Let x_1, x_2, ldots, x_6 be the roots of the | vedclass

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Question

(c)

We have,

$x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$

$\Rightarrow x^6\left(1+\frac{2}{x}+\left(\frac{2}{x}\right)^2+\left(\frac{2}{x}\right)

Vedclass · Accepted Answer

$\left|x_i\right|=2$ for all values of $i$

Vedclass · Answer

(c)

We have,

$x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$

$\Rightarrow x^6\left(1+\frac{2}{x}+\left(\frac{2}{x}ight)^2+\left(\frac{2}{x}ight)^3+\left(\frac{2}{x}ight)^4ight.$ $\left.+\left(\frac{2}{x}ight)^5+\left(\frac{2}{x}ight)^6ight)=0$

$\Rightarrow \frac{1-\left(\frac{2}{x}ight)^7}{1-\frac{2}{x}}=0 \quad\left[x^6 
eq 0ight]$

$\Rightarrow \quad\left(\frac{2}{x}ight)^7=1 \Rightarrow x^7=2^7$

$x=2$

$	herefore\left|x_iight|=2$ for all value of $i$

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

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