Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

If $\alpha ,\beta$ are the roots of the above equation, then it is given that

$\alpha ={\beta }^n$

Now

$\frac{c}{a}=\alpha .\beta$

$={\beta }^n.\beta$

$={\beta }^{n+1}$.

Hence

$c=a{\beta }^{n+1}$...(i)

And

$\frac{-b}{a}=\alpha +\beta$

$=\beta +{\beta }^n$

Therefore

$b=-a(\beta +{\beta }^n)$ ...(ii)

Therefore

$(a{c}^n{)}^{1/n+1}+(a^{n}c{)}^{1/n+1}+b$

$=(a.(a{\beta }^{n+1}{)}^n{)}^{1/n+1}+\left(a^n\right(a{\beta }^{n+1}){)}^{1/n+1}+b$

$=\left(\right(a{\beta }^n{)}^{n+1}{)}^{1/n+1}+\left(\right(a\beta {)}^{n+1}{)}^{1/n+1}-a(\beta +{\beta }^n)$

$=a\left({\beta }^n\right)+a\left(\beta \right)-a(\beta +{\beta }^n)$

$=0$