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quadratic equation

jacks

Well-known member
Apr 5, 2012
226
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
Jan 26, 2012
890
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

Well-known member
Apr 5, 2012
226
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
Jan 26, 2012
890
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