F(f(x))= (x^4)-4(x^2)+2, find f(x).

[GreenTea09]

Homework Statement

http://postimg.org/image/48919pl1x/

Homework Equations

f(f(x))= (x^4)-4(x^2)+2,

The Attempt at a Solution

[f(x)]^2=
(x^4)-4(x^2)+2=
[(x-2)^2]-(4x^2)+2

i can't seem to square root (x^4)-4(x^2)+2...

 

[Saitama]

Homework Statement

http://postimg.org/image/48919pl1x/

Homework Equations

f(f(x))= (x^4)-4(x^2)+2,

The Attempt at a Solution

[f(x)]^2=
(x^4)-4(x^2)+2=
[(x-2)^2]-(4x^2)+2

i can't seem to square root (x^4)-4(x^2)+2...

The question states that f(x) is a polynomial. Use this.

http://en.wikipedia.org/wiki/Polynomial

 

[GreenTea09]

ok,i tried this solution
let f(x)=a(x^2)+bx+c
(a(x^2)+bx+c)*(a(x^2)+bx+c) =(x^4)-4(x^2)+2

comparing the coefficient of x^2
2c+b^2=-4 ---eqn(2)
comparing the coefficient of constant,
i get a c value of sqrt2 and -sqrt2
which i try to fit into eqn(2) but dosent work out..

 

[arildno]

f(f(x)) must also be evaluated on the bf(x)+c terms!

 

[Mark44]

Homework Statement

http://postimg.org/image/48919pl1x/

The image link is broken.

Homework Equations

f(f(x))= (x^4)-4(x^2)+2,

The Attempt at a Solution

[f(x)]^2=

No one else pointed this out, so I will. f(f(x)) is not the same as [f(x)]².

(x^4)-4(x^2)+2=
[(x-2)^2]-(4x^2)+2

i can't seem to square root (x^4)-4(x^2)+2...

That's not relevant here, since you are not being asked to solve for f(x) in the equation [f(x)]² = ...

What you're working with is a composite of f with itself -- (f o f)(x) --, not the square of f(x).