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 Three Root Trial (Posted on 2016-09-10)
Determine all possible values of a real number C such that the equation

(x2 - 2Cx - 4(C2 + 1))(x2 - 4x - 2C(C2 + 1)) = 0

has precisely three distinct roots.

 No Solution Yet Submitted by K Sengupta No Rating

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 solution Comment 1 of 1
One possibility is for each quadratic to share a root.

Multiplying (x-p)(x-q) and (x-q)(x-r) give equations for c in terms of (p,q,r) with solutions c=3 and c=-1.

The only other possibility is for one quadratic to have a repeated root.

Say the first quadratic has the dual root p.  Since the discriminant is postive, p is real.  But it's necessary too for p^2=-4(c^2+1) which is impossible.

If it's the second quadratic with the repeated root then 2p=4 and p^2=-2c(c^2+1).  So p=2 and substituting gives a cubic equation with a real root c=-1 by inspection and the remaining roots imaginary.

When c=3 (p,q,r)=(10,-4,-6).

When c=-1 (p,q,r)=(2,1+sqrt(17),1-sqrt(17)).

 Posted by xdog on 2016-09-11 09:45:48

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