Question

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

• A. $-b$
• B. $b$
• C. $-b^{\frac{1}{n+1}}$
• D. $b^{\frac{1}{n+1}}$

Answer 1

The correct option is A

$-b$

Let $\alpha ,{\alpha }^n$ be the two roots of $a{x}^2+bx+c=0$.

Then $\alpha +{\alpha }^n=-\frac{b}{a},$
and $\alpha .{\alpha }^n=\frac{c}{a}\Rightarrow \alpha =(\frac{c}{a}{)}^{\frac{1}{n+1}}$

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

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