(1+(1/x^2))^2+(4/x^2)-4 find the square root by division and factorization? the answer must be in alpha numri

Answer 1

Change it a little:
[(x^2+1)/x^2]^2 + 4/(x^2) -4
(x^2+1)^2/x^4 + 4/x^2 - 4
[(x^2+1)^2 + 4x^2 - 4x^4]/(x^4)
[-3x^4 + 6x^2 + 1]/(x^4)
[(-3x^2+1)^2]/(x^4)
Now that is the factored function in terms of a square
Taking the square root:
(-3x^2+1)/(x^2) = -3 + 1/x^2

Answer 2

1

Answer 3

I have already answered this question, however, here it is again:
(1+1/x^2)^2 + 4/x^2 - 4 = 1 + 2/x^2 + 1/x^4 + 4/x^2 - 4 = 9 + 6/x^2 + 1/x^4 -12 = (1/x^2 + 3)^2 - 12 = 0. Therefore 1/x^2 = sqrt(12) - 3 and x = (+/-) 1/sqrt(sqrt(12) - 3) = (+/-) 1.468.