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

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- May 23rd 2010, 05:43 AMLukybearDouble 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, 06:02 AMDinkydoe
It's not hard to see that it's a root of in the first place.

Now you're left with showing that it's also a root of

(wich is using the abc-formula or plugging in in f'(x)) - May 24th 2010, 02:02 AMLukybear
Sorry, im not following. Could you please expand on that?

- May 24th 2010, 02:30 AMmr 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 where is the repeated root and 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:

and your job is to solve for a in terms of p, q and r. - May 24th 2010, 03:35 AMLukybear
- May 24th 2010, 04:11 AMDinkydoe
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

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

Instead of trying to solve the value yourself.

To get back to my post. A root of an equation is a double root when it's also a root of . 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, 06:55 AMJester
Try this. First we'll number our equations

First eliminate from and giving

then from and giving

then from and giving

Now can you manipulate these three new equations as to get all by itself? ( no or ) - May 24th 2010, 09:24 AMArchie Meade
The 3 resulting equations should be

- May 24th 2010, 10:39 PMLukybear
- May 25th 2010, 03:16 AMArchie 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

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

Substituting back into f(x)=0

Again using the value for , we get

- May 25th 2010, 03:52 AMDinkydoe
I see Archie Meade allready outlined a way to do it:

The reason why this works:

If is a double root of , then it's also a root of

Proof:

With the chain rule: if then . Hence . - May 25th 2010, 05:18 AMLukybear
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, 08:15 AMJester
- May 25th 2010, 11:10 AMArchie Meade
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 and to be left with , the double root.

Using the derivative, we proceed to eliminate and to be left with x,

because this x is the double root, corresponding to f(x)=0 and f'(x)=0.

for the double root

hence we now have a linear expression for at the double root .....you could use

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

becomes

We want x and f(x) still contains but since the value of x is the x co-ordinate of the double root.

hence

is

for the x-value of the double root.

We just replace all terms at any stage by the value of 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.