polynomials q (1 Viewer)

[Jaydels]

This q is the last one in exercise 4.3 from cambridge, if someone could help me with it i would be very grateful.......

18. The equation x^3+qx+r=o has roots a,b and c. Find expressions for
a) a^2 + b^2 + c^2
b) a^3 + b^3 + c^3
c) a^5 + b^5 + c^5

The only one I'm stuck on is c). It looks like maybe you're supposed to use the results from the previous two to get an expression for it, but it turns out messy and is much too complicated for me

 

[JamiL]

times everything by x^2
so when u did
a^3=r-aq
there for
a^5=(r-aq)^2
do that to all n then add, n bob is ur uncle

 

[Jaydels]

so just squaring the expression for the sum of the roots cubed works? I wasn't sure if it would

 

[turtle_2468]

it doesn't work - he's made at least 2 mistakes... (it would be -r-aq for instance)
you know the complicated thing you got by multiplying the results of the first two parts? Let the answer to c be z, and the answer to a), b) be x, y respectively. Then z=xy-(a^3b^2+a^3c^2+b^3a^2+b^3c^2+c^3a^2+c^3b^2). This looks really complicated, but notice that from the original equation we know that a+b+c=0. Therefore if we take a^3b^2+b^3a^2 from the messy bit above it simplifies to: a^2b^2(a+b)=a^2b^2(-c)=-a^2b^2c. Do this to the other two "pairs" as well and you end up with -(a^2b^2c+a^2bc^2+ab^2c^2) in the brackets, which simplifies to:
-abc(ab+ac+bc)=rq.
Hope that made sense..

 

[Slidey]

This q is the last one in exercise 4.3 from cambridge, if someone could help me with it i would be very grateful.......

18. The equation x^3+qx+r=o has roots a,b and c. Find expressions for
a) a^2 + b^2 + c^2
b) a^3 + b^3 + c^3
c) a^5 + b^5 + c^5

The only one I'm stuck on is c). It looks like maybe you're supposed to use the results from the previous two to get an expression for it, but it turns out messy and is much too complicated for me

My method:

x^3=-qx-r
x^5=-qx^3-rx^2
Sub in x= a, b, c and add each result:
a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)

But first:
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=-2q
and for a^3+b^3+c^3:
x^3=-qx-r
a^3+b^3+c^3=-q(a+b+c)-r=-r

Therefore,
a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)
a^5+b^5+c^5=-q.-r - r.-2q=rq

Answer:
a^5+b^5+c^5 = rq #

More systematic (can be used to evaluate a^n+b^n+c^n... if you have enough patience) than Turtle's method.

 

[turtle_2468]

Yeah... he's right

 

[DistantCube]

My method:

x^3=-qx-r
x^5=-qx^3-rx^2
Sub in x= a, b, c and add each result:
a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)

But first:
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=-2q
and for a^3+b^3+c^3:
x^3=-qx-r
a^3+b^3+c^3=-q(a+b+c)-r=-r

Therefore,
a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)
a^5+b^5+c^5=-q.-r - r.-2q=rq

Answer:
a^5+b^5+c^5 = rq #

More systematic (can be used to evaluate a^n+b^n+c^n... if you have enough patience) than Turtle's method.

That's pretty, I like it.

 

[Jaydels]

thanks peoples

 

[Trev]

That's pretty, I like it.

Haha, you sound like my maths teacher. Referring to equation after equation as 'pretty'... sometimes going as far as (he was referring to a complex number graph question with circles in it) as 'beautiful', he's a champ teacher lol!