If one root of ax²+bx+c=0 is the n-th power of the other, show that (acⁿ)^1/(n+1)+(aⁿc)^1/(n+1)+b=0.

Answer: Proved. Roots α,αⁿ: αⁿ⁺¹= ca and α+αⁿ=- ba. Multiplying the sum by a gives (aⁿc)^1/(n+1)+(acⁿ)^1/(n+1)=-b. JEE Advanced 1983 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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[JEE Advanced 1983] if one root of ax 2 bx c 0 is the n th power

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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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VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

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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