Question

If one root of the quadratic equation ${ax}^2+bx+c=0$ is equal to the $n^{th}$ power of the other, then show that
${\left({ac}^n\right)}^{\frac{1}{n+1}}+{\left(a^{n}c\right)}^{\frac{1}{n+1}}+b=0$

Answer 1

Given that $\beta ={\alpha }^n$
Product of roots $=\alpha \beta =\alpha {\alpha }^n={\alpha }^{n+1}=\frac{c}{a}$
$\Rightarrow \alpha ={\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}$
Substituting $\alpha$ in ${ax}^2+bx+c=0$
$a{\left(\frac{c}{a}\right)}^{\frac{2}{n+1}}+b{\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}+c=0$
${\left({ac}^n\right)}^{\frac{1}{n+1}}+{\left(a^{n}c\right)}^{\frac{1}{n+1}}+b=0$
Ans: 1