Waste less time on Facebook — follow Brilliant.

please help needed ...thanks:)

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
2 years ago

No vote yet
1 vote


Sort by:

Top Newest

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. Aditya Raut · 2 years ago

Log in to reply

@Aditya Raut Bingo you got it right aditya..thanks well for ur convenience the problem is from hall &knight Rajat Bisht · 2 years ago

Log in to reply

Log in to reply


Problem Loading...

Note Loading...

Set Loading...