The number of real solutions of the equation 3 ( x 2 + 1 x 2 ) - 2 ( x + 1 x ) + 5 = 0 , is [2023]

Question

The number of real solutions of the equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is [2023]

Smart Achievers · Accepted Answer

(4)

$Given, 3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$

$\Rightarrow 3 [{(x + \frac{1}{x})}^{2} - 2] - 2 (x + \frac{1}{x}) + 5 = 0$

$\Rightarrow 3 [y^{2} - 2] - 2 y + 5 = 0$ $(Let y = x + \frac{1}{x})$

$\Rightarrow 3 y^{2} - 6 - 2 y + 5 = 0 \Rightarrow 3 y^{2} - 2 y - 1 = 0$

$\Rightarrow y = 1, \frac{- 1}{3}$ $But y \in (- \infty, - 2] \cup [2, \infty)$

$Hence, no real solution.$

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