please help needed ...thanks:)

If(ax2+bx+c)y+a′x2+b′x+c′=0(ax^{2}+bx+c)y+a'x^{2}+b'x+c'=0 find the condition that xx may be a rational function of yy

#Algebra #HelpMe! #Quadratic

Note by Rajat Bisht
6 years, 3 months ago

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Comments

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

(ay+a′)x2+(by+b′)x+(cy+c′)=0(ay+a')x^2+(by+b')x+(cy+c') =0

Solving the quadratic, its roots will be

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

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

Thus I think, conditions will be as following -

(i)(b2−4ac)y2−2(2ac′+2a′c−bb′)y+(b′2−4a′c′)=0(i) \quad (b^2-4ac)y^2-2(2ac'+2a'c-bb')y+(b'^2-4a'c')=0 ... i.e. giving 2 values of yy

(ii)(ii)\quad Comparing the above quadratic with kx2+lx+m=0kx^2+lx+m=0, condition for perfect square is (l2k)2=mk  ⟹  l2=4mk  ⟹  (2ac′+2a′c−bb′)2=(b2−4ac)(b′2−4a′c′)\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 - 6 years, 3 months ago

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

Rajat Bisht - 6 years, 3 months ago
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