Question

if p=root 3 minus root 5 upon 3 + root 5 and q is equal to 3 + root 5 upon 3 minus root root 5 find the value of p square + q square

Answer 1

$p = \frac{3-\sqrt5}{3+\sqrt5}$

$q = \frac{3+\sqrt5}{3-\sqrt5}$

$p^2 + q^2 = (\frac{3-\sqrt5}{3+\sqrt5})^2+(\frac{3+\sqrt5}{3-\sqrt5} )^2$

Cross multiply

$p^2 + q^2 = \frac{(3-\sqrt5)^4+(3+\sqrt5)^4}{(3+\sqrt5)^2(3-\sqrt5)^2}$

$p^2 + q^2 = \frac{[(3-\sqrt5)^2-(3+\sqrt5)^2]^2+2(3-\sqrt5)^2(3+\sqrt5)^2}{(3+\sqrt5)^2(3-\sqrt5)^2}$

What I actually did in the above step is

$a^4+b^4 = (a^2)^2+(b^2)^2 = (a^2-b^2)^2+2a^2b^2$

Coming back to the question

$p^2 + q^2 = \frac{[(3-\sqrt5)^2-(3+\sqrt5)^2]^2}{(3+\sqrt5)^2(3-\sqrt5)^2}+2$

$p^2 + q^2 = \frac{[6*(-2\sqrt5)]^2}{(9-5)^2}+2$

$p^2 + q^2 = \frac{144*5}{16}+2$

$p^2+q^2 = 45+2$

$\huge\black\boxed{p^2+q^2 = 47}$