The number of real solutions of the equation $3\left(x^2 + \dfrac{1}{x^2}\right) - 2\left(x + \dfrac{1}{x}\right) + 5 = 0$, is
Step-by-Step Solution
Key Concept: Let $t = x + \frac{1}{x}$, so $x^2 + \frac{1}{x^2} = t^2 - 2$. The equation becomes $3(t^2-2) - 2t + 5 = 0$, i.e., $3t^2 - 2t - 1 = 0$.
$3t^2 - 2t - 1 = 0 \Rightarrow (t-1)(3t+1) = 0 \Rightarrow t = 1$ or $t = -\frac{1}{3}$. Since $|x + \frac{1}{x}| \geq 2$ for real $x \neq 0$, neither $t=1$ nor $t=-\frac{1}{3}$ yields real solutions. Answer: 0.
Correct Answer: 2