Complex Numbers And Quadratic Equations question 177
Question: If one root of the quadratic equation $ ax^{2}+bx+c=0 $ is equal to the nth power of the other root, then the value of $ {{(ac^{n})}^{\frac{1}{n+1}}}+{{(a^{n}c)}^{\frac{1}{n+1}}}= $ [IIT 1983]
Options:
A) $ b $
B) - b
C) $ {b^{\frac{1}{n+1}}} $
D) $ -{b^{\frac{1}{n+1}}} $
Show Answer
Answer:
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 $ {{( \frac{c}{a} )}^{\frac{1}{n+1}}}+{{( \frac{c}{a} )}^{\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}}}+{{(ac^{n})}^{\frac{1}{n+1}}}=-b $ .
BETA
