Roots of $f(x)$ Quadratic Equation: $1x^2+2x+3=0$

[juantheron]

if $f(x)$ be a continuous 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]

if $f(x)$ be a continuous 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

 

[juantheron]

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]

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