polynomials Qs~~~~ (1 Viewer)

[Hikari Clover]

1. the roots of x^3+px^2+qx+r=0 form an AP.
prove that 2p^3+27r-9pq=0

2. 1)express (tan 4a) in terms of (tan a)
2) hence solve

plz help:wave:

 

[beentherdunthat]

lol, Neap question?

I saw a question like this today, but cbf'd to try it. Damns. COuld've hinted soemthing at least.

 

[HSCsimulator]

1.

let the roots be (a-1), a, (a+1)

sum of roots: -b/a

therefore, (a-1)+a+(a+1) = -p
3a=-p
a=-p/3

when a=-p/3,

(-p/3)^3 + p(-p/3)^2 + q (-p/3) + r = 0

algebraic manipulation of the above ^ gets 2p^3+27r-9pq=0

 

[HSCsimulator]

2. 1)

tan4a
= tan(2a+2a)
= (2tan2a) / (1-tan²2a) [using the double angle formula expansion]
= [2(2tana/1-tan²a)] / (1-tan²2a) [using the double angle formula expansion on the numerator)
= [2(2tana/1-tan²a)] / (1-{2tana/1-tan²a}²) [using the double angle formula expansion on the denominator]

and i think its algebraic manipulation from there

cant understand the 2nd q

 

[Hikari Clover]

回复: Re: polynomials Qs~~~~

1.

let the roots be (a-1), a, (a+1)

sum of roots: -b/a

therefore, (a-1)+a+(a+1) = -p
3a=-p
a=-p/3

when a=-p/3,

(-p/3)^3 + p(-p/3)^2 + q (-p/3) + r = 0

algebraic manipulation of the above ^ gets 2p^3+27r-9pq=0

how....how could u let the roots be a+1 and a-1 ?? at least use a+d and a-d,isn't it?

 

[Hikari Clover]

回复: Re: polynomials Qs~~~~

2. 1)

tan4a
= tan(2a+2a)
= (2tan2a) / (1-tan²2a) [using the double angle formula expansion]
= [2(2tana/1-tan²a)] / (1-tan²2a) [using the double angle formula expansion on the numerator)
= [2(2tana/1-tan²a)] / (1-{2tana/1-tan²a}²) [using the double angle formula expansion on the denominator]

and i think its algebraic manipulation from there

cant understand the 2nd q

dont no how to type square root in my computer

 

[Hikari Clover]

回复: Re: polynomials Qs~~~~

here is the 2(2)

 

[fishy89sg]

Re: 回复: Re: polynomials Qs~~~~

how....how could u let the roots be a+1 and a-1 ?? at least use a+d and a-d,isn't it?

it still comes to the same thing:O

cause when you do sum of the roots, the "d" disappears

 

[fishy89sg]

Re: 回复: Re: polynomials Qs~~~~

here is the 2(2)

i only think i can solve that if the root 3's werent there lol

 

[Hikari Clover]

回复: Re: 回复: Re: polynomials Qs~~~~

use this identity to solve 2(2)

what i knew is to let a=pi/24.........

 

[martinb]

Solution to question 2 attached

 

[ssglain]

Re: 回复: Re: polynomials Qs~~~~

how....how could u let the roots be a+1 and a-1 ?? at least use a+d and a-d,isn't it?

it still comes to the same thing:O
cause when you do sum of the roots, the "d" disappears

Be careful with that. It doesn't really matter for this question to let the roots be a - 1, a, a + 1 because of cancellations, but don't ever do this if the question asks to find the roots. You can never assume that the arithmetc mean is 1 out of your imagination.

Q2 might have been easier using complex numbers.