SOLUTION: Hi, I'm trying to find all real solutions to this equation: 4x^-4-16x^-2+4=0 I have: 4/x^4-16/x^2+4=0 multiplying by x^4: 4-16x^2+4=0 4x^4-16x^2+4=0 4(x^4-4x^2+1) = 0 (

Question 31323This question is from textbook College Algebra
: Hi,
I'm trying to find all real solutions to this equation:
4x^-4-16x^-2+4=0
I have:
4/x^4-16/x^2+4=0
multiplying by x^4:
4-16x^2+4=0
4x^4-16x^2+4=0
4(x^4-4x^2+1) = 0 (unfactorable?)
a=1;b=4;c=1
Using the quadratic equation I get:
-4+-sq root of 3, but this doesn't seem right.
Thanks for all your help! This question is from textbook College Algebra

Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
you had an error with your signs half way through...

this doesn't factorise simply. So lets use the quadratic formula.

First however, let -->

so,

so, or

Hence, from , we get:
or
or

jon.

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