If alpha_1, alpha_2, alpha_3, alpha_4, alpha_ | vedclass

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(C) The given equation is $x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0$. By inspection,$x = 1$ is a root,so $\alpha_1 = 1$. Dividing the polynomial by $(x-1

Vedclass · Accepted Answer

$7$

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(C) The given equation is $x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0$. By inspection,$x = 1$ is a root,so $\alpha_1 = 1$. Dividing the polynomial by $(x-1)$,we get the reduced equation: $x^4 - 4x^3 + 5x^2 - 4x + 1 = 0$. Dividing by $x^2$,we have $(x^2 + \frac{1}{x^2}) - 4(x + \frac{1}{x}) + 5 = 0$. Let $x + \frac{1}{x} = a$. Then $x^2 + \frac{1}{x^2} = a^2 - 2$. Substituting this,we get $(a^2 - 2) - 4a + 5 = 0$,which simplifies to $a^2 - 4a + 3 = 0$. Solving for $a$,we get $(a - 3)(a - 1) = 0$,so $a = 1$ or $a = 3$. For $a = 1$,$x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0$. The roots are $\alpha_2, \alpha_3$. Since $x^2 - x + 1 = 0$,we have $\frac{1}{x^2} = x - 1$. Thus $\frac{1}{\alpha_2^2} + \frac{1}{\alpha_3^2} = (\alpha_2 + \alpha_3) - 2 = 1 - 2 = -1$. For $a = 3$,$x + \frac{1}{x} = 3 \Rightarrow x^2 - 3x + 1 = 0$. The roots are $\alpha_4, \alpha_5$. Since $x^2 - 3x + 1 = 0$,we have $\frac{1}{x^2} = 3x - 1$. Thus $\frac{1}{\alpha_4^2} + \frac{1}{\alpha_5^2} = 3(\alpha_4 + \alpha_5) - 2 = 3(3) - 2 = 7$. Finally,$\frac{1}{\alpha_1^2} + \frac{1}{\alpha_2^2} + \frac{1}{\alpha_3^2} + \frac{1}{\alpha_4^2} + \frac{1}{\alpha_5^2} = \frac{1}{1^2} + (-1) + 7 = 1 - 1 + 7 = 7$.

Column-$I$	Column-$II$
$A$. $\alpha+\beta+\gamma$	$(1)$ $-\frac{43}{4}$
$B$. $\alpha^2+\beta^2+\gamma^2$	$(2)$ $-\frac{7}{8}$
$C$. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$	$(3)$ $86$
$D$. $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta}$	$(4)$ $0$
	$(5)$ $10$

If $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$ are the roots of $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$,then $\frac{1}{\alpha_1^2}+\frac{1}{\alpha_2^2}+\frac{1}{\alpha_3^2}+\frac{1}{\alpha_4^2}+\frac{1}{\alpha_5^2}=$

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