First slide

Theory of equations

Question

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

Moderate

A

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

$a / a^{'} = c / c^{'}$
C

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

${aa}^{'} + {cc}^{'} = 1$

Solution

Given $({ax}^{2} + c) y + (a^{'} x^{2} + c^{'}) = 0$
or $x^{2} (ay + a^{'}) + (cy + c^{'}) = 0$
If x is rational, then the discriminant of the above equation
must be a perfect square. Hence,
$0 - 4 (ay + a^{'}) (cy + c^{'})$ must be a perfect square
$\Rightarrow - {acy}^{2} - ({ac}^{'} + a^{'} c) y - a^{'} c^{'}$
$\Rightarrow {({ac}^{'} + a^{'} c)}^{2} - 4 {aca}^{'} c^{'} = 0$ $[∵ D = 0]$
$\begin{array}{l} \Rightarrow {({ac}^{'} - a^{'} c)}^{2} = 0 \\ \Rightarrow {ac}^{'} = a^{'} c \\ \Rightarrow \frac{a}{a^{'}} = \frac{c}{c^{'}} \end{array}$

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