A) \[b\]
B) - b
C) \[{{b}^{\frac{1}{n+1}}}\]
D) \[-{{b}^{\frac{1}{n+1}}}\]
Correct Answer: B
Solution :
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\[{{({{a}^{n}}c)}^{\frac{1}{n+1}}}+{{(a{{c}^{n}})}^{\frac{1}{n+1}}}=-b\].
You need to login to perform this action.
You will be redirected in
3 sec
