If 2x = 3 + $\sqrt{7}$, find the value of : $4x^2 + \dfrac{1}{x^2}$.

Given: 2x = 3 + $\sqrt{7}$

⇒ x = $\dfrac{3 + \sqrt{7}}{2}$

Squaring both the sides, we get:

⇒ x² = $\dfrac{(3 + \sqrt{7})}{2^2}^2$

⇒ 4x² = 3² + 2 x 3 x $\sqrt{7} + (\sqrt{7})^2$

⇒ 4x² = 9 + 6 $\sqrt{7}$ + 7

⇒ 4x² = 16 + 6 $\sqrt{7}$

Now,

$\dfrac{1}{4x^2} = \dfrac{1}{16 + 6\sqrt{7}}\\[1em] \Rightarrow\dfrac{1}{4x^2} = \dfrac{1 \times (16 - 6\sqrt{7c})}{(16 + 6\sqrt{7}) \times (16 - 6\sqrt{7})}\\[1em] = \dfrac{16 - 6\sqrt{7}}{16^2 - (6\sqrt{7})^2}\\[1em] = \dfrac{16 - 6\sqrt{7}}{256 - 252}\\[1em] = \dfrac{16 - 6\sqrt{7}}{4}\\[1em] \Rightarrow \dfrac{1}{x^2} = \dfrac{4(16 - 6\sqrt{7})}{4}\\[1em] = 16 - 6\sqrt{7}$

Now,

$4x^2 + \dfrac{1}{x^2} = 16 + 6 \sqrt{7} + 16 - 6 \sqrt{7}\\[1em] = 32$

Hence, the value of $4x^2 + \dfrac{1}{x^2} = 32$.