Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ ...

Question

Problem : Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ and $f(0) =2+\sqrt{3}+\sqrt{5}$ Prove that the equation $f(2009)x^2 +2f(0)x +f(2009)=0$ has only rational roots.

Solution :

Let us assume the function f(x) be $2+(\sqrt{3}+\sqrt{5})(x+1)$ which satisfies $f(0) =2+\sqrt{3}+\sqrt{5}$

now $f(2009)x^2 =(2010((\sqrt{3}+\sqrt{5})x^2$ .....(1)

Also $2f(0)x =2(2+\sqrt{3}+\sqrt{5})x$ ...(2)

f$(2009)=(2010)(\sqrt{3}+\sqrt{5})$ ......(3)

Now solving for $f(2009)x^2 +2f(0)x +f(2009)=0$ by putting the values of $f(2009)x^2 , 2f(0)x ; f(2009)$ from 1,2 & 3 I am getting roots which are irrational please suggest some other method for this ... thanks....

Answer 1

A continuous function that takes only irrational values on an interval must be constant. Hence $f(2009)=f(0)\ne0$ and the quadratic is equivalent to $x^2+2x+1=0$ (with double root $x=-1$.

Answer 2

A continuous function only taking irrational values is constant, hence $f(2009)x^2 + 2f(0)x + f(2009)=0$ simplifies to $x^2+2x+1=0$ or $(x+1)^2=0$.