Solveeit Logo

Question

Mathematics Question on Quadratic Equations

If one root of the quadratic equation ax2+bx+c=0a x^{2}+b x+c=0 is equal to nth n^{\text {th }} power of the other root, then the value of: ann1C1n1+cnn1a1n1a^{\frac{n}{n-1}} C^{\frac{1}{n-1}}+c^{\frac{n}{n-1}} a^{\frac{1}{n-1}} is equal to

A

bb

B

b-b

C

1bn+1\frac{1}{b^{n+1}}

D

bn+1-b^{n+1}

Answer

b-b

Explanation

Solution

Let one root be α\alpha
Then the other root is αn\alpha^{ n }
So,product of roots =ca=\frac{ c }{ a }
(α)(αn)=ca\therefore(\alpha)\left(\alpha^{ n }\right)=\frac{ c }{ a }
αn+1=ca\therefore \alpha^{ n +1}=\frac{ c }{ a }
α=(ca)1n+1\therefore \alpha=\left(\frac{ c }{ a }\right)^{\frac{1}{n+1}}...(1)
sum of roots =ba=-\frac{b}{a}
α+αn=ba\therefore \alpha+\alpha^{ n }=-\frac{ b }{ a }
Substituting the value of α\alpha from equation (1), we get
(ca)1n+1+(ca)nn+1=ba\therefore\left(\frac{ c }{ a }\right)^{\frac{1}{n+1}}+\left(\frac{ c }{ a }\right)^{\frac{ n }{n+1}}=-\frac{ b }{ a }
ann+1C1n+1+a1n+1Cnn+1=b\therefore a ^{\frac{n}{n+1}} C ^{\frac{1}{n+1}}+ a ^{\frac{1}{n+1}} C ^{\frac{n}{n+1}}=- b