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Question

If (ax2+bx+c)y+ax2+bx+c=0, find the condition that x may be a rational function of y.

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Solution

(ax2+bx+c)y+ax2+bx+c=0

For, y is a rational function,
y=(ax2+bx+c)ax2+bx+c

The denominator, ax2+bx+c0

xb±b24ac2a

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