Question

If one root of the quadratic equation $a{x}^2$+bx+c=0 is equal to the nth 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.
• C.
• D.

Answer 1

The correct option is B

Let $\alpha ,{\alpha }^n$ be the two roots.

Then $\alpha +{\alpha }^n=-\frac{b}{a},\alpha {\alpha }^n=\frac{c}{a}$

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

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