Question

If one root of $a{x}^2+bx+c=0$ is equal to $n^{th}$ power of the other, then $(a{c}^n{)}^{\frac{1}{(n+1)}}+(a^{n}c{)}^{\frac{1}{(n+1)}}$ is equal to

• A. $1$
• B. $-1$
• C. $b$
• D. $-b$

Answer 1

The correct option is D $-b$

Let the roots of the equation $a{x}^2+bx+c=0$ be $\alpha$ and $(\alpha {)}^n$
Such that, $\alpha +(\alpha {)}^n=\frac{-b}{a}$
$\alpha .(\alpha {)}^n=\frac{c}{a}$

$\alpha ={\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}$
Therefore, ${\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}+{\left(\frac{c}{a}\right)}^{\frac{n}{n+1}}=\frac{-b}{a}$

$a{\left(\frac{1}{a}\right)}^{\frac{1}{n+1}}(c^{\frac{1}{n+1}}+\frac{c^n}{a^n})=-b$

$\frac{{\left(\frac{1}{a}\right)}^{\frac{n}{n+1}}(c^{\frac{1}{n+1}}{a}^n+c^n)}{a^n}=-b$

${\left(\frac{1}{a}\right)}^{\frac{1}{n+1}}(c^{\frac{1}{n+1}}{a}^n+c^n)=-b$

$(c{a}^n{)}^{\frac{1}{n+1}}+(c^{n}a{)}^{\frac{1}{n+1}}=-b$