1. ## roots help

The equation x^4 -5x^3 -9x^2 +ax +b =0 has a triple root. Find a and b and hence all roots of this equation.

2. ## Re: roots help

This is a fourth degree polynomial, so it has four roots. Three of those are identical, so there is only one other unique root.

Firstly, express the polynomial as a product of its factors.

(note that the polynomial is monic, so we may write $(x-\alpha)$ instead of $(kx-\alpha)$)

$x^4 -5x^3 -9x^2 +ax +b \equiv (x-\alpha)^3(x-\beta)$

Because these two expressions are identically equal, you may differentiate both sides.

3. ## Re: roots help

Originally Posted by fan96
This is a fourth degree polynomial, so it has four roots. Three of those are identical, so there is only one other unique root.

Firstly, express the polynomial as a product of its factors.

(note that the polynomial is monic, so we may write $(x-\alpha)$ instead of $(kx-\alpha)$)

$x^4 -5x^3 -9x^2 +ax +b \equiv (x-\alpha)^3(x-\beta)$

Because these two expressions are identically equal, you may differentiate both sides.
I differentiate both sides and apply P'(x)=P''(x)=0 ?

5. ## Re: roots help

Sorry for the messy handwriting (haven't written in months)

Basically you're right, proving that the P(x) = P'(x) = P''(x) = 0 and you'll end up usually with two roots. By solving for the right one and verifying it using the original equation, you are able to work out for the other variables and hence, solve the equation.

6. ## Re: roots help

Here's another way to do it.

$x^4 -5x^3 -9x^2 +ax +b \equiv (x-\alpha)^3(x-\beta)$

Taking the second derivative of both sides (this is easy for the RHS if you keep factoring out $(x-\alpha)$)

\begin{aligned}12x^2 - 30x - 18 & \equiv 4(x-\alpha)^2 + (4x-3\beta-\alpha)2(x-\alpha) \\ 6(2x^2-5x-3) & \equiv 6(x-\alpha)(2x-\alpha-\beta)\end{aligned}

Factorising the LHS,

$6(x-3)(2x+1) & \equiv 6(x-\alpha)(2x-\alpha-\beta)$

Equating both sides gives us $\alpha = 3, \, \beta = -4$.
__________________________________________________ __________

Substituting back into the original, we get

$\begin{cases}\phantom{-} 3a + b = \phantom{-}135 \\ -4a + b = -432\\ \end{cases}$

Solving, $a = 81, \, b = -108$
__________________________________________________ __________

Alternatively, because this is a monic polynomial, the constant is the product of all the roots, so $b = 3\times 3\times 3 \times -4 = -108$

We can then solve for $a$ simply by substituting $b$ into the original polynomial

\begin{aligned} 3a -243 &= 0 \\ a&= 81 \end{aligned}

7. ## Re: roots help

I'm pretty sure we can't find differentiate it 2 times since that's a 4U theorem, especially since this is a 3U question, markers may not acknowledge it. It may be faster but I wouldn't risk it. I'll try work out something that doesn't involve that

8. ## Re: roots help

FINALLY, TOOK ME FOREVER

$x^4 -5x^3 -9x^2 +ax +b$

Sum of roots:

$-\frac{b}{a}=5$

Sum of pairs:

$\frac{c}{a}=-9$

Since the polynomial is monic and we are given it has a triple root

$x^4 -5x^3 -9x^2 +ax +b = (x-\alpha)^3(x-\beta)$

Therefore we can make two new equations:

\begin{aligned}3\alpha+\beta=5 (sum of roots)\\\beta=5-3\alpha \\ 3\alpha^2+3\alpha\beta=-9 (sum of pairs) \\ \end{aligned}

Substitute and expand:

\begin{aligned}3\alpha^2+3\alpha(5-3\alpha)=9\\-6\alpha+15\alpha=-9\\-6\alpha^2+15\alpha+9=0\\2\alpha^2-5\alpha-3=0\end{aligned}

Factorising:

\begin{aligned}2\alpha^2-5\alpha-3=0\\(\alpha-3)(2\alpha+1)=0\\\alpha=3,-\frac{1}{2}\end{aligned}

Check:

\begin{aligned}\alpha=-\frac{1}{2} \\ 3(-\frac{1}{2})+\beta=5 \\ \beta=\frac{13}{2}\end{aligned}

Therefore, false

\begin{aligned}\alpha=3\\3(3)+\beta=5\\\beta=-4\end{aligned}

Therefore, $\alpha=3 and \beta=-4$

Sum of triplets:

$a=-[\alpha^3+3\alpha^2\beta]=-[(3)^3+3(3)^2(-4)]=81$

Product of roots:

$b=\alpha^3\beta=(3)^3(-4)=-108$

9. ## Re: roots help

Originally Posted by darkk_blu
I'm pretty sure we can't find differentiate it 2 times since that's a 4U theorem, especially since this is a 3U question, markers may not acknowledge it. It may be faster but I wouldn't risk it. I'll try work out something that doesn't involve that
You mean the multiple root theorem?

It's related to my method (from the observation that the triple root $(x-\alpha)$ remains a root of the second derivative), but the theorem itself isn't actually being used at all. It's only differentiation by the product rule and equating identities. Both of those are 2U methods, just applied at a higher level.

Also, as a side note I've heard that 4U content is accepted in 3U for full marks (and vice versa for 3U and 2U), but the only caveat is that an error means a zero, since the working out is not "valid". If you prove the content before using it I think it's okay though.

10. ## Re: roots help

Uhhh... guys, can you check this out?
http://m.wolframalpha.com/input/?i=e...%28x-13%2F2%29
http://m.wolframalpha.com/input/?i=e...5E3%28x%2B4%29
There are 2 solutions both with triple roots

11. ## Re: roots help

Originally Posted by darkk_blu
Uhhh... guys, can you check this out?
http://m.wolframalpha.com/input/?i=e...%28x-13%2F2%29
http://m.wolframalpha.com/input/?i=e...5E3%28x%2B4%29
There are 2 solutions both with triple roots
I made an error on this line.

$6(x-3)(2x+1) \equiv 6(x-\alpha)(2x-\alpha-\beta)$

Swapping around the factors on the RHS actually yields another solution:

$6(x-3)(2x+1) \equiv 6(2x-\alpha-\beta)(x-\alpha)$

$6(x-3)(2x+1) \equiv 6\left(x-\frac{\alpha}{2}-\frac{\beta}{2}\right)(2x-2\alpha)$

$-2\alpha = 1 \implies \alpha = -\frac{1}{2}$

$\frac{\alpha}{2}+\frac{\beta}{2} = 3 \implies \beta=\frac{13}{2}$

Since the question didn't say that the coefficients had to be integers, this is also a valid solution. Very interesting... I didn't consider the possibility that two sets of solutions existed.

12. ## Re: roots help

Originally Posted by fan96
You mean the multiple root theorem?

It's related to my method (from the observation that the triple root $(x-\alpha)$ remains a root of the second derivative), but the theorem itself isn't actually being used at all. It's only differentiation by the product rule and equating identities. Both of those are 2U methods, just applied at a higher level.

Also, as a side note I've heard that 4U content is accepted in 3U for full marks (and vice versa for 3U and 2U), but the only caveat is that an error means a zero, since the working out is not "valid". If you prove the content before using it I think it's okay though.
Yep, it can be used, very useful for integration and motion. But 4U methods are used within reason. For example, don't be using l'hopital for limits, only to check your answer.

13. ## Re: roots help

Originally Posted by fan96
I made an error on this line.

$6(x-3)(2x+1) \equiv 6(x-\alpha)(2x-\alpha-\beta)$

Swapping around the factors on the RHS actually yields another solution:

$6(x-3)(2x+1) \equiv 6(2x-\alpha-\beta)(x-\alpha)$

$6(x-3)(2x+1) \equiv 6\left(x-\frac{\alpha}{2}-\frac{\beta}{2}\right)(2x-2\alpha)$

$-2\alpha = 1 \implies \alpha = -\frac{1}{2}$

$\frac{\alpha}{2}+\frac{\beta}{2} = 3 \implies \beta=\frac{13}{2}$

Since the question didn't say that the coefficients had to be integers, this is also a valid solution. Very interesting... I didn't consider the possibility that two sets of solutions existed.
Yeah man, I thought there was only one solution because the question subtly implied it, I got two solutions then looked at everyone else’s and was like, yeah, I’m probably being dumb so I also took one solution, I happened to be rethinking about the question after school so I brought it up again. lol sorry

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