SOLUTION: One root of the equation x2+px+q=0 is the cube of the reciprocal of the other. Show that q4+q(2q-p2)+1=0

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Question 696766: One root of the equation x2+px+q=0 is the cube of the reciprocal of the other. Show that q4+q(2q-p2)+1=0
Answer by MathGeek87(9) About Me  (Show Source):
You can put this solution on YOUR website!
x1=(-p-sqrt(p^2-4*q))/2
x2=(-p+sqrt(p^2-4*q))/2
solve
q^4+q*(2*q-p^2)+1=0
p=(q^2 + 1)/q^(1/2)
OR
p= -(q^2 + 1)/q^(1/2)
sub p=(q^2 + 1)/q^(1/2) in x1 and x2 you get:
x1 =

- ((q^2 + 1)^2/q - 4*q)^(1/2)/2 - (q^2 + 1)/(2*q^(1/2))


x2 =

((q^2 + 1)^2/q - 4*q)^(1/2)/2 - (q^2 + 1)/(2*q^(1/2))
sub x1 in
x^2+p*x+q
it is =0
sub x2 in
x^2+p*x+q
it is =0
then q^4+q*(2*q-p^2)+1=0
you get the same result when use:
p= -(q^2 + 1)/q^(1/2)