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w^4-2w^2-2=0
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will my answers have the sqrt symbol in it?
If w^2 = x, then: x^2 - 2x - 2 = 0 x^2-2x = 2 x^2-2x + 1 = 2 + 1 (x-1)^2 = 3 x-1 = + or - sqrt(3) x = + or - sqrt(3) + 1 w^2 = + or - sqrt(3) + 1 w = + or - sqrt(+ or - sqrt(3) + 1)
\[\sqrt{1+\sqrt{3}},-\sqrt{1+\sqrt{3}}\]
Crazy solution, but that should be correct
ok thats what i came up with, it that some what right?
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That is the correct solution symbolically.
substitution w^2=t with condition t greater than or equal to zero...you get t^2-2t-2=0 solve it for t: |dw:1329521526971:dw|
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