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If one root of the ax^(2) + bx + c = 0 i...

If one root of the `ax^(2) + bx + c = 0` is equal to nth power of the other root, then the value of `(ac^(n))^((1)/(n+1)) + (a^(n)c)^((1)/(n+1))` equal :

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Let `alpha` be one root of the equation `ax^(2)+bx+c=0`
then other root be `alpha^(n)`
`:.alpha +alpha^(n)=-b/a` ……..i
and `alpha. alpha^(n)=c/a`
`impliesalpha^(n+1)=c/a`
`impliesalpha=(c/a)^(1/(n+1))`
`:.` From Eq. (i) we get
`(c/a)^(1/(n+1))+(c/a)^(n/(n+1))=-b/a`
`implies(c)^(1/(n+1)).a^(-1/(n+1)+1)+(c^(n))^(1/(n+1)).a^(-n/(n+1)+1)+b=0`
`implies1/c^(1/(n+1)).a^(n/(n+1))+(c^(n))^(1/(n+1)).a^(1/(n+1))+b=0`
`=(a^(n)c)^(1/(n+1))+(c^(n)a)^(1/(n+1)+b=0`
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