Polynomial question- please help (1 Viewer)

[NizDiz]

The equation x^3 + px + q = 0 has roots a, b and gamma ( you can just let is be c).

Find the cubic equation with roots 1/a + 1/b, 1/b + 1/c, 1/a + 1/c

 

[HeroicPandas]

The equation x^3 + px + q = 0 has roots a, b and gamma ( you can just let is be c).

Find the cubic equation with roots 1/a + 1/b, 1/b + 1/c, 1/a + 1/c

impossibru!

with roots only 1/a + 1/b?? 1/b+1/c etc??

are u asking for equation with roots: 1/a, 1/b, 1/c?

 

[braintic]

The equation x^3 + px + q = 0 has roots a, b and gamma ( you can just let is be c).

Find the cubic equation with roots 1/a + 1/b, 1/b + 1/c, 1/a + 1/c

See attachment: View attachment Misc 1.pdf

 

[Makematics]

Heroic, why u say it cant be done? I think this is right, just rearrange the last part to get the answer. I honestly cbf doing that haha...

 

[Makematics]

See attachment: View attachment 28309

but that isnt what the question asked for?

 

[asianese]

I got the final answer to be:

(from makematics approach - which was my initial approach as well)

(probs wrong)

 

[asianese]

Bolded of the question NizDiz posted - look its impossibru, unless it was a massive typo or SUPER-CONFUSION

The question as i predicted with my first post wants to find an equation with roots as reciprocals (1/a, 1/b, 1/c)

An alternative to ur method Makematics:

Normal equation has roots: a, b, c
For new equation with roots: 1/a, 1/b, 1/c

Let y = 1/x

So, x = 1/y
Sub this into original equation:

(1/y)^3 + p(1/y) + q = 0

1 + py^2 + qy^3 = 0

So new equation with roots: 1/a, 1/b, 1/c is:





ANOTHER Interesting thought: reciprocal polynomials http://en.wikipedia.org/wiki/Reciprocal_polynomial

this is how u check if ur answer is correct or incorrect


Summary:
i dont know what this method is called but it is MARVELOUS

Revise:


P(x) has roots a, b, c
Find the equation with roots: a^2, b^2, c^2

It's not impossibru. Why are you rejecting this answer? Makematics has a logical sequence of steps. Perhaps you have no been exposed to such examples, which is a pity. This is just another variation on roots and polynomials.

 

[asianese]

Bolded of the question NizDiz posted - look its impossibru, unless it was a massive typo or SUPER-CONFUSION

The question as i predicted with my first post wants to find an equation with roots as reciprocals (1/a, 1/b, 1/c)

An alternative to ur method Makematics:

It's not 'impossibru'. I don't think its a typo.

Your alternative is for a different question, not the one he is looking for.

 

[asianese]

ANOTHER Interesting thought: reciprocal polynomials or something (from the internet

if u want polynomial with roots reciprocated, u just swap around the coefficients of the polynomial (a good example is NizDiz's question)

this is how u check if ur answer is correct or incorrect

That is ONLY for the rare case that the roots in question are in fact For all other cases this generality doesn't apply.

 

[HeroicPandas]

It's not 'impossibru'. I don't think its a typo.

Your alternative is for a different question, not the one he is looking for.

yes.....i see

 

[Makematics]

wait heroic how is it that you havent come across this method of finding equations with related roots? it's a pretty standard procedure in 4U

 

[Makematics]

I got the final answer to be:

(from makematics approach - which was my initial approach as well)

(probs wrong)

yup all good i finished off mine and got the same as you.

 

[Joelolol]

yup all good i finished off mine and got the same as you.

Yup same here

 

[HeroicPandas]

wait heroic how is it that you havent come across this method of finding equations with related roots? it's a pretty standard procedure in 4U

who said i never came across that method?

i misread the question and thought it was a typo

 

[Makematics]

who said i never came across that method?

i misread the question and thought it was a typo

oh ok my bad just a misunderstanding!

 

[HeroicPandas]

oh ok my bad just a misunderstanding!

lol its ok! misreading questions is fun...