Given: #(y + 2/y)^2 + 3y +6/y = 4#
This is one way to solve. Use #(a+b)^2 = a^2+2ab+ b^2#
#y^2 + 2cancel(y)(2/cancel(y))+4/y_2 + 3y +6/y = 4#
#y^2 + 4+4/y_2 + 3y +6/y = 4#
Multiply both sides by #y^2# to eliminate the fractions:
#y^4 + 4y^2 + 4 + 3y^3 +6y = 4y^2#
Add like terms and put in descending order:
#y^4 + 3y^3 + 6y + 4 = 0#
Factor:
Can't use group factoring.
Use #(y^2 +ay + b)(y^2 + cy + d) = y^4 + 3y^3 + 6y + 4 #
#y^4 + (a+c)y^3 + (d+ac+b)y^2 + (ad + bc)y + bd = y^4 + 3y^3 + 6y + 4 #
Solve the system:
#a + c = 3" "# the coefficient of the #y^3# term
#d + ac + b = 0" "# because there is no #y^2# term
#ad + bc = 6" "# the coefficient of the #y# term
#bd = 4#
Start with the possibilities for #bd = (2, 2), (4, 1), (1, 4)#
If #b = 2, d = 2#, then from the 2nd equation: #ac = -4#
Try #a = -1, c = 4" "# works for all equations!
Factored: #" "(y^2 - y + 2)(y^2 + 4y + 2) = 0#
Solve each trinomial either by completing the square or using the quadratic formula:
#y^2 - y + 2 = 0; " "y^2 + 4y + 2 = 0#
#y = (1+- sqrt(1-4(1)(2)))/2; " " y = (-4+- sqrt(16-4(1)(2)))/2#
#y = (1+- sqrt(7)i)/2; " " y = -2 +-sqrt(8)/2 = -2 +- sqrt(2)#