Sum of the moduli of the complex roots of the | vedclass

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(D) Given equation: $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ Let $t = x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2} = t^2-2$. The equation becomes $(t^2-2

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(D) Given equation: $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ Let $t = x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2} = t^2-2$. The equation becomes $(t^2-2)-5t+6=0$,which simplifies to $t^2-5t+4=0$. Factoring gives $(t-4)(t-1)=0$,so $t=4$ or $t=1$. Case $1$: $x+\frac{1}{x}=4 \Rightarrow x^2-4x+1=0$. The discriminant $D = 16-4 = 12 > 0$,so roots are real. Case $2$: $x+\frac{1}{x}=1 \Rightarrow x^2-x+1=0$. The discriminant $D = 1-4 = -3 The roots are $x = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}$. Let the roots be $\alpha = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\beta = \frac{1}{2} - i\frac{\sqrt{3}}{2}$. The modulus of each complex root is $|\alpha| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$. Similarly,$|\beta| = 1$. The sum of the moduli is $|\alpha| + |\beta| = 1 + 1 = 2$.

Sum of the moduli of the complex roots of the equation $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ is

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