Let x₁, x₂, ldots, xt be roots of polynomial equation x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0 . T

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4-2.Quadratic Equations and Inequations

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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,

$\left|x_i\right|=2$ for exactly one value of $i$

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

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

$\left|x_i\right|=2$ for no value of $i$

(KVPY-2017)

Solution

(c)

We have,

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

$\Rightarrow x^6\left(1+\frac{2}{x}+\left(\frac{2}{x}\right)^2+\left(\frac{2}{x}\right)^3+\left(\frac{2}{x}\right)^4\right.$ $\left.+\left(\frac{2}{x}\right)^5+\left(\frac{2}{x}\right)^6\right)=0$

$\Rightarrow \frac{1-\left(\frac{2}{x}\right)^7}{1-\frac{2}{x}}=0 \quad\left[x^6 \neq 0\right]$

$\Rightarrow \quad\left(\frac{2}{x}\right)^7=1 \Rightarrow x^7=2^7$

$x=2$

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

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