1. ## Can't get my head around this

Hi i'm having some trouble explaining one thing.
We know x=2 and x=4 are the only solutions to f(x)= 0. Solve the equations f(4x) = 0 - Please explain!
This is what's written in my practice questions. Can someone help my explain this.

The teacher said it was pretty simple once you figured it out.

2. Originally Posted by pandemic
Hi i'm having some trouble explaining one thing.

This is what's written in my practice questions. Can someone help my explain this.

The teacher said it was pretty simple once you figured it out.
If x=2 and x=4 are roots of f(x)=0, then

$f(x)=(x-2)(x-4)$

$f(x)=x^2-6x+8$

$f(4x)=(4x)^2-6(4x)+8$

$f(4x)=16x^2-24x+8$

$f(4x)=2x^2-3x+1$

$f(4x)=(2x-1)(x-1)$

$f(4x)=0$

$(2x-1)(x-1)=0$

$2x-1=0 \ \ or \ \ x-1=0$

$x=\frac{1}{2} \ \ or \ \ x = 1$

3. Hello !

@ masters : but huuum... we do not know if f is a polynomial... and actually, if x=2 and x=4 and f a polynomial, then f(x)=a(x-2)(x-4) where a is a non-zero constant.

@ the OP :
"x=2 and x=4 are the only solutions to f(x)=0" this can be rewritten :

$f(x)=0 \Longleftrightarrow x=2 \text{ or } x=4$

So to the solution f(4x), let $t=4x$ for more convenience and less confusing (it doesn't change anything if we call x t or whatever) :
$f(t)=0 \implies t= ??$

From the first equivalence, we have $f(t)=0 \Longleftrightarrow t=2 \text{ or } t=4$

That is to say $4x=2 \text{ or } 4x=4$, that is to say $x=\frac 12 \text{ or } x=1$

4. Thanks alot for a quick response and a good one.