Solve :

$9(x^2 + \dfrac{1}{x^2}) - 9(x + \dfrac{1}{x}) - 52 = 0$

Let $x + \dfrac{1}{x} = a$ ……..(i)

Squaring, both sides we get,

$\Rightarrow x^2 + \dfrac{1}{x^2} + 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 - 2 …….(ii)$

Substituting the values from equations (i) and (ii) we get,

⇒ 9(a² - 2) - 9a - 52 = 0

⇒ 9a² - 18 - 9a - 52 = 0

⇒ 9a² - 9a - 70 = 0

⇒ 9a² - 30a + 21a - 70 = 0

⇒ 3a(3a - 10) + 7(3a - 10) = 0

⇒ (3a + 7)(3a - 10) = 0

⇒ (3a + 7) = 0 or 3a - 10 = 0      [Zero product rule]

⇒ 3a = -7 or 3a = 10

⇒ $a = -\dfrac{7}{3} \text{ or } a = \dfrac{10}{3}$

Considering a = $\dfrac{10}{3}$ we get,

$\Rightarrow x + \dfrac{1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow 3(x^2 + 1) = 10x \\[1em] \Rightarrow 3x^2 + 3 = 10x \\[1em] \Rightarrow 3x^2 - 10x + 3 = 0 \\[1em] \Rightarrow 3x^2 - 9x - x + 3 = 0 \\[1em] \Rightarrow 3x(x - 3) - 1(x - 3) = 0 \\[1em] \Rightarrow (3x - 1)(x - 3) = 0 \\[1em] \Rightarrow 3x - 1 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow 3x = 1 \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{1}{3} \text{ or } x = 3.$

Considering a = $-\dfrac{7}{3}$ we get,

$\Rightarrow x + \dfrac{1}{x} = -\dfrac{7}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = -\dfrac{7}{3} \\[1em] \Rightarrow 3(x^2 + 1) = -7x \\[1em] \Rightarrow 3x^2 + 3 = -7x \\[1em] \Rightarrow 3x^2 + 7x + 3 = 0$

Comparing 3x² + 7x + 3 = 0 with ax² + bx + c = 0 we get,

a = 3, b = 7 and c = 3.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\Rightarrow x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)} \\[1em] = \dfrac{-7 \pm \sqrt{49 - 36}}{6} \\[1em] = \dfrac{-7 \pm \sqrt{13}}{6}.$

Hence, x = 3, $\dfrac{1}{3}, \dfrac{-7 \pm \sqrt{13}}{6}.$