Question

If $\alpha and\beta$ are the zeroes of the polynomial $a{x}^2+bx+c$, then find the polynomial whose zeroes are $2\alpha +3\beta ,3\alpha +2\beta$.

Answer 1

$\alpha and\beta$ are the zeroes of the polynomial $a{x}^2+bx+c$

Sum of the zeroes = $(3\alpha +2\beta )+(2\alpha +3\beta )=5(\alpha +\beta )$

Product of zeroes = $(3\alpha +2\beta )\times (2\alpha +3\beta )=(6{\alpha }^2+9\alpha \beta +4\alpha \beta +6{\beta }^2)=(13\alpha \beta +6{\alpha }^2+6{\beta }^2)$

Now, the polynomial having zeroes $(3\alpha +2\beta )and(2\alpha +3\beta )$ is

$x^2+5(\alpha +\beta )x+(13\alpha \beta +6{\alpha }^2+6{\beta }^2)$

$=x^2+5(-\frac{b}{a})\times +(13\times \frac{c}{a}+6\{{\alpha }^2+{\beta }^2\})[{\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta ]=x^2+5(-\frac{b}{a})\times +(13\times \frac{c}{a}+6\{{(-\frac{b}{a})}^2-2x\frac{c}{a}\})=\frac{1}{a^2}(a^2{x}^2-5abx-6b^2+11ac)$