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,
$(a x^{2} + c) y + (a^{'} x^{2} + c^{'}) = 0 \Rightarrow x^{2} (a y + a^{'}) + (c y + c^{'}) = 0$
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. $- a c y^{2} - (a c^{'} + a^{'} c) y - a^{'} c^{'}$ must be a perfect square
$\Rightarrow {(a c^{'} + a^{'} c)}^{2} - 4 a c a^{'} c^{'} = 0$ [ $∴$ Disc $= 0$ ]
$\Rightarrow {(a c^{'} - a^{'} c)}^{2} = 0 \Rightarrow a c^{'} = a^{'} c \Rightarrow \frac{a}{a^{'}} = \frac{c}{c^{'}}$

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