• # question_answer 13) If one root of the quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to the nth power of the other root, then the value of ${{(a{{c}^{n}})}^{\frac{1}{n+1}}}+{{({{a}^{n}}c)}^{\frac{1}{n+1}}}=$ [IIT 1983] A) $b$ B) - b C) ${{b}^{\frac{1}{n+1}}}$ D) $-{{b}^{\frac{1}{n+1}}}$

Let $\alpha ,{{\alpha }^{n}}$ be the two roots. Then $\alpha +{{\alpha }^{n}}=-b/a,\alpha {{\alpha }^{n}}=c/a$ Eliminating$\alpha$, we get ${{\left( \frac{c}{a} \right)}^{\frac{1}{n+1}}}+{{\left( \frac{c}{a} \right)}^{\frac{n}{n+1}}}=-\frac{b}{a}$ Þ $a.{{a}^{-\ \frac{1}{n+1}}}.{{c}^{\frac{1}{n+1}}}+a.{{a}^{-\frac{n}{n+1}}}.{{c}^{\frac{n}{n+1}}}=-b$ or${{({{a}^{n}}c)}^{\frac{1}{n+1}}}+{{(a{{c}^{n}})}^{\frac{1}{n+1}}}=-b$.