How to do this (2 Viewers)

[coca cola]

@, $, % are the roots of the equation x^3 - bx^2 + cx - d = 0, find an expression in terms of b, c, d for:

(@$)^3 + (@%)^3 + (%$)^3

I don't understand Geha's solution in his 50 tips.

He's solutions goes like this:

i.e. (@$)^3 = b(@$)^2 - c(@$) + d... etc the usual steps after that.

But how could you substitute a product of 2 roots into the equation? How could you assume the products of the 2 roots is also a root to the original equation?

Its on Pg 112 if you have the book.

 

[coca cola]

And this, how does:

-cos(5pi/7) + cos(4pi/7) + cos(6pi/7) = -sin(2pi/7) + sin(3pi/7) + sin(pi/7)

Pg 152 of Phoenix By Topic, solution to Polynomial number 4, iv).

 

[ngai]

cos x = sin(pi-x)

therefore cos 0 = sin pi
and hence 1 = 0
yay

 

[Estel]

On inspection, I think I may have made a mistake

 

[ben_ratus]

-cos(5pi/7) + cos(4pi/7) + cos(6pi/7) = -sin(2pi/7) + sin(3pi/7) + sin(pi/7)

Sin(x) = Cos (pi/2 - x)
he just did that

 

[mojako]

>> But how could you substitute a product of 2 roots into the equation? <<
No you could not.

 

[coca cola]

umm so there is something wrong with these solutions and its not just me, but how odd, how could Geha do a question completely wrong, its not like its a simple typo.

 

[mojako]

have you written down the complete Q (on BoS)?
including all the lead-ups and part you don't think are really relevant?
there might be some conditions for the roots which are given on previous parts...

 

[coca cola]

Sin(x) = Cos (pi/2 - x)
he just did that

I don't think so, because if you did that you'll get a different solution. I thought it couldn't be wrong because the question states to show the result and the solution showed that result, but in a odd way.

 

[coca cola]

have you written down the complete Q (on BoS)?
including all the lead-ups and part you don't think are really relevant?
there might be some conditions for the roots which are given on previous parts...

Oh ok, the entire question is:

Tip 27 Example 2

@, $, % are the roots of the equation x^3 - bx^2 + cx - d = 0, find an expression in terms of b, c, d for:

i) @^2 + $^2 + %^2

ii) @^3 + $^3 + %^3

iii) (1 + @^3)(1 + $^3)(1 + %^3)

its just part iii), when you expand it you'll get something with this (@$)^3 + (@%)^3 + (%$)^3. And Geha states that the best way to solve this is to substitute the product of the 2 roots back into the equation, like what you do with if its a single root. Then he proceeds by doing it that way.

 

[mojako]

And this, how does:

-cos(5pi/7) + cos(4pi/7) + cos(6pi/7) = -sin(2pi/7) + sin(3pi/7) + sin(pi/7)

Pg 152 of Phoenix By Topic, solution to Polynomial number 4, iv).

by calculator, this isn't correct.

 

[mojako]

I personally don't think you can sub @$ etc.

Do it like ngai suggested somewhere down this thread.

The approaches I was thinking of are
1.
Find an equation whose roots are @$, @%, $%, call them y_1, y_2, y_3.
then from that equation you can find y_1^3 + y_2^3 + y_3^3

How do we find that equation?
Say we need to find an eqn with roots @^2 etc, and want to write that equation in y just to make things different.
we write "a (root y)^3 - b (root y)^2 + c (root y) - d = 0"
Formally we find root y by letting y = x^2 --> x = root y (or minus root y... you can also use that).

Now, to find an equation with roots @$, @%, $%, we need to find a common expression for them. @$ is "@$% / %", @% is "@$% / $" etc.
so each of them is "@$% divided by @, $, or %"
but we know @$% = d (product of roots of the original eqn)
so our new roots is d / @, d / $, d / %.
so we let y = d / x --> x = d / y
and continue from there.

(This is similar to ngai's but his looks simpler)

2.
Alternatively, find an eqn whose roots are 1 + @^3 etc.
If you can transform the new eqn into a cubic though (I don't think you can).

 

[Logix]

i think putting it as @$%/% would work.
However, for the 2nd method that u stated, u need to get it back into the form ax^3+bx^2+cx+d, dont u? i dont think all subsitutions of x can get back into that form, especially if u sub x=cube root of (x-1)

 

[coca cola]

If you do this, can you tell me if the result turns out to be the same as Geha's solution?

In the book the solution is: (@$)^3 + (@%)^3 + (%$)^3 = b(c^2) -2(b^2)d - c^2 + d

Sorry but I don't quite understand your method.

 

[CrashOveride]

Is Geha's book anygood? Is it just worked solutions from past (how many years??) HSC papers grouped into topic by topic ? Inform me some1 thnx

 

[mojako]

i think putting it as @$%/% would work.
However, for the 2nd method that u stated, u need to get it back into the form ax^3+bx^2+cx+d, dont u? i dont think all subsitutions of x can get back into that form, especially if u sub x=cube root of (x-1)

yea.. I dont think it works, hehe...
(cubic root)^3 + (cubic root)^2 + (cubic root) + constant cant easily be made to look like a cubic.
but something like (cubic root)^3 + (cubic root) + constant can.

 

[ngai]

how could Geha do a question completely wrong, its not like its a simple typo.

nobody's perfect

if u want to see if his solution is wrong, then just create a cubic by expanding (x-1)(x-2)(x-3), which has roots 1,2,3, and see if u get contradiction

 

[mojako]

nobody's perfect

if u want to see if his solution is wrong, then just create a cubic by expanding (x-1)(x-2)(x-3), which has roots 1,2,3, and see if u get contradiction

ok.. he's wrong.. but how do you do part (iii) in post #10?
can we do that kind of question?

 

[Jase]

I don't see why you cant do it like the way its layed out.

if A is a root and B is a root. Then AB is also a solution. as in, AB divides into P(x). Hence any product of two roots works with P(x). So just do that,,

or you could expand it all and transmogrify everything until you end up with the correct forms. If you really want.

 

[ngai]

@, $, % are the roots of the equation x^3 - bx^2 + cx - d = 0
iii) (1 + @^3)(1 + $^3)(1 + %^3)

expand, and i like to use a,b,c as the roots
and i'll rename the equation to be x^3 - px^2 + qx - r = 0
1 + a^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ca)^3 + (abc)^3
and (ab)^3 = (abc)^3/c^3, (bc)^3 = ... etc
now, u know 1
u know a^3 + b^3 + c^3
u know (abc)^3
so all u needa know is (1/a^3) + (1/b^3) + (1/c^3)
x^3 - px^2 + qx - r = 0 has roots a,b,c
so rx^3 - qx^2 + px - 1 = 0 has roots 1/a, 1/b, 1/c (let y=1/x, so x = 1/y, etc etc)
and then u can get (1/a^3) + (1/b^3) + (1/c^3) using the same way u did part i and ii for the other eqn