Question

if alpha beta gamma are the roots of the eqauation x^3+px^2+qx+r=0 then sigma (alpa-beta)^2

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\alpha\;and\;\beta\;}\textsf{roots of the equation}\;\mathsf{x^3+px^2+qx+r=0}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{\sum\,(\alpha-\beta)^2}$

$\underline{\textbf{Solution:}}$

$\mathsf{Since\;\alpha\;and\;\beta\;}\textsf{roots of the equation}\;\mathsf{x^3+px^2+qx+r=0,}$

$\textsf{we have}$

$\mathsf{\alpha+\beta+\gamma=\dfrac{-p}{1}=-p}$--------(1)

$\mathsf{\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{q}{1}=q}$--------(2)

$\mathsf{\alpha\,\beta\,\gamma=\dfrac{-r}{1}=-r}$--------(3)

$\mathsf{Now,}$

$\mathsf{\sum\,(\alpha-\beta)^2}$

$\mathsf{=(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2}$

$\textsf{Using the identity,}\;\boxed{\bf\,(a-b)^2=a^2+b^2-2ab}$

$\mathsf{=(\alpha^2+\beta^2-2\,\alpha\,\beta)+(\beta^2+\gamma^2-2\,\beta\,\gamma)+(\gamma^2+\alpha^2-2\,\gamma\,\alpha)}$

$\mathsf{=2(\alpha^2+\beta^2+\gamma^2)-2(\alpha\,\beta+\beta\,\gamma+\gamma\,\alpha)}$

$\mathsf{=2[(\alpha^2+\beta^2+\gamma^2)-(\alpha\,\beta+\beta\,\gamma+\gamma\,\alpha)]}$

$\textsf{Using the identity,}$

$\boxed{\bf\,(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)}$

$\mathsf{=2[(\alpha+\beta+\gamma)^2-2(\alpha\,\beta+\beta\,\gamma+\gamma\,\alpha)-(\alpha\,\beta+\beta\,\gamma+\gamma\,\alpha)]}$

$\mathsf{=2[(\alpha+\beta+\gamma)^2-3(\alpha\,\beta+\beta\,\gamma+\gamma\,\alpha)]}$

$\textsf{Using equations (1) and (2), we get}$

$\mathsf{=2[(-p)^2-3q]}$

$\mathsf{=2[p^2-3q]}$

$\mathsf{=2\,p^2-6\,q}$

$\implies\boxed{\mathsf{\sum\,(\alpha-\beta)^2=2\,p^2-6\,q}}$

$\underline{\textbf{Find more:}}$

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