Question

If alpha and beta are zeros of 3x2 -4x+1 find polynomial whose zeros are alpha square/beta and beta square/alpha

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Answer 1

$</p><p></p><p>\underline{\textbf{\color{cyan} Given \; Question\; Is }} \\ </p><p></p><p>\alpha \;\; and \; \beta \; are \; the \; zeros \; of \;\\</p><p></p><p>{\color{green}p(x)=3x^2-4x+1 } \\ </p><p></p><p>{\color{red}Find\:g(x)\;whose\;zeros\;are\; \frac{\alpha^2}{\beta^2} \; \; And \; \; \frac{\beta^2}{\alpha^2}}\\ \\</p><p></p><p>\underline{\underline{\textit{Answer}}} \\</p><p></p><p></p><p>{\color{green}p(x)=3x^2-4x+1 } \\ </p><p></p><p>\alpha + \beta = \frac{4}{3} \\</p><p>And \\</p><p>\alpha * \beta = \frac{1}{3} \\ </p><p></p><p>\alpha^2 + \beta^2 = (\alpha + \beta)^2 -2\alpha \beta \\ </p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 -2*\left(\frac{2}{3}\right) \\</p><p></p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{16}{9}\right) -\left(\frac{4}{3}\right) \\</p><p></p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{16}{9}\right) -\left(\frac{4*3}{3*3}\right) \\</p><p></p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{16}{9}\right) -\left(\frac{12}{9}\right) \\</p><p></p><p></p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{(16-12)}{9}\right) \\</p><p></p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{4}{9}\right)\\</p><p></p><p></p><p>\alpha^2 + \beta^2 = \left(\frac{2}{3}\right)\\</p><p></p><p></p><p>\underline{{\color{cyan}Squaring\;On\;Both\; Sides}} \\</p><p></p><p></p><p>\left(\alpha^2 + \beta^2 \right)^2= \left(\frac{2}{3}\right)^2\\</p><p></p><p></p><p></p><p>(\alpha^4 + \beta^4) -2\alpha^2 \beta^2 = \left(\frac{4}{9}\right)\\</p><p></p><p></p><p>(\alpha^4 + \beta^4) -2*\left(\frac{1}{9}\right) = \left(\frac{4}{9}\right)\\</p><p></p><p></p><p></p><p>(\alpha^4 + \beta^4) -\frac{2}{9}= \left(\frac{4}{9}\right)\\</p><p></p><p></p><p>(\alpha^4 + \beta^4) = \left(\frac{4}{9}\right)+ \frac{2}{9} \\</p><p></p><p></p><p>(\alpha^4 + \beta^4) = \frac{2}{3} \\</p><p></p><p></p><p>{\color{green}The\; Required\;g(x)=x^2-\left(\frac{(\alpha^4+\beta^4)}{\alpha^2 \beta^2}\right)+1} \\</p><p></p><p></p><p>\therefore\;\;{\color{cyan}g(x)=x^2-6x+1} ✓✓</p><p></p><p></p><p>$