Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then ${(\frac{1}{p} + \frac{1}{q})}^{- 2}$ ${\frac{}{} \frac{}{}}^{}$ is equal to……….

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Vishal Baghel | Contributor-Level 10

3 months ago

$? p + q = 3$ ….(i)

and $p^{4} + q^{4}$ $^{}^{}$ = 369 ….(ii)

${{(p + q)}^{2} - 2 p q}^{2} - 2 p^{2} q^{2} = 369$

or ${(9 - 2 p q)}^{2} - 2 {(p q)}^{2} = 369$ $^{}^{}$

or ${(p q)}^{2} - 18 p q - 144 = 0$ $^{}$

$∴ p q = - 6 o r 24$

But pq 24 is not possible

$∴ p q = - 6$

$H e n c e, {(\frac{1}{p} + \frac{1}{q})}^{- 2} = {(\frac{p q}{p + q})}^{2} = {(- 2)}^{2} = 4$

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Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then ${(\frac{1}{p} + \frac{1}{q})}^{- 2}$ ${\frac{}{} \frac{}{}}^{}$ is equal to……….

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Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then ${(\frac{1}{p} + \frac{1}{q})}^{- 2}$ ${\frac{}{} \frac{}{}}^{}$ is equal to……….