Say $y = 3\sqrt{2} - 4$, find $y^2 + \frac{1}{y^2}$.

$y = 3\sqrt{2} - 4$

Find $y^2 + \dfrac{1}{y^2}$.

---

Using the identity:

$y^2 + \dfrac{1}{y^2} = \left(y + \dfrac{1}{y}\right)^2 - 2$

---

$\dfrac{1}{y} = \dfrac{1}{3\sqrt{2} - 4}$

$= \dfrac{1}{3\sqrt{2} - 4} \cdot \dfrac{3\sqrt{2} + 4}{3\sqrt{2} + 4}$

$= \dfrac{3\sqrt{2} + 4}{(3\sqrt{2} - 4)(3\sqrt{2} + 4)}$

$= \dfrac{3\sqrt{2} + 4}{(3\sqrt{2})^2 - 4^2}$

$= \dfrac{3\sqrt{2} + 4}{18 - 16}$

$= \dfrac{3\sqrt{2} + 4}{2}$

---

$y + \dfrac{1}{y} = (3\sqrt{2} - 4) + \dfrac{3\sqrt{2} + 4}{2}$

$= 3\sqrt{2} - 4 + \dfrac{3\sqrt{2}}{2} + 2$

$= 3\sqrt{2} + \dfrac{3\sqrt{2}}{2} - 2$

$= \dfrac{6\sqrt{2} + 3\sqrt{2}}{2} - 2$

$= \dfrac{9\sqrt{2}}{2} - 2$

$= \dfrac{9\sqrt{2} - 4}{2}$

---

$y^2 + \dfrac{1}{y^2} = \left(y + \dfrac{1}{y}\right)^2 - 2$

$= \left(\dfrac{9\sqrt{2} - 4}{2}\right)^2 - 2$

$= \dfrac{(9\sqrt{2} - 4)^2}{4} - 2$

$(9\sqrt{2} - 4)^2 = (9\sqrt{2})^2 - 2(9\sqrt{2})(4) + 4^2$

$= 162 - 72\sqrt{2} + 16$

$= 178 - 72\sqrt{2}$

Therefore:

$y^2 + \dfrac{1}{y^2} = \dfrac{178 - 72\sqrt{2}}{4} - 2$

$= \dfrac{178 - 72\sqrt{2} - 8}{4}$

$= \dfrac{170 - 72\sqrt{2}}{4}$