First slide

Theory of equations

Question

$If (a x^{2} + c) y + (a^{'} x^{2} + c^{'}) = 0 and x is a rational function of y and a c is negative, then$

Moderate

A

$a c^{'} + a^{'} c = 0$
B

$\frac{a}{a^{'}} = \frac{c}{c^{'}}$
C

$a^{2} + c^{2} = a^{' 2} + c^{' 2}$
D

$a a^{'} + c c^{'} = 1$

Solution

$We have : (a x^{2} + c) y + (a^{'} x^{2} + c^{'}) = 0 or x^{2} (a y + a^{'}) + (c y + c^{'}) = 0 . If x is rational, then the discriminant of the$
above equation must be a perfect square.
$i.e. 0 - 4 (a y + a^{'}) (c y + c^{'}) must be a perfect square$
$i.e. - a c y^{2} - (a c^{'} + a^{'} c) y - a^{'} c = 0 [∵ Disc. = 0] - - - - (1)$
$\Rightarrow {(a c^{'} + a^{'} c)}^{2} - 4 a c a^{'} c^{'} = 0 w e g e t p e r f e c t s q u a r e w h e n r o o t s a r e e q u a l a n d r o o t s a r e e q u a l i f d i s c r i m i n a n t o f (1) i s z e r o$
$\Rightarrow {(a c^{'} - a^{'} c)}^{2} = 0 \Rightarrow a c^{'} = a^{'} c \Rightarrow \frac{a}{a^{'}} = \frac{c}{c^{'}}$

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