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Question Number 20296 by Tinkutara last updated on 25/Aug/17

If (m_{r}, \frac{1}{m_{r}}); r = 1, 2, 3, 4 be four pairs

of values of x and y satisfy the equation

x^{2} + y^{2} + 2 g x + 2 f y + c = 0, then prove

that m_{1} . m_{2} . m_{3} . m_{4} = 1 .

Answered by ajfour last updated on 25/Aug/17

{(m_{r} + g)}^{2} + {(\frac{1}{m_{r}} + f)}^{2} = g^{2} + f^{2} - c

m_{r}^{2} {(m_{r} + g)}^{2} - m_{r}^{2} (g^{2} + f^{2} - c)

+ {({f m}_{r} + 1)}^{2} = 0

\Rightarrow m_{r} a r e r o o t s o f a b o v e e q u a t i o n

w h o s e c o n s t a n t t e r m i s 1 a n d

c o e f f i c i n t o f m_{r}^{4} i s a l s o 1 .

s o m_{1} . m_{2} . m_{3} . m_{4} = 1

Commented by Tinkutara last updated on 25/Aug/17

Thank you very much Sir!

Answered by Tinkutara last updated on 25/Aug/17

Let f (x) = x^{4} + 2 {g x}^{3} + {c x}^{2} + 2 f x + 1

= x^{2} (x^{2} + \frac{1}{x^{2}} + 2 g x + \frac{2 f}{x} + c)

f (x) has roots m_{1}, m_{2}, m_{3}, m_{4} so

m_{1} m_{2} m_{3} m_{4} = 1 by Vieta .