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 B

-b

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

Then $\alpha +{\alpha }^n=-\frac{b}{a}\cdots \left(i\right),\alpha \times {\alpha }^n=\frac{c}{a}$

$\Rightarrow {\alpha }^{n+1}=\frac{c}{a}\Rightarrow \alpha =(\frac{c}{a}{)}^{\frac{1}{n+1}}\cdots \left(ii\right)$
Substituting $\left(ii\right)$ in $\left(i\right),$ we get:
$(\frac{c}{a}{)}^{\frac{1}{n+1}}+(\frac{c}{a}{)}^{\frac{n}{n+1}}=-\frac{b}{a}$
$\Rightarrow a\cdot (a{)}^{-\frac{1}{n+1}}\cdot (c{)}^{\frac{1}{n+1}}+(a.a{)}^{-\frac{n}{n+1}}\cdot c^{\frac{n}{n+1}}=-b$

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