Question

If $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+c=0$, then find the equation whose roots are given by
i) $\alpha +\frac{1}{\beta },\beta +\frac{1}{\alpha }$
ii) ${\alpha }^2+2,{\beta }^2+2$

Answer 1

$a{x}^2+bx+c=0$

$\alpha$ & $\beta$ are roots of above eqn

$\alpha +\beta =\frac{-b}{a}$

$\alpha \beta =\frac{c}{a}$

$\alpha +\frac{1}{\alpha },\beta +\frac{1}{\beta }$ roots of other eqn

sum $=\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha }$

$=\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }$

$=\frac{{\alpha }^2+1}{\alpha }+\frac{{\beta }^2+1}{\beta }$

$=\frac{{\alpha }^2\beta +\beta +\alpha {\beta }^2+\alpha }{\alpha \beta }$

$=\frac{\alpha \beta (\alpha +\beta )+(\alpha -\alpha )}{\alpha \beta }=\frac{\frac{c}{a}\left(\frac{-b}{a}\right)+\frac{-b}{a}}{c/a}$

$=\frac{-b/a(c/a+1)}{c/a}$

$=\frac{-b(c+a)}{c}$

Product $=(\alpha +\frac{1}{\beta })({\beta }^2-\frac{1}{\alpha })$

$=\left(\frac{\alpha \beta +1}{\beta }\right)\left(\frac{\alpha \beta +1}{\alpha }\right)$

$=\frac{(\alpha \beta +1{)}^2}{\alpha \beta }=\frac{(c/a+1{)}^2}{c/a}$

$\Rightarrow \frac{(c+a{)}^2}{ca}$

eqn $\Rightarrow x^2-(-b\frac{(c+a)}{c})x+\frac{(c+a{)}^2}{ca}=0$

${\alpha }^2+2,{\beta }^2+2$

Sum $={\alpha }^2+2+{\beta }^2+2$

$=(\alpha +\beta {)}^2+4-2\alpha \beta$

$={(-\frac{b}{a})}^2-2\left(\frac{c}{a}\right)+4$

$=\frac{b^2}{a^2}-\frac{2c}{a}+4=\frac{b^2-2ac+4a^2}{a^2}$

Product $=({\alpha }^2+2)({\beta }^2+2)$

$={\alpha }^2{\beta }^2+2({\alpha }^2+{\beta }^2)+4$

$=(\alpha \beta {)}^2+2((\alpha +\beta {)}^2-2\alpha \beta )+4$

$={\left(\frac{c}{a}\right)}^2+2({\left(\frac{-b}{a}\right)}^2-2\left(\frac{c}{a}\right))+4$

$=\frac{c^2}{a^2}+2(\frac{b^2}{a^2}-\frac{2c}{a})+4$

$=\frac{c^2+2b^2-4ac+4a^2}{a^2}$

$\therefore$ eqn $\Rightarrow x^2-\left(\frac{b^2-2ac+4a^2}{a^2}\right)x+\left(\frac{c^2+2b^2-4ac+4a^2}{a^2}\right)$

eqn $\Rightarrow \left(a^2\right)x^2-(b^2-2ac+4a^2)x+(c^2+2b^2-4ac+4a^2)=0$