#f=X^4+3X^3+aX^2-3X+1;ainRR#. Which are values of #a# for which #x_1^2+x_2^2+x_3^2+x_4^2=12#? Where #x_1,x_2,x_3,x_4# are roots of #f(x)=0#.

1 Answer
George C.
Jun 16, 2017

#a = -3/2#

Explanation:

Given:

#f(x) = x^4+3x^3+ax^2-3x+1#

#color(white)(f(x)) = (x-x_1)(x-x_2)(x-x_3)(x-x_4)#

#color(white)(f(x)) = x^4-(x_1+x_2+x_3+x_4)x^3+(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)x^2-(x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2)x+x_1x_2x_3x_4#

Equating coefficients, we have:

#{(x_1+x_2+x_3+x_4=-3),(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=a),(x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2=3),(x_1x_2x_3x_4=1):}#

Note that:

#9 = (-3)^2#

#color(white)(9) = (x_1+x_2+x_3+x_4)^2#

#color(white)(9) = x_1^2+x_2^2+x_3^2+x_4^2+2(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)#

#color(white)(9) = 12+2a#

Hence:

#2a = 9-12 = -3#

So:

#a = -3/2#

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