Question

Let $p$ and $q$ be two real numbers such that $p+q=3$ and $p^4+q^4=369$. Then ${\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}$ is equal to_____.

Answer 1

Answer: 4

$p+q=3p^4+q^4=369$
${\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}$
$(p+q{)}^2=9$
$p^2+q^2=9-2pq$
$\frac{1}{{\left(\frac{1}{p}+\frac{1}{q}\right)}^2}=\frac{(qp{)}^2}{(q+p{)}^2}=\frac{(qp{)}^2}{9}$
$p^4+q^4={\left(p^2+q^2\right)}^2-2p^2{q}^2$
$369=(9-2pq{)}^2-2(pq{)}^2$
$369=81+4p^2{q}^2-36pq-2p^2{q}^2$
$288=2p^2{q}^2-36pq$
$144=p^2{q}^2-18pq$
$(pq{)}^2-2\times 9\times pq+9^2=144+9^2$
$(pq-9{)}^2=225$
$pq-9=\pm 15$
$pq=\pm 15+9$
$pq=24,-6$
(24 is rejected because $p^2+q^2=9-2pq$ is negative)
$\frac{(qp{)}^2}{9}=\frac{1(-6{)}^2}{9}=4$