Surprisingly hard poly Q (2 Viewers)

[azureus88]

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

what would be a good way to approach this question? i tried using the relationship between coefficients but didnt seem to get the answer after several attempts. thanks in advance

 

[Trebla]

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

what would be a good way to approach this question? i tried using the relationship between coefficients but didnt seem to get the answer after several attempts. thanks in advance

You'll need a few tricks up your sleeve to solve this with minimum mess...

Let the roots be α, β, γ and δ
We're given that αβ = γδ

Sum of roots:
[1] α + β + γ + δ = 11/6

Sum of roots in triplets:
αβγ + αβδ + αγδ + βγδ = -22/6 = -11/3
αβ(γ + δ) + γδ(α + β) = -11/3
But αβ = γδ:
=> αβ(α + β + γ + δ) = -11/3
But sum of roots is α + β + γ + δ = 11/6
=> 11αβ/6 = -11/3
=> αβ = -2
=> γδ = -2

Sum of roots in pairs:
αβ + αγ + αδ + βγ + βδ + γδ = -26/6 = -13/3
-4 + αγ + αδ + βγ + βδ = -13/3
α(γ + δ) + β(γ + δ) = -1/3
[2] (α + β)(γ + δ) = -1/3

Let p = α + β and q = γ + δ, then equations [1] and [2] reduce to:
p + q = 11/6
pq = -1/3
Hence we know that p and q are the roots of the quadratic equation:
z² - 11z/6 - 1/3 = 0
=>6z² - 11z - 2 = 0
=> (6z + 1)(z - 2) = 0
z = 2 or - 1/6
So let p = 2 and q = - 1/6 (doesn't matter which way round you choose)
Hence:
α + β = 2
γ + δ = -1/6

Putting our useful equations together:
α + β = 2
αβ = -2
and
γ + δ = -1/6
γδ = -2

Let's resolve the α and β first:
We know that α and β are the roots of the quadratic
z² - 2z - 2 = 0
(z - 1)² = 3
z = 1±√3
Hence α = 1+√3 and β = 1 - √3

Now the γ and δ:
We know that γ and δ are the roots of the quadratic
z² + z/6 - 2 = 0
6z² + z - 12 = 0
(2z + 3)(3z - 4) = 0
z = -3/2 or 4/3
Hence γ = -3/2 and δ = 4/3

So our roots are: 1+√3, 1-√3, -3/2 and 4/3

 

[azureus88]

wow, nicely done. so how do you appraoch these questions? do u just try 2 figure it out along the way or do u come up with a general plan first.

 

[tommykins]

most of the time they're simpler than that. but if they give you a condition about the roots, use the relationship between the coefficients (sum/products etc.)

 

[gurmies]

oh god..was this from the fitzpatrick 3 unit book? I did the same question. Ultimate ball-kicker. Kudos Trebla

 

[azureus88]

i got another question 2 ask u guys. "show that if the polynomials x^3 +ax^2 - x+b and x^3 + bx^2 -x+a have a common factor of degree 2, then a+b=0

 

[gurmies]

Suppose:

x^3 + ax^2 -x + b = (x^2 + px + q)(x+d) [1]

x^3 + bx^2 -x + a = (x^2 + px + q)(x+e) [2]

Now, [1] - [2]

x^2 (a-b) + (b-a) = (x^2 + px + q)(d-e)

(a-b)(x^2-1) = (x^2 + px + q)(d-e)

= (d-e)x^2 + (d-e)px + (d-e)q

From this, we can deduce that: (a-b) = (d-e)

Now since the expression on the left contains no monic x, we can conclude that either d = e, or p = 0. If d = e, then the polynomials are the same, i.e. a+b cannot = 0. So for the sake of the question, we assume that p = 0.

So,

(a-b)(x^2-1) = (d-e)(x^2 + q)

Therefore, q = -1, to suit the LHS.

x^3 + ax^2 - x + b = (x^2-1)(x+d) = x^3 + dx^2 -x -d.

Therefore, a = d and b = -d

Adding the two expressions we get a+b = 0

Might I also ask where you got this question from? I haven't ever seen anything like it before

 

[azureus88]

first question is from fitz and second question from cambridge

 

[gurmies]

oh ok, thanks