Question

The equation ${\left(x-3\right)}^9+{\left(x-3^2\right)}^9+.....+{\left(x-3^9\right)}^9=0$ has

• A. All the roots real
• B. One real and eight imaginary roots
• C. Real roots necessay $x=3,3^2,....,3^9$
• D. None of these

Answer 1

The correct option is B

One real and eight imaginary roots

Explanation for the correct option:

Determine the nature of the roots of the given equation.

An equation ${\left(x-3\right)}^9+{\left(x-3^2\right)}^9+.....+{\left(x-3^9\right)}^9=0$ is given.

Assume that, $f\left(x\right)={\left(x-3\right)}^9+{\left(x-3^2\right)}^9+.....+{\left(x-3^9\right)}^9$

Differentiate the given equation with respect to $x$.

$$\begin{gathered} f\text{'}\left(x\right)=9{\left(x-3\right)}^8+9{\left(x-3^2\right)}^8+.....+9{\left(x-3^9\right)}^8 \\ \Rightarrow f\text{'}\left(x\right)=9\left[{\left(x-3\right)}^8+{\left(x-3^2\right)}^8+.....+{\left(x-3^9\right)}^8\right] \end{gathered}$$

Since, $f\text{'}\left(x\right)>0$ which means that $f\left(x\right)$ is an increasing function. so, the $f\left(x\right)$ cut the $x-$axis only once.

Therefore, the given equation has only one real root and eight imaginary roots.

Hence, the option $B$ is correct .