# gce further mathematics

posted by
**lee** on
.

The roots of the equation x^4 - 3x^2 + 5x - 2 = 0 are a, b, c, d. By the relation y=x^2, or otherwise, show that a^2, b^2, c^2, d^2 are the roots of the equation y^4 - 6y^3 + 5y^2 - 13y + 4 = 0.

Let y = x^2 and substitute that into the x polynomial.

y^2 - 3y + 5 sqrt y -2 = 0

y^2 -3y -2 = - 5 sqrt y

Square both sides

y^4 - 6y^3 + 9 y^2 -4y^2 +12y + 4 = 25 y

y^4 - 6y^3 +5 y^2 -13y +4 = 0

Values of a, b, c and d that were roots of the original (x) equation must have their squares also be solutions of the y equation.

If a r is a root of a polynomial p(x) then (x-r)|p(x)

Thus the first equation is

(1) p(x)=(x-a)(x-b)(x-c)(x-d)=x^4-3x^2+5x-2=0

Suppose you now multiply p(x)(x+a)(x+b)(x+c)(x+d). You would get

(2) (x^2-a^2)(x^2-b^2)(x^2-c^2)(x^2-d^2)=

(y-a^2)(y-b^2)(y-c^2)(y-d^2)=

y^4-6y^3+5y^2-13y+4=0

= your second equation.

What this means is that if you were to use the relation y=x^2 in the second one and divide it by the first you shoud get

(x+a)(x+b)(x+c)(x+d)

Try that and see what you get. If it is the case, then the conclusion follows.

I just realized you don't know the roots. When you do the steps in the previous post you should get

(x+a)(x+b)(x+c)(x+d)=x^4-3x^2-5x-2

If you get this then the conclusion follows. Multiply these terms to see how the coefficients are related to (1) in the previous post.

the 2nd part of the question asks : find the value of s~2 where s~n = a^n + b^n + c^n + d^n. and hence show that s~8 = 6(s~6) - 5(s~4) + 62 . how do i do this?

That's just substituting n=2 to get

s-2=a^2+b^2+c^2+d^2

In the original equation we have

x^4+px^3+qx^2+rx+s p=0,q=3,r=5,s=-2

(-p)^2-2q=a^2+b^2+c^2+d^2=2*-3=-6

I think the second part is done recursively but I don't see how just at the moment. It might also make use of the result from the first part. I'll check it in just a bit.

What course are you taking where these question arise? It looks like modern algebra to me. Just wonderin'

I'm aware of using a formula of Newton to find the sum of n-th powers so I think that's what may be expected here.

I'm interested to know what you're studying so I know what can be taken as known. If you are studying theory of equations, roots of polys, symmetric functions, etc, then there should be something on Newton's recursive formula.

BTW, what is s supposed to be? we have s-n=sum of n-th powers. What is s here?

I do not know what s stands for.I'm studying gce a level further mathematics

Okay, do you see the thread with Jack above? We're discussing the same problem. I think I'm having some trouble with the notation right now.

I think s-n is just the sum of the n-th powers of the roots of the poly., but I'm not sure how the result follows even if that's the case.

yes, i'm trying to follow the problem with Jack now

lee, i know you!

you know my bro jackson?!