If #x+1/x = sqrt(3)# then what is #x^9+x^2+2# ?

1 Answer
George C.
May 13, 2017

#x^9+x^2+2 = 5/2+-(sqrt(3)/2-1)i#

Explanation:

If #x > 0# then #x+1/x >= 2#

If #x < 0# then #x+1/x <= -2#

So there are no real values of #x# satisfying:

#x+1/x = sqrt(3) ~~ 1.732#

We can find complex solutions, by first multiplying through by #x# to get a quadratic:

#x^2+1 = sqrt(3)x#

Subtract #sqrt(3)x# from both sides to get:

#x^2-sqrt(3)x+1 = 0#

This has roots given by the quadratic formula:

#x = (sqrt(3)+-sqrt((sqrt(3))^2-4(1)(1)))/(2(1))#

#color(white)(x) = (sqrt(3)+-sqrt(3-4))/2#

#color(white)(x) = sqrt(3)/2+-1/2i#

#color(white)(x) = cos(pi/6)+-isin(pi/6)#

#color(white)()#
If #x = cos(pi/6)+isin(pi/6)# then we find:

#x^9+x^2+2 = cos((3pi)/2)+isin((3pi)/2)+cos(pi/3)+isin(pi/3)+2#

#color(white)(x^9+x^2+2) = 0 -i+1/2+sqrt(3)/2i+2#

#color(white)(x^9+x^2+2) = 5/2+(sqrt(3)/2-1)i#

If #x = cos(pi/6)-isin(pi/6)# then we find:

#x^9+x^2+2 = cos((3pi)/2)-isin((3pi)/2)+cos(pi/3)-isin(pi/3)+2#

#color(white)(x^9+x^2+2) = 0 +i+1/2-sqrt(3)/2i+2#

#color(white)(x^9+x^2+2) = 5/2-(sqrt(3)/2-1)i#

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