[JEE Advanced 1983] If one root of $ax^2+bx+c=0$ is the $n$-th power of the other, show that $(ac^n)^{1/(n+1)}+(a^nc)^{1/(n+1)}+b=0$.
If one root of $ax^2+bx+c=0$ is the $n$-th power of the other, show that $(ac^n)^{1/(n+1)}+(a^nc)^{1/(n+1)}+b=0$.
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✓ Vidaara Team
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Answer: Proved.
Roots $\alpha,\alpha^n$: $\alpha^{n+1}=\frac ca$ and $\alpha+\alpha^n=-\frac ba$. Multiplying the sum by $a$ gives $(a^nc)^{1/(n+1)}+(ac^n)^{1/(n+1)}=-b$.
JEE Advanced 1983 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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