Let p and q be two real numbers such that p + q =3 and p4+q4=369 Then 1p+1q-2 is equal to ___________.

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

Solution:

$g i v e n t h a t p + q = 3, p^{4} + q^{4} = 369; (p + q)^{2} = 9 p^{2} + q^{2} = 9 - 2 p q \frac{1}{{(\frac{1}{p} + \frac{1}{q})}^{2}} = \frac{(q p)^{2}}{(q + p)^{2}} = \frac{(q p)^{2}}{9}; n o w p^{4} + q^{4} = {(p^{2} + q^{2})}^{2} - 2 p^{2} q^{2} \Rightarrow 369 = (9 - 2 p q)^{2} - 2 (p q)^{2} \Rightarrow 369 = 81 + 4 p^{2} q^{2} - 36 p q - 2 p^{2} q^{2} \Rightarrow 288 = 2 p^{2} q^{2} - 36 p q \Rightarrow 144 = p^{2} q^{2} - 18 p q (p q)^{2} - 2 \times 9 \times p q + 9^{2} = 144 + 9^{2} (p q - 9)^{2} = 225 p q - 9 = \pm 15 p q = \pm 15 + 9 p q = 24, - 6$

(24 is rejected because $p^{2} + q^{2} = 9 - 2 p q$ is negative)

$\frac{(q p)^{2}}{9} = \frac{16 \times 16}{9} = 4$

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