If alpha, beta are the roots of x^2 - px + q | vedclass

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(A) Given that $\alpha, \beta$ are roots of $x^2 - px + q = 0$,we have $\alpha + \beta = p$ and $\alpha\beta = q$.Given that $\alpha', \beta'$ are roo

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$2\{p^2 - 2q + p'^2 - 2q' - pp'\}$

Vedclass · Answer

(A) Given that $\alpha, \beta$ are roots of $x^2 - px + q = 0$,we have $\alpha + \beta = p$ and $\alpha\beta = q$.Given that $\alpha', \beta'$ are roots of $x^2 - p'x + q' = 0$,we have $\alpha' + \beta' = p'$ and $\alpha'\beta' = q'$.Let $S = (\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$.Expanding the terms: $S = (\alpha^2 - 2\alpha\alpha' + \alpha'^2) + (\beta^2 - 2\beta\alpha' + \alpha'^2) + (\alpha^2 - 2\alpha\beta' + \beta'^2) + (\beta^2 - 2\beta\beta' + \beta'^2)$.Grouping the terms: $S = 2(\alpha^2 + \beta^2) + 2(\alpha'^2 + \beta'^2) - 2\alpha'(\alpha + \beta) - 2\beta'(\alpha + \beta)$.$S = 2(\alpha^2 + \beta^2) + 2(\alpha'^2 + \beta'^2) - 2(\alpha' + \beta')(\alpha + \beta)$.Using the identities $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q$ and $\alpha'^2 + \beta'^2 = (\alpha' + \beta')^2 - 2\alpha'\beta' = p'^2 - 2q'$.Substituting these values: $S = 2(p^2 - 2q) + 2(p'^2 - 2q') - 2(p')(p)$.$S = 2\{p^2 - 2q + p'^2 - 2q' - pp'\}$.

If $\alpha, \beta$ are the roots of $x^2 - px + q = 0$ and $\alpha', \beta'$ are the roots of $x^2 - p'x + q' = 0$,then the value of $(\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$ is

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