Prove that there doesn't exist a set of three different quadratic trinomials f(x), g(x), h(x) such that
the solutions of the equation f(x)=g(x) are 1 and 4,
the solutions of the equation g(x)=h(x) are 2 and 5, and
the solutions of the equation h(x)=f(x) are 3 and 6.
Let
f(x)-g(x)=p(x-1)(x-4)=px^2-5px+4p
g(x)-h(x)=q(x-2)(x-5)=qx^2-7qx+10q
f(x)-h(x)=r(x-3)(x-6)=rx^2-9rx+18r
for some values of (p,q,r)
Adding the first two equations gives
f(x)-h(x)=(p+q)x^2+(-5p-7q)x+(4p+10q)
Which together with the third equation yields the system
p+q=r
-5p-7q=-9r
4p+10q=18r
The only solution is (0,0,0) which implies f(x)=g(x)=h(x). This is clearly impossible.
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Posted by Jer
on 2021-03-24 11:54:48 |
Solution
