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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