Question 70342
    m^4  + 4 m^2 -12 =0
  Now find the roots of 12 
 (6,2) (4,3) In this case since product of two roots is negative one of the roots is negative. The difference between the two is 4.
 So the possible set of the roots is (6 , -2).

     m^4 + 6 m^2 -2m^2 -12 =0
     m^2(m^2+ 6) -2(m^2 + 6) = 0
     (m^2-2)* (m^2+6) =0  hence (m^2 -2) =0  or (m^ + 6) =0
     m^2 = 2 or m^2 = -6
     m= + or - sqrt(2)  or m^2 = i^2 * 6
     m= + or - i * sqrt(6)