Question

Let $x_1,x_2,.....,x_6$ be the roots of the polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$. Then.

• A. $|x_i|=2$ for exactly one value of $i$
• B. $|x_i|=2$ for exactly two values of $i$
• C. $|x_i|=2$ for all values of $i$
• D. $|x_i|=2$ for no value of $i$

Answer 1

Answer: (C)

$|x_i|=2$ for all values of $i$

Given equation is $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$

Rearranging $64+32x+{16}^2+8x^3+4x^4+2x^5+x^6=0$

Looking at the polynomial we can see the terms are in geometric progression with first term as $64$ and $7$ terms in the GP and common ratio as $\frac{x}{2}$.

So, we can write using the sum of n terms of a G.P I.E $a\times \frac{r^n-1}{r-1}$, where $a$ is the first term of the GP and $r$ is the common difference.

Hence, we can write

$64\times \left(\frac{(\frac{x}{2}{)}^7-1}{\frac{x}{2}-1}\right)=0$

or $(\frac{x}{2}{)}^7-1=0$

$x=2$ is the only solution for this equation

Hence, $\left|x_i\right|=2$ for all $i=1,2,3,4,5,6,$