If p = $\dfrac{2-\sqrt{5}}{2+\sqrt{5}}$ and q = $\dfrac{2+\sqrt{5}}{2-\sqrt{5}}$ find the values of :

(i) p + q

(ii) p - q

(iii) p² + q²

(iv) p² - q²

(i)

$p+q = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} +\dfrac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2-\sqrt{5})^2 + (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\dfrac{4 + 5 - 4\sqrt{5} + 4 + 5 + 4\sqrt{5}}{(2)^2 -(\sqrt{5})^2} \\[1.5em] \bold{p+q = -18} \qquad \text{….(i)} \\[1.5em]$

$(\text{ii})$

$p-q = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} -\dfrac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2-\sqrt{5})^2 - (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\dfrac{4 + 5 - 4\sqrt{5} - 4 - 5 - 4\sqrt{5}}{(2)^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\bold {p - q = 8\sqrt{5}} \qquad \text{….(ii)}$

$(\text{iii})$

$(p+q)^2 = (p)^2 + (q)^2 +2pq \\[1.5em] \Rightarrow(p)^2+(q)^2 =(p+q)^2 -2pq \qquad \text{….(iii)} \\[1.5em] \Rightarrow pq = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} ×\dfrac{2+\sqrt{5}} {2-\sqrt{5}} = 1 \qquad \text{….(iv)} \\[1.5em]$

substituting value of (i) and (iv) in (iii) :

$\bold{p^2+q^2 = (-18)^2 - 2 = 324-2 = 322}$

$(\text{iv})$

$(p)^2 - (q)^2 = (p+q)(p-q) \qquad \text{….(v)}$

Using (i) and (ii) in (v) we get,

$\bold{p^2 - q^2 = -18 × 8\sqrt{5} = -144\sqrt{5}}$