The equation #x^4 -2x^3-3x^2+4x-1=0# has four distinct real roots #x_1,x_2,x_3,x_4# such that #x_1<x_2<x_3<x_4# and product of two roots is unity, then the value of #x_1x_2 + x_1x_3 + x_2x_4 + x_3x_4#?

1 Answer
Cesareo R.
Mar 8, 2018

#-3#

Explanation:

Expanding

#(x+x_1)(x+x_2)(x+x_3)(x+x_4)# and comparing we have

#{(x_1x_2x_3x_4=-1), (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4=4), (x_1 x_2 + x_1 x_3 + x_2 x_3 + x_1 x_4 + x_2 x_4 + x_3 x_4=-3),(x_1 + x_2 + x_3 + x_4=-2):}#

Analyzing now

#x_1 x_2 + x_1 x_3 + x_2 x_3 + x_1 x_4 + x_2 x_4 + x_3 x_4=x_1x_2 + x_1x_3 + x_2x_4 + x_3x_4+(x_2x_3+x_1x_4)#

Choosing #x_1x_4=1# follows #x_2x_3 = -1# (see the first condition)

hence

#x_1x_2 + x_1x_3 + x_2x_4 + x_3x_4+(x_2x_3+x_1x_4) = -3# or

#x_1x_2 + x_1x_3 + x_2x_4 + x_3x_4 = -3-(x_2x_3+x_1x_4)=-3#

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