Solve the equation #3x^4-25x^3+50x^2-50x+12=0# given that the product of two of its roots is 2?

1 Answer
Aug 30, 2017

See below.

Explanation:

We know that #x_1x_2x_3x_4=12#

now making #x_1x_2 = 2# then #x_3x_4=6#

Observing #3 x^4 - 25 x^3 + 50 x^2 - 50 x + 12 = 0# by inspection we find that #x = 6# is a root then making #x_1=6# we obtain #x_2 = 1/3# and a we can verify #x=1/3# is also a root. So we have

#3 x^4 - 25 x^3 + 50 x^2 - 50 x + 12 = 3(x-6)(x-1/3)(x^2+alpha x+beta)#

so grouping coefficients we have

#{(6 beta=12), (19 beta - 6 alpha= 50), (19 alpha - 3 beta=-44), (3 alpha=-6):}#

and solving we have

#alpha=-2, beta=2# then

#3 x^4 - 25 x^3 + 50 x^2 - 50 x + 12 = 3(x-6)(x-1/3)(x^2-2 x+2)#

and

#x^2-2x+2=0# has two complex roots

#x = 1 pm i#