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 (

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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 (      Log On


   

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

+4x%5E-4-16x%5E-2%2B4=0+
+4%2Fx%5E4-16%2Fx%5E2%2B4=0+
+4-16x%5E2%2B4x%5E4=0+
+4x%5E4-16x%5E2%2B4=0+
+x%5E4-4x%5E2%2B1=0+

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

First however, let y=x%5E2 --> +y%5E2-4y%2B1=0+

so, +y+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29+
+y+=+%28-%28-4%29+%2B-+sqrt%28%28-4%29%5E2+-+4%281%29%281%29%29%29%2F%282%281%29%29+
+y+=+%284+%2B-+sqrt%2816+-+4%29%29%2F%282%29+
+y+=+%284+%2B-+sqrt%2812%29%29%2F%282%29+
+y+=+%284+%2B-+sqrt%284%2A3%29%29%2F%282%29+
+y+=+%284+%2B-+sqrt%284%29sqrt%283%29%29%2F%282%29+
+y+=+%284+%2B-+2sqrt%283%29%29%2F%282%29+
+y+=+%282%282+%2B-+sqrt%283%29%29%29%2F%282%29+
+y+=+2+%2B-+sqrt%283%29+

so, +y+=+2+%2B+sqrt%283%29+ or +y+=+2+-+sqrt%283%29+

Hence, from y=x%5E2, we get:
+x+=+sqrt%282+%2B+sqrt%283%29%29+ or +x+=+-sqrt%282+%2B+sqrt%283%29%29+
+x+=+sqrt%282+-+sqrt%283%29%29+ or +x+=+-sqrt%282+-+sqrt%283%29%29+

jon.