Double Root of a Polynomial

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May 23rd 2010, 04:43 AM
Lukybear

Double Root of a Polynomial

If x^3 + 3px^2 + 3qx + r = 0 has a double root, show that the double root must be (pq-r)/(2q-2p^2)
May 23rd 2010, 05:02 AM
Dinkydoe

It's not hard to see that it's a root of $f(x)=0$ in the first place.

Now you're left with showing that it's also a root of $f'(x)=3x^2+6px+3q=0$

(wich is using the abc-formula or plugging in $x=\frac{pq-r}{2q-2p^2}$ in f'(x))
May 24th 2010, 01:02 AM
Lukybear

Sorry, im not following. Could you please expand on that?
May 24th 2010, 01:30 AM
mr fantastic

Quote:

Originally Posted by Lukybear

If x^3 + 3px^2 + 3qx + r = 0 has a double root, show that the double root must be (pq-r)/(2q-2p^2)

Since you didn't post this in the calculus subforum I assume that a calculus approach cannot be used.

Note that your cubic must have the form $(x - a)^2 (x - b)$ where $x = a$ is the repeated root and $x = b$ is the third root. So you can expand this expression and equate the coefficients of the powers of x to the given cubic. This will give 3 equations:

$3p = -b - 2a$

$3q = 2ab + a^2$

$r = -a^2 b$

and your job is to solve for a in terms of p, q and r.
May 24th 2010, 02:35 AM
Lukybear

Quote:

Originally Posted by mr fantastic

Since you didn't post this in the calculus subforum I assume that a calculus approach cannot be used.

Note that your cubic must have the form $(x - a)^2 (x - b)$ where $x = a$ is the repeated root and $x = b$ is the third root. So you can expand this expression and equate the coefficients of the powers of x to the given cubic. This will give 3 equations:

$3p = -b - 2a$

$3q = 2ab + a^2$

$r = -a^2 b$

and your job is to solve for a in terms of p, q and r.

Yep, Ive gotten to that part.

But in simultaneous equations, I cannot isolate the a. If you could give me guidance that would be great.

Also, is it too late to migrate this post to the calculus forum? I would like to see a calculus approach if possible.
May 24th 2010, 03:11 AM
Dinkydoe

To be honoust, I don't think this approach will make things easier.

It's a pretty tough system to crack, especially considering we must find the expression $a= \frac{pq-r}{2q-2p^2}$

Another way ofcourse would be to fill in the the given expression for " $a$ ", show that it satisfies:

$3p=-b-2a$
$3q=2ab+a^2$
$r=-a^2b$

Instead of trying to solve the value $a$ yourself.

To get back to my post. A root $x=a$ of an equation $f(x)=0$ is a double root when it's also a root of $f'(x)=0$ . This fact can be shown quite easily, but I'm not sure if you may use this. It would make your work a lot easier.
May 24th 2010, 05:55 AM
Danny

Try this. First we'll number our equations

$ 1)\;\; 2a + b + 3p = 0 $
$ 2)\;\; a^2 + 2ab - 3 q = 0 $
$ 3)\;\;a^2b - r = 0 $

First eliminate $b$ from $(1)$ and $(2)$ giving

$3a^2 + 6ap + 3q = 0,$

then from $(1)$ and $(3)$ giving

$ 2a^3 + 3a^2p + r = 0, $

then from $(2)$ and $(3)$ giving

$ a^3 - 3aq + 2r = 0. $

Now can you manipulate these three new equations as to get $a$ all by itself? ( ${\it i.e.}$ no $a^2$ or $a^3$ )
May 24th 2010, 08:24 AM
Archie Meade

The 3 resulting equations should be

$3a^2+6ap+3q=0$

$2a^3+3a^2p+r=0$

$a^3-3aq-2r=0$
May 24th 2010, 09:39 PM
Lukybear

Quote:

Originally Posted by Dinkydoe

To be honoust, I don't think this approach will make things easier.

It's a pretty tough system to crack, especially considering we must find the expression $a= \frac{pq-r}{2q-2p^2}$

Another way ofcourse would be to fill in the the given expression for " $a$ ", show that it satisfies:

$3p=-b-2a$
$3q=2ab+a^2$
$r=-a^2b$

Instead of trying to solve the value $a$ yourself.

To get back to my post. A root $x=a$ of an equation $f(x)=0$ is a double root when it's also a root of $f'(x)=0$ . This fact can be shown quite easily, but I'm not sure if you may use this. It would make your work a lot easier.

Thank you guys very much. Ive finally got the answer.

But I am still very interested in the calculus approach. Could you please expand on that? How would you do that.
May 25th 2010, 02:16 AM
Archie Meade

Quote:

Originally Posted by Lukybear

Thank you guys very much. Ive finally got the answer.

But I am still very interested in the calculus approach. Could you please expand on that? How would you do that.

At a double root, the curve has either a local maximum or local minimum.
We don't know which, but we know that the derivative is zero there.
The derivative is not zero at the other root.

Hence we solve for

$x^3+3px^2+3qx+r=0$

$3x^2+6px+3q=0$

re-arranging the 2nd equation for x^2, we get

$x^2=-\frac{6px+3q}{3}=-2px-q$

Substituting back into f(x)=0

$x(-2px-q)+3p(-2px-q)+3qx+r=0$

$-2px^2-qx-6p^2x-3pq+3qx+r=0$

Again using the value for $x^2$ , we get

$2p(2px+q)-qx-6p^2x-3pq+3qx+r=0$

$4p^2x+2pq-qx-6p^2x-3pq+3qx+r=0$

$x\left(4p^2-q-6p^2+3q\right)+2pq-3pq+r=0$

$x\left(-2p^2+2q\right)-pq+r=0$

$x=\frac{pq-r}{2q-2p^2}$
May 25th 2010, 02:52 AM
Dinkydoe

I see Archie Meade allready outlined a way to do it:

The reason why this works:

If $a$ is a double root of $f(x)=0$ , then it's also a root of $f'(x)=0$

Proof:
With the chain rule: if $f(x)=(x-a)^2(x-b)$ then $f'(x)=(x-a)^2+2(x-a)(x-b)$ . Hence $f'(a)=0$ .
May 25th 2010, 04:18 AM
Lukybear

Quote:

Originally Posted by Archie Meade

At a double root, the curve has either a local maximum or local minimum.
We don't know which, but we know that the derivative is zero there.
The derivative is not zero at the other root.

Hence we solve for

$x^3+3px^2+3qx+r=0$

$3x^2+6px+3q=0$

re-arranging the 2nd equation for x^2, we get

$x^2=-\frac{6px+3q}{3}=-2px-q$

Substituting back into f(x)=0

$x(-2px-q)+3p(-2px-q)+3qx+r=0$

$-2px^2-qx-6p^2x-3pq+3qx+r=0$

Again using the value for $x^2$ , we get

$2p(2px+q)-qx-6p^2x-3pq+3qx+r=0$

$4p^2x+2pq-qx-6p^2x-3pq+3qx+r=0$

$x\left(4p^2-q-6p^2+3q\right)+2pq-3pq+r=0$

$x\left(-2p^2+2q\right)-pq+r=0$

$x=\frac{pq-r}{2q-2p^2}$

Thank you very much.

I did realize before all that you had put forward. But just didnt make the connection.

Can i just ask how is the equation able to accommodate for the sub of x^2 twice? I've never seen this type of solving method before.

Wouldn't there be like a "redudancy" in the equation?
May 25th 2010, 07:15 AM
Danny

Quote:

Originally Posted by Danny

Try this. First we'll number our equations

$ 1)\;\; 2a + b + 3p = 0 $
$ 2)\;\; a^2 + 2ab - 3 q = 0 $
$ 3)\;\;a^2b {\color{red}{ \;+\;}} r= 0 $

First eliminate $b$ from $(1)$ and $(2)$ giving

$3a^2 + 6ap + 3q = 0,$

then from $(1)$ and $(3)$ giving

$ 2a^3 + 3a^2p {\color{red}{ \;-\;}} r = 0, $

then from $(2)$ and $(3)$ giving

$ a^3 - 3aq {\color{red}{ \;-\;}} 2r = 0. $

Now can you manipulate these three new equations as to get $a$ all by itself? ( ${\it i.e.}$ no $a^2$ or $a^3$ )

Yes, I there is a typo (in red above) as pointed out by Archie.
May 25th 2010, 10:10 AM
Archie Meade

Quote:

Originally Posted by Lukybear

Thank you very much.

I did realize before all that you had put forward. But just didnt make the connection.

Can i just ask how is the equation able to accommodate for the sub of x^2 twice? I've never seen this type of solving method before.

Wouldn't there be like a "redudancy" in the equation?

By virtue of the fact that we are looking for the double root,
we can utilise the derivative which is a quadratic.

The second root does not satisfy .... derivative = 0
the double root and one other value of x satisfies this (if the cubic has both a local max and local min as in this case if f(x) has no triple root).
The double root is one of the two values of x for which the derivative is zero.

However, only the double root satisfies both f(x)=0 and f'(x)=0.

Notice the way Danny solved the equations....
We are looking for the double root "a".
The 3 equations all contain the 2nd root "b", which is a fly in the ointment,
hence he proceeded to eliminate "b" as we had a system of simultaneous equations.
He also eliminated $a^3$ and $a^2$ to be left with $a$ , the double root.

Using the derivative, we proceed to eliminate $x^3$ and $x^2$ to be left with x,
because this x is the double root, corresponding to f(x)=0 and f'(x)=0.

$3x^2+6px+3q=0$ for the double root

hence we now have a linear expression for $x^2$ at the double root .....you could use $u=x^2$

$3x^2=-6px-3q\ \Rightarrow\ x^2=-2px-q$

Therefore we can rewrite f(x)=0 for the double root

$f(x)=(x)x^2+(3p)x^2+3qx+r=0$

becomes

$f(x)=x(-2px-q)+3p(-2px-q)+3qx+r=0$

We want x and f(x) still contains $x^2$ but $x^2=-2px-q$ since the value of x is the x co-ordinate of the double root.

hence

$f(x)=(-2p)x^2-qx-6p^2x-3pq+3qx+r=0$

is $f(x)=(-2p)(-2px-q)-qx-6p^2x-3pq+3qx+r=0$

for the x-value of the double root.
We just replace all $x^2$ terms at any stage by the value of $x^2$ given by the derivative, until the only x terms left are multiples of x.

I was thinking "graph" as the double root is on the x-axis and the x-axis is the tangent to the curve there.

Dinkydoe showed nicely how to work with it without that necessity.