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# If $\alpha$ and $\beta$ are the roots of the equation $4{x}^{2}-5x+2=0$, find the equation whose roots are $\alpha +\frac{1}{\alpha }\mathrm{and}\beta +\frac{1}{\beta }$

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Solution

## $\mathrm{We}\mathrm{have}4{x}^{2}-5x+2=0\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{comparing}\mathrm{this}\mathrm{equation}\mathrm{with}a{x}^{2}+bx+c=0,\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\mathit{}a=4,b=-5,c=2\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{\alpha }+\mathrm{\beta }=\frac{-b}{a}\mathrm{and}\mathrm{\alpha \beta }=\frac{c}{a}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{we}\mathrm{get}\mathrm{\alpha }+\mathrm{\beta }=-\left(\frac{-5}{4}\right)=\frac{5}{4}...\left(1\right)$ $\mathrm{\alpha \beta }=\frac{2}{4}=\frac{1}{2}$ $\mathrm{Let}{\mathrm{\alpha }}_{1}=\mathrm{\alpha }+\frac{1}{\mathrm{\alpha }}\mathrm{and}{\mathrm{\beta }}_{1}=\mathrm{\beta }+\frac{1}{\mathrm{\beta }}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{\mathrm{\alpha }}_{1}+{\mathrm{\beta }}_{1}=\mathrm{\alpha }+\frac{1}{\mathrm{\alpha }}+\mathrm{\beta }+\frac{1}{\mathrm{\beta }}\phantom{\rule{0ex}{0ex}}=\left(\mathrm{\alpha }+\mathrm{\beta }\right)+\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\left(\frac{\frac{5}{4}}{\frac{1}{2}}\right)\left[\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{5}{2}=\frac{15}{4}$ ${\alpha }_{1}{\beta }_{1}=\left(\alpha +\frac{1}{\alpha }\right)\left(\beta +\frac{1}{\beta }\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{\alpha }^{2}+1}{\alpha }\right)\left(\frac{{\beta }^{2}+1}{\beta }\right)\phantom{\rule{0ex}{0ex}}=\frac{{\alpha }^{2}{\beta }^{2}+{\alpha }^{2}+{\beta }^{2}+1}{\alpha \beta }\phantom{\rule{0ex}{0ex}}=\frac{{\left(\frac{1}{2}\right)}^{2}+{\left(\frac{5}{4}\right)}^{2}-\left(2×\frac{1}{2}\right)+1}{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{4}+\frac{25}{16}}{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=\frac{29}{8}$ $\text{W}e\mathrm{know}\mathrm{that}\mathrm{if}{\mathrm{\alpha }}_{1}\text{and}{\mathrm{\beta }}_{1}\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{a}\mathrm{quadratic}\mathrm{equation},\mathrm{then}\mathrm{the}\mathrm{quadratic}\phantom{\rule{0ex}{0ex}}\mathrm{equation}\mathrm{is}\phantom{\rule{0ex}{0ex}}{x}^{2}-\left({\alpha }_{1}+{\beta }_{1}\right)x+{\alpha }_{1}{\beta }_{1}=0\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituing}{\alpha }_{1}+{\beta }_{1}=\frac{15}{4}\mathrm{and}{\alpha }_{1}{\beta }_{1}=\frac{29}{8},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{x}^{2}-\frac{15}{4}x+\frac{29}{8}=0\phantom{\rule{0ex}{0ex}}⇒8{x}^{2}-30x+29=0$

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