• May 6th 2013, 11:25 PM
Cake
Hey Guys,
I am a bit confused on what to do for this problem. If I can get some help I'd appreciate it.

Suppose that f(x) = x4 + 7x2+ 12. Find the values of x such that

a. f(x) = 12
b. f(x) = 6

I figured part b already but I am having trouble on part (a) which is

12 = x4 + 7x2+ 12
-12 -12

Now I'm left with x4 + 7x2

I use "Let u = x2 , Let u2 = x4"

u2 + u

Do I get a third term by completing the square method? I am lost with this step.
• May 6th 2013, 11:58 PM
ibdutt
It is not must that quadratic polynomial should always have three terms. It may be a monomial / binomial or a trinomial.
Now you have u(u+ 7u ) = 0
In second case we will have
u^2 + 7u + 6 = 0
• May 7th 2013, 10:54 AM
mathguy25
What is the quadratic formula? What are the values of a, b, and c?
• May 7th 2013, 05:49 PM
Cake
Quote:

Originally Posted by ibdutt
It is not must that quadratic polynomial should always have three terms. It may be a monomial / binomial or a trinomial.
Now you have u(u+ 7u ) = 0
In second case we will have
u^2 + 7u + 6 = 0

Hey ibdutt,

You are correct :D
Honestly, I did factor the problem and ended up with u(u + 7u) = 0 but I thought it looked wrong and I wasn't used to seeing a quadratic polynomial with two terms. Thanks for clearing things up for me!

Cheers!
• May 7th 2013, 05:51 PM
Cake