2, without the eight

$\noindent Let the roots be $\alpha, \alpha, \alpha ^{-1}, \alpha ^{-1}$. From sum of roots: $\alpha + \alpha + \alpha ^{-1} + \alpha ^{-1} = - A \Rightarrow -A = 2 \left(\alpha + \alpha ^{-1}\right) \quad (1)$.$

$\noindent From sum of pairs of roots, $\alpha^2 + 1 + 1 + 1 + 1 + \alpha^{-2} = B \Rightarrow B = \alpha^2+ \alpha ^{-2} + 4 \quad (2)$.$

$\noindent Using the identity $\alpha^2 + \alpha ^{-2} = \left(\alpha + \alpha ^{-1}\right)^2 -2$, and in view of (1), we have $B = \left(-\frac{A}{2}\right)^2 -2 + 4 = \frac{A^2}{4} + 2$. Multiplying through by $4$, we have, $4B = A^2 + 8$, as required.$

leehuan · Jul 14, 2016

Thanks lol I messed up my sum of two roots at a time

Polynomials question (1 Viewer)

leehuan

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InteGrand

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leehuan

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