If $(a x^{2} + b x + c) y + a^{'} x^{2} + b^{'} x + c^{'} = 0$ , then the condition that x may be rational function of y is

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Solution

The correct option is A
The given equation can be written as
$(a y + a^{'}) x^{2} + (b y + b^{'}) x + (c y + c^{'}) = 0$
The condition that x may be rational function of y is,
$(b y + b^{'})^{2} - 4 (a y + a^{'}) (c y + c^{'})$ is a perfect square that is,
$(b^{2} - 4 a c) y^{2} + (2 b b^{'} - 4 a c^{'} - 4 a^{'} c) y + b^{' 2} - 4 a^{'} c^{'}$ is a perfect square
$\Rightarrow D = 0$ that is,
$4 (b b^{'} - 2 a c^{'} - 2 a^{'} c)^{2} - 4 (b^{2} - 4 a c) (b^{' 2} - 4 a^{'} c^{'}) = 0$
or $(a c^{'} + a^{'} c) - 4 a a^{'} c c^{'} = a b b^{'} c + a^{'} b b^{'} c - a^{'} b b^{'} c - a^{'} c^{'} b^{2} - a c b^{' 2}$
or $(a c^{'} - a^{'} c)^{2} = (a b^{'} - a^{'} b) (b c^{'} - b^{'} c)$

Suggest Corrections

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If a root of the equation $a x^{2} + b x + c = 0$ be reciprocal of the equation then $a^{'} x^{2} + b^{'} x + c^{'} = 0$ , then

a x^{2} + b x + c = 0

a^{'} x^{2} + b^{'} x + c^{'} = 0

\frac{a}{c^{'}} = \frac{b}{b^{'}} = \frac{c}{a^{'}}

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