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

Note by Rajat Bisht
3 years, 4 months ago

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Here's one of the things that I could instantaneously think of, maybe it's one way...

$$(ay+a')x^2+(by+b')x+(cy+c') =0$$

Solving the quadratic, its roots will be

$$x=\dfrac{-by-b' \pm \sqrt{(by+b')^2-4(ay+a')(cy+c')}}{2ay+2a'}$$

For $$x$$ to be rational function of $$y$$, the term $$(by+b')^2-4(ay+a')(cy+c')$$ should be perfect square, giving square root as a linear in $$y$$ , or it could be zero.

Thus I think, conditions will be as following -

$$(i) \quad (b^2-4ac)y^2-2(2ac'+2a'c-bb')y+(b'^2-4a'c')=0$$ ... i.e. giving 2 values of $$y$$

$$(ii)\quad$$ Comparing the above quadratic with $$kx^2+lx+m=0$$, condition for perfect square is $$\Bigl( \dfrac{l}{2k} \Bigr)^2 = \dfrac{m}{k} \implies l^2=4mk \implies (2ac'+2a'c-bb')^2 = (b^2-4ac)(b'^2-4a'c')$$

@Rajat Bisht if you have something like answer key to check. please tell me if this is right.

- 3 years, 4 months ago

Bingo you got it right aditya..thanks well for ur convenience the problem is from hall &knight

- 3 years, 4 months ago