# Can't get my head around this

• September 9th 2008, 12:12 PM
pandemic
Can't get my head around this
Hi i'm having some trouble explaining one thing.
Quote:

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.
• September 9th 2008, 12:33 PM
masters
Quote:

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

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

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

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

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

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

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

$f(4x)=0$

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

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

$x=\frac{1}{2} \ \ or \ \ x = 1$
• September 9th 2008, 12:40 PM
Moo
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 $x=\frac 12 \text{ or } x=1$
• September 9th 2008, 02:16 PM
pandemic
Thanks alot for a quick response and a good one.