polynomials question (1 Viewer)

c0okies

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If alpha and beta are roots of the equation x^2 +mx +n=0, find the roots of nx^2+(2n-m^2)x+n=0 in terms of alpha and beta.

[darn this is probably real simple but i havent got the hang of how to do these ones]

ok here are a few more that i ended up not being able to do...

18. If the roots of the equation x^3+px^2+qx+r=0 form consecutive terms of a gemetric sequence, prove that q^3=p^3r.
Show that this condition is satisfied for the equation 8x^3-100x^2+250x-125=0 and solve this equation.

22.If alpha, beta, delta are the roots of 3x^3+8x^2-1=0 find the value of

(beta +1/delta)(delta +1/alpha)(alpha +1/beta)

i did this question but ive continuously gotten 2/3.. can someone tell me what ive done wrong? because the answer was -6.
 
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c0okies

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If alpha and beta are roots of the equation x^2 +mx +n=0, find the roots of nx^2+(2n-m^2)x+n=0 in terms of alpha and beta.

[darn this is probably real simple but i havent got the hang of how to do these ones]

thanx
 

c0okies

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If alpha and beta are roots of the equation x^2 +mx +n=0, find the roots of nx^2+(2n-m^2)x+n=0 in terms of alpha and beta.

[darn this is probably real simple but i havent got the hang of how to do these ones]

thanx ahead for any
 

Trev

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x²+mx+n=0; let roots be a, b.
a+b=-m
ab=n

nx²+(2n-m²)x+n=0;
(using quadratic formula)
x = {m²+2n +/- √[(2n-m²)²-4n²]}/2n
= {(a+b)²+2ab +/- m√[(a+b)²-4ab]}/2ab.
take out the other m (yeah, i'm lazy to do it there) and then expand and simplify the rest.

Now you can delete the other threads.
 

c0okies

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can someone do this question please? ive gone a bit mental over it because i spent ages on it ><

Solve the question 6x^4-11x^3-26x^2+22x+24=0 given that the product of two of the roots is equal to the product of the other two.

[ you're not permitted to solve it by dividing the roots or any guess-and-check way; you have to solve it algebraically - which is a bugger]

thanx ahead for any help ;]
 
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Dreamerish*~

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6x4 - 11x3 - 26x2 + 22x + 24 = 0

Let the roots be α, β, γ and δ. From the question, αβ = γδ.


α + β + γ + δ = -b/a = 11/6 ... (1)

αβ + αγ + αδ + βγ + βδ + γδ = c/1 = -26/6 = -13/3 ... (2)

αβγ + βγδ + γδα + δαβ = -d/a = -22/6 = -11/3 ... (3)

αβγδ = e/a = 24/6 = 4 ... (4)


From (4), we can see that αβγδ = 4.

Since αβ = γδ, we can deduce that (αβ)2 = 4, and (γδ)2 = 4.

Therefore αβ = +2, and γδ = +2.

You can take these values (use plus or minus 2, trial and error) and put them back into (2) and (3).

That's a start. See how you go. :)
 

c0okies

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polynomials question from fitzpatrick

hi ive done this question several times and for some reason i cant get to the answer, and when i get a quadratic for alpha, the discriminant is a negative..

can someone please help me?

18. Solve the equation 6x^4-29x^3+40x^2-7x-12=0, the product of two of the roots being two.
 

Slidey

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Re: polynomials question from fitzpatrick

Uhg an algebra bash. First inspect the function: (2x-3) is a moderately easy factor to find (because P(3/2)=0). With this, test f(4/3) to ensure that 3/2 isn't the root which multiplies to give 2 (it isn't). Now, you've reduced it down to: p(x)=3x^3-10x^2+5x+4.

From here, it is slightly easier:
a+b+c=10/3
ab+ac+bc=5/3
abc=-4/3

Let ab=2,
abc=2c=-2/3, therefore: c=-1/3. Which isn't true, hence the question is incorrect and no two roots multiply to give 2.

Anticlimactic, huh?
 
P

pLuvia

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Re: polynomials question from fitzpatrick

22.If alpha, beta, delta are the roots of 3x3+8x2-1=0 find the value of

(β +1/γ)(γ +1/α)(α +1/β)

Let the roots be α, β, γ

sum of roots = α + β + γ = -8/3
sum of paired roots = αβ + αγ + βγ = 0
product of roots = αβγ = 1/3

Expanding this (β +1/γ)(γ +1/α)(α +1/β) we get

(αβγ) + (α+β+γ) + [(αβ + αγ + βγ)/αβγ] + (1/αβγ)

sub in the roots

(1/3) + (-8/3) + 0 + 3= 2/3

Hope that helped :D

Sometimes the answers in Fitzpatrick are wrong ;)
 
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