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

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