Roots of polynomials

[Guest]

f(x) = kx^2 + 3x - 11
g(x) = mx^3 - 2x^2 + 3x - 9

Given that the sum of the roots of f is equal to the products of the roots of g and g has at least one root on the imaginary axis, solve f(x)=0 and g(x)=0 completely.

I got as far as k = -m/3
and if two of the roots of g are in the form p \pm qi and the remaining real root = \lambda, then (p^2+q^2) \lambda = 9/m
Also the roots of f are (-3 +/- \sqrt{9+44k}) / (2k)

[HallsofIvy]

Given that [tex]f(x)= kx^2- 3x- 11[/tex] it follows that the sum of the zeros of f is -3/k.

Given that [tex]g(x)= mx^3- 2x^2- 3x- 9[/tex] it follows that the product of the root g is -9/m.

Since the sum of the roots of f is the product of the roots of g then, yes, -3/k= -9/m so -3m= -9k and m= 3k.

Now, why do you say " if two of the roots of g are in the form p \pm qi"? The problem say that "g has at least one root on the imaginary axis", not just that two of roots are complex!

Are we to assume that m is real? If so then any non-real root of g has its conjugate as another root of g so if "g has at least one root on the imaginary axis", say, x= ai, then x= -ai is also a root and [tex]g(x)= m(x- ai)(x+ ai)(x- r)= m(x^2+ a^2)(x- r)= m(x^3- rx^2+ a^2x- a^2r)= mx^3- 2x^3- 3x- 9[/tex]. So we must have [tex]-mr= -2[/tex], [tex]ma^2= -3[/tex], and [tex]-ma^2r= -9[/tex], three equations to solve for m, a, and r.

[Guest]

Hi,

What exam paper is this from? Could you send a link to the paper/mark scheme?