### Welcome to our community

#### jacks

##### Well-known member
if $f(x)$ be a continous and assumes only rational values so that $f(2010) =1.$ then roots of

the equation $f(1)x^2 + 2f(2)x + 3f(3) =0$ are

#### CaptainBlack

##### Well-known member
if $f(x)$ be a continous and assumes only rational values so that $f(2010) =1.$ then roots of

the equation $f(1)x^2 + 2f(2)x + 3f(3) =0$ are
Continuous and rational implies that $$f(x)$$ is a constant.

CB

• jacks

#### jacks

##### Well-known member
means $x^2+2x+3=0\Leftrightarrow (x+1)^2+2>0\forall x\in \mathbb{R}$

Means no real Roots.

but I did not understand the line if $f(x)$ is Conti. and assume only rational values .then it must be Constant

Thanks

#### CaptainBlack

##### Well-known member
means $x^2+2x+3=0\Leftrightarrow (x+1)^2+2>0\forall x\in \mathbb{R}$

Means no real Roots.

but I did not understand the line if $f(x)$ is Conti. and assume only rational values .then it must be Constant

Thanks
If $$f(x)$$ is continuous it satisfies the intermediate value principle, that is $$f(x)$$ takes on all values between $$f(a)$$ and $$f(b)$$ for any distinct reals $$a$$ and $$b$$.

We are told that $$f(a)$$ and $$f(b)$$ are rational, and if they are not equal there is an irrational $$\rho$$ between them and a $$c \in (a,b)$$ such that $$f(c)=\rho$$ which contradicts $$f(x)$$ only taking rational values, so for any two real $$a, b$$ $$f(a)=f(b)$$ hence $$f(x)$$ is a constant.

CB

• HallsofIvy