Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$, then find the equation whose roots are

A) $\frac{1}{a^2},\frac{1}{{\beta }^2}$

B) $\frac{1}{a\alpha +\beta },\frac{1}{a\beta +b}$

C) $\alpha +\frac{1}{\beta },\beta +\frac{1}{\alpha }$

Answer 1

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

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

(i) equation whose roots are $\frac{1}{{\alpha }^2},\frac{1}{{\beta }^2}$

Sum of roots $=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{(\alpha \beta {)}^2}=\frac{(\alpha +\beta {)}^2-2\alpha \beta )}{(\alpha \beta {)}^2}$

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

product $=\frac{1}{{\alpha }^2}\frac{1}{{\beta }^2}=\frac{a^2}{c^2}$

$\therefore$ equation is $c^2{x}^2+(2ac-b^2)x+a^2=0$

(ii) $\frac{1}{a\alpha +\beta },\frac{1}{a\beta +\alpha }$

Sum $\Rightarrow \frac{a\beta +a\alpha +\beta }{a^2\alpha \beta +{\alpha }^{2}a+a{\beta }^2+\alpha \beta }$

$\Rightarrow \frac{(\alpha +\beta )(1+a)}{(a^2+1)\left(\alpha \beta \right)+a\left(\alpha \beta \right)}$

$\Rightarrow \frac{(1+a)\left(\frac{b}{a}\right)}{(a^2+1)\frac{c}{a}+a(\frac{b^2}{a^2}-\frac{2c}{a})}$

$\Rightarrow \frac{-(1+a)b}{(a^2+1)c+a(b^2-2ac)}$

$\Rightarrow \frac{-(1+a)b}{c(1+a^2)+a{b}^2}$

Product $\frac{a}{c(1-a^2)+a{b}^2}$

$\therefore \left(c\right(1-a^2)+a{b}^2)x^2+(1+a)bx+a=0$

(iii) $\alpha +\frac{1}{\beta },\beta +\frac{1}{\alpha }$

Sum $\Rightarrow \frac{\alpha \beta +1}{\beta }+\frac{\alpha \beta +1}{\alpha }$

$\Rightarrow \frac{{\alpha }^2\beta \left(\alpha \beta \right)+({\alpha }^2+\beta }{\alpha \beta }$

$\Rightarrow \frac{-b}{a}+\frac{-b}{c}$

$-b\left[\frac{a+c}{ac}\right]$

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

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

$\Rightarrow (\alpha +\beta )+\frac{\alpha \beta }{\alpha \beta }$

$\left[\frac{ac}{a+c}\right]x^2+bx+(a+c)=0$