How to solve x^4+ab=(1+ab)x^2 ?

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Move terms to the lefthand side

`x^4 -(1+ab)x^2+ ab = 0`

Let `u = x^2`

Then we have

`u^2 - (1+ab)u +ab = 0`

Now solve this quadratic using the quadratic formula:

`u = (-B +- sqrt(B^2 -4AC))/(2A)`

where `A = 1`, `B = -(1+ab)`  and `C = ab`

Then solutions are

`u = (1+ab +- sqrt((1+ab)^2 -4ab))/2`

` = (1+ab +- sqrt(a^2b^2 + 2ab + 1 -4ab))/2`

` = (1 +ab +- sqrt(a^2b^2 -2ab + 1))/2a`

` = (1+ab +-sqrt((ab -1)^2))/2`

`= (1+ab +- (ab-1))/2`

` `` `` = (ab)/2`  or 1

`implies u = x^2 = (ab)/2`  or 1

`implies x = +-sqrt((ab)/2)`  or `+-1`

If `ab <0` then two of these roots will be complex

x = -sqrt(ab/2) or sqrt(ab/2) or 1 or -1

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