Question

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

Answer 1

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`