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Question Number 138766 by nadovic last updated on 18/Apr/21

$$\mathrm{If}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$3x^2+x+2=0,\mathrm{find}\mathrm{the}\mathrm{equation}\mathrm{whose}$$$$\mathrm{roots}\mathrm{are}\frac{1}{{\alpha }^2}\mathrm{and}\frac{1}{{\beta }^2}\mathrm{and}\mathrm{show}\mathrm{that}$$$$27{\alpha }^4=11\alpha +10.$$

Commented by MJS_new last updated on 18/Apr/21

$$x^2+px+q=0\mathrm{with}\mathrm{roots}\alpha ,\beta$$$$\Rightarrow$$$$x^2+\frac{2q-p^2}{q^2}x+\frac{1}{q^2}=0\mathrm{with}\mathrm{roots}\frac{1}{{\alpha }^2},\frac{1}{{\beta }^2}$$

Answered by bramlexs22 last updated on 18/Apr/21

$$3x^2+x+2=0\to \{\begin{array}{l}\alpha \\ \beta \end{array}$$$$\Rightarrow 3{\left(\sqrt{x}\right)}^2+\sqrt{x}+2=0$$$$\Rightarrow \sqrt{x}=-2-3x$$$$\Rightarrow x=4+12x+9x^2$$$$\Rightarrow 9x^2+11x+4=0\to \{\begin{array}{l}{\alpha }^2\\ {\beta }^2\end{array}$$$$and4x^2+11x+9=0\to \{\begin{array}{l}\frac{1}{{\alpha }^2}\\ \frac{1}{{\beta }^2}\end{array}$$

Commented by Rasheed.Sindhi last updated on 18/Apr/21

$$\Rightarrow x=4+12x+3x^2$$$$\stackrel{?}{\Rightarrow }3x^2+11x+4=0\to \{\begin{array}{l}{\alpha }^2\\ {\beta }^2\end{array}$$$$$$

Commented by bramlexs22 last updated on 18/Apr/21

$$typo$$

Answered by Rasheed.Sindhi last updated on 18/Apr/21

$$\alpha +\beta =-1/3,\alpha \beta =2/3$$$$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}$$$$=\frac{{(-1/3)}^2-2\left(2/3\right)}{{\left(2/3\right)}^2}=\frac{\frac{1}{9}-\frac{4}{3}}{\frac{4}{9}}$$$$=\frac{1-12}{4}=-\frac{11}{4}$$$$\frac{1}{{\alpha }^2}.\frac{1}{{\beta }^2}=\frac{1}{{\left(\alpha \beta \right)}^2}=\frac{1}{{\left(2/3\right)}^2}=\frac{9}{4}$$$$x^2-(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2})x+\frac{1}{{\alpha }^2}.\frac{1}{{\beta }^2}=0$$$$x^2-(-\frac{11}{4})x+\frac{9}{4}=0$$$$4x^2+11x+9=0$$$$$$

Answered by physicstutes last updated on 18/Apr/21

$$\alpha +\beta =-\frac{b}{a}=-\frac{1}{3}$$$$\alpha \beta =\frac{c}{a}=\frac{2}{3}$$$$\mathrm{sum}\mathrm{of}\mathrm{roots}:\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}=\frac{\left(\frac{1}{9}\right)-\frac{4}{3}}{\frac{4}{9}}=\frac{-11}{4}$$$$\mathrm{product}\mathrm{of}\mathrm{roots}:\left(\frac{1}{{\alpha }^2}\right)\left(\frac{1}{{\beta }^2}\right)=\frac{1}{{\left(\alpha \beta \right)}^2}=\frac{9}{4}$$$$\mathrm{new}\mathrm{equation}:x^2+\frac{11}{4}x+\frac{9}{4}=0\Rightarrow 4x^2+11x+9=0$$

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