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Question Number 10742 by okhema last updated on 24/Feb/17

$$lettherootsoftheequation2x^3-5x^2+4x+6=0$$$$be\alpha ,\beta and\gamma .$$$$i)statethevaluesof\alpha +\beta +\gamma ,\alpha \beta +\alpha \gamma +\beta \gamma and\alpha \beta \gamma .$$$$ii)henceorotherwisedetermineanequationwithintegercoefficientswhichasroots\frac{1}{{\alpha }^2},\frac{1}{{\beta }^2},and\frac{1}{{\gamma }^2}$$$$$$

Answered by sandy_suhendra last updated on 24/Feb/17

$$\left(\mathrm{i}\right)\mathrm{a}=2,\mathrm{b}=-5,\mathrm{c}=4\mathrm{and}\mathrm{d}=6$$$$\alpha +\beta +\gamma =\frac{-\mathrm{b}}{\mathrm{a}}=\frac{5}{2}$$$$\alpha \beta +\alpha \gamma +\beta \gamma =\frac{\mathrm{c}}{\mathrm{a}}=2$$$$\alpha \beta \gamma =\frac{-\mathrm{d}}{\mathrm{a}}=-3$$

Answered by sandy_suhendra last updated on 24/Feb/17

$$\mathrm{ii})$$$$\mathrm{let}{\mathrm{x}}_1=\frac{1}{{\alpha }^2},{\mathrm{x}}_2=\frac{1}{{\beta }^2},{\mathrm{x}}_3=\frac{1}{{\gamma }^2}$$$${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$$$$=\frac{{\beta }^2{\gamma }^2+{\alpha }^2{\gamma }^2+{\alpha }^2{\beta }^2}{{\left(\alpha \beta \gamma \right)}^{2_{}}}$$$$=\frac{{(\alpha \beta +\alpha \gamma +\beta \gamma )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )}{{\left(\alpha \beta \gamma \right)}^2}$$$$=\frac{2^2-2(-3)(2.5)}{{(-3)}^2}$$$$=\frac{19}{9}=\frac{-\mathrm{b}}{\mathrm{a}}$$$${\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_1{\mathrm{x}}_3+{\mathrm{x}}_2{\mathrm{x}}_3=\frac{1}{{\alpha }^2{\beta }^2}+\frac{1}{{\alpha }^2{\gamma }^2}+\frac{1}{{\beta }^2{\gamma }^2}$$$$=\frac{{\gamma }^2+{\beta }^2+{\alpha }^2}{{\alpha }^2{\beta }^2{\gamma }^2}$$$$=\frac{{(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\alpha \gamma +\beta \gamma )}{{\left(\alpha \beta \gamma \right)}^2}$$$$=\frac{{(2.5)}^2-2\left(2\right)}{{(-3)}^2}$$$$=\frac{1}{4}=\frac{\mathrm{c}}{\mathrm{a}}$$$${\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3=\frac{1}{{\alpha }^2}\times \frac{1}{{\beta }^2}\times \frac{1}{{\gamma }^2}=\frac{1}{{\left(\alpha \beta \gamma \right)}^2}$$$$=\frac{1}{{(-3)}^2}=\frac{1}{9}=\frac{-\mathrm{d}}{\mathrm{a}}$$$$\mathrm{the}\mathrm{new}\mathrm{equation}:$$$${\mathrm{x}}^3-\left(\frac{-\mathrm{b}}{\mathrm{a}}\right){\mathrm{x}}^2+\left(\frac{\mathrm{c}}{\mathrm{a}}\right)\mathrm{x}-\left(\frac{-\mathrm{d}}{\mathrm{a}}\right)=0$$$${\mathrm{x}}^3-\frac{19}{9}{\mathrm{x}}^2+\frac{1}{4}\mathrm{x}-\frac{1}{9}=0....\left(\times 36\right)$$$${36\mathrm{x}}^3-{76\mathrm{x}}^2+9\mathrm{x}-4=0$$

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