Let $p$ and $q$ be two real numbers such that $p+q=3$ and $p^{4}+q^{4}=369$. Then $\left(\frac{1}{p}

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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

Professor PC Thomas Classes · Answer

(4) : Given, $p+q=3$\ $\Rightarrow \quad(p+q)^{2}=9$\ $\Rightarrow p^{2}+q^{2}+2 p q=9$\ $\Rightarrow p^{2}+q^{2}=9-2 p q$\ $\Rightarrow \quad p^{2}+q^{2}=9-2 p q$\ $\Rightarrow \quad\left(p^{2}+q^{2}ight)^{2}=(9-2 p q)^{2}$\ $\Rightarrow p^{4}+q^{4}+2 p^{2} q^{2}=81-36 p q+4 p^{2} q^{2}$\ $\Rightarrow \quad 369=81-36 p q+2 p^{2} q^{2}$\ $\Rightarrow \quad 2 p^{2} q^{2}-36 p q-288=0$\ $\Rightarrow p^{2} q^{2}-18 p q-144=0$\ $\Rightarrow \quad p^{2} q^{2}-24 p q+6 p q-144=0$\ $\Rightarrow p q(p q-24)+6(p q-24)=0$\ $\Rightarrow \quad(p q+6)(p q-24)=0$\ $\Rightarrow p q=-6,24$\ (Ignore 24)\ So, $\left(\frac{1}{p}+\frac{1}{q}ight)^{-2}=\left(\frac{p q}{p+q}ight)^{2}=\left(\frac{-6}{3}ight)^{2}=4$\