The number of real solutions of the equation | vedclass

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(B) Given equation: $3(x^2 + \frac{1}{x^2}) - 2(x + \frac{1}{x}) + 5 = 0$Let $t = x + \frac{1}{x}$. Then $x^2 + \frac{1}{x^2} = t^2 - 2$.Substituting

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$0$

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(B) Given equation: $3(x^2 + \frac{1}{x^2}) - 2(x + \frac{1}{x}) + 5 = 0$Let $t = x + \frac{1}{x}$. Then $x^2 + \frac{1}{x^2} = t^2 - 2$.Substituting this into the equation: $3(t^2 - 2) - 2t + 5 = 0$$3t^2 - 6 - 2t + 5 = 0$$3t^2 - 2t - 1 = 0$Factoring the quadratic: $3t^2 - 3t + t - 1 = 0 \Rightarrow 3t(t - 1) + 1(t - 1) = 0$$(3t + 1)(t - 1) = 0$,so $t = 1$ or $t = -\frac{1}{3}$.Case $1$: $x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 Case $2$: $x + \frac{1}{x} = -\frac{1}{3} \Rightarrow 3x^2 + x + 3 = 0$. The discriminant $D = (1)^2 - 4(3)(3) = 1 - 36 = -35 Thus,the number of real solutions is $0$.

The number of real solutions of the equation $3(x^2 + \frac{1}{x^2}) - 2(x + \frac{1}{x}) + 5 = 0$ is:

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