Let p and q be two real numbers such that p+q | vedclass

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(C) Given $p+q=3$ and $p^{4}+q^{4}=369$.We need to find $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2} = \left(\frac{p+q}{pq}\right)^{-2} = \frac{(pq)^{2}

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$4$

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(C) Given $p+q=3$ and $p^{4}+q^{4}=369$.We need to find $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2} = \left(\frac{p+q}{pq}\right)^{-2} = \frac{(pq)^{2}}{(p+q)^{2}} = \frac{(pq)^{2}}{3^{2}} = \frac{(pq)^{2}}{9}$.We know that $p^{2}+q^{2} = (p+q)^{2}-2pq = 9-2pq$.Also,$p^{4}+q^{4} = (p^{2}+q^{2})^{2}-2p^{2}q^{2} = (9-2pq)^{2}-2(pq)^{2} = 369$.Expanding this,we get $81+4(pq)^{2}-36pq-2(pq)^{2} = 369$.$2(pq)^{2}-36pq+81 = 369 \implies 2(pq)^{2}-36pq-288 = 0$.Dividing by $2$,we get $(pq)^{2}-18pq-144 = 0$.Factoring the quadratic equation,$(pq-24)(pq+6) = 0$.So,$pq=24$ or $pq=-6$.If $pq=24$,then $p^{2}+q^{2} = 9-2(24) = 9-48 = -39$,which is impossible for real numbers $p$ and $q$.Thus,$pq=-6$.Substituting $pq=-6$ into our expression: $\frac{(pq)^{2}}{9} = \frac{(-6)^{2}}{9} = \frac{36}{9} = 4$.

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

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