Question

Match the following:For the quadratic equation $a{x}^2+bx+c=0$ with roots $\alpha$ and $\beta$.

Answer 1

(i) $\alpha =\beta$
$\Rightarrow$ Two roots are equal, then $b^2=4ac$
(ii) $\alpha =2\beta$
We have $\alpha +\beta =\frac{-b}{a};\alpha \beta =\frac{c}{a}$
$\Rightarrow 3\beta =\frac{-b}{a},2{\beta }^2=\frac{c}{a}$
$\Rightarrow \beta =\frac{-b}{3a}$
$\Rightarrow 2(\frac{-b}{3a}{)}^2=\frac{c}{a}$
$\Rightarrow 2b^2=9ac$
(iii)$\alpha =3\beta$
$\alpha +\beta =4\beta =\frac{-b}{a}$
$\alpha \beta =3{\beta }^2=\frac{c}{a}$
$\therefore 3(\frac{-b}{4a}{)}^2=\frac{c}{a}$
$\Rightarrow 3b^2=16ac$
(iv)$\alpha ={\beta }^2$
$\alpha +\beta =\beta +{\beta }^2=\frac{-b}{a}$
$\alpha \beta ={\beta }^3=\frac{c}{a}$
$\therefore (\frac{c}{a}{)}^{\frac{1}{3}}+(\frac{c}{a}{)}^{\frac{2}{3}}+\frac{b}{a}=0$
$\Rightarrow (a{c}^2{)}^{1/3}+(a^{2}c{)}^{1/3}+b=0$