Do there exist quadratic polynomials P(x) and Q(x) with real coeffcients such that the polynomial P(Q(x)) has precisely the zeros x = 2, x = 3, x =5 and x = 7?
Let P(x)=ax^2+bx+c and Q(x)=dx^2+ex+f and the polynomial with the given zeros be (x-2)(x-3)(x-5)(x-7)*Z. Expanding both gives the equation
ad^2x^4+2adex^3+(ae^2+2adf+bd)x^2+(2aef+be)x+(af^2+bf+c) = (x^4-17x^3+101x^2-247x+210)*Z
But we can drop the Z which is just a vertical stretch that can be applied to P if solutions are found.
Equating terms of equal degree in x gives the system
[4] ad^2=1
[3] 2ade=-17
[2] ae^2+2adf+bd=101
[1] 2aef+be=-247
[0] af^2+bf+c=210
[2] and [1] together form a linear system in f and b. Solving each for b gives
b=-2af+(101-ae^2)/d
b=-2af-247/e
This system is two parallel lines and will give no solutions unless
Using [4] and [3] we see a=1/d^2 and e=-17d/2
Substituting these into the above gives the equation
115/(4d)=494/d
Which only has the solution d=0.
This makes a=1/d^2 undefined and so no two quadratic functions exist.
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Posted by Jer
on 2020-12-04 10:47:21 |
Solution