If x^3 + 3px^2 + 3qx + r = 0 has a double root, show that the double root must be (pq-r)/(2q-2p^2)
It's not hard to see that it's a root of $\displaystyle f(x)=0$ in the first place.
Now you're left with showing that it's also a root of $\displaystyle f'(x)=3x^2+6px+3q=0$
(wich is using the abc-formula or plugging in $\displaystyle x=\frac{pq-r}{2q-2p^2}$ in f'(x))
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 $\displaystyle (x - a)^2 (x - b)$ where $\displaystyle x = a$ is the repeated root and $\displaystyle x = b$ 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:
$\displaystyle 3p = -b - 2a$
$\displaystyle 3q = 2ab + a^2$
$\displaystyle r = -a^2 b$
and your job is to solve for a in terms of p, q and r.
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 $\displaystyle a= \frac{pq-r}{2q-2p^2}$
Another way ofcourse would be to fill in the the given expression for "$\displaystyle a$", show that it satisfies:
$\displaystyle 3p=-b-2a$
$\displaystyle 3q=2ab+a^2$
$\displaystyle r=-a^2b$
Instead of trying to solve the value $\displaystyle a$ yourself.
To get back to my post. A root $\displaystyle x=a$ of an equation $\displaystyle f(x)=0$ is a double root when it's also a root of $\displaystyle f'(x)=0$. 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.
Try this. First we'll number our equations
$\displaystyle
1)\;\; 2a + b + 3p = 0
$
$\displaystyle
2)\;\; a^2 + 2ab - 3 q = 0
$
$\displaystyle
3)\;\;a^2b - r = 0
$
First eliminate $\displaystyle b$ from $\displaystyle (1)$ and $\displaystyle (2)$ giving
$\displaystyle 3a^2 + 6ap + 3q = 0,$
then from $\displaystyle (1)$ and $\displaystyle (3)$ giving
$\displaystyle
2a^3 + 3a^2p + r = 0,
$
then from $\displaystyle (2)$ and $\displaystyle (3)$ giving
$\displaystyle
a^3 - 3aq + 2r = 0.
$
Now can you manipulate these three new equations as to get $\displaystyle a$ all by itself? ( $\displaystyle {\it i.e.}$ no $\displaystyle a^2$ or $\displaystyle a^3$ )
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
$\displaystyle x^3+3px^2+3qx+r=0$
$\displaystyle 3x^2+6px+3q=0$
re-arranging the 2nd equation for x^2, we get
$\displaystyle x^2=-\frac{6px+3q}{3}=-2px-q$
Substituting back into f(x)=0
$\displaystyle x(-2px-q)+3p(-2px-q)+3qx+r=0$
$\displaystyle -2px^2-qx-6p^2x-3pq+3qx+r=0$
Again using the value for $\displaystyle x^2$, we get
$\displaystyle 2p(2px+q)-qx-6p^2x-3pq+3qx+r=0$
$\displaystyle 4p^2x+2pq-qx-6p^2x-3pq+3qx+r=0$
$\displaystyle x\left(4p^2-q-6p^2+3q\right)+2pq-3pq+r=0$
$\displaystyle x\left(-2p^2+2q\right)-pq+r=0$
$\displaystyle x=\frac{pq-r}{2q-2p^2}$
I see Archie Meade allready outlined a way to do it:
The reason why this works:
If $\displaystyle a $ is a double root of $\displaystyle f(x)=0$, then it's also a root of $\displaystyle f'(x)=0$
Proof:
With the chain rule: if $\displaystyle f(x)=(x-a)^2(x-b)$ then $\displaystyle f'(x)=(x-a)^2+2(x-a)(x-b)$. Hence $\displaystyle f'(a)=0$.
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 $\displaystyle a^3$ and $\displaystyle a^2$ to be left with $\displaystyle a$, the double root.
Using the derivative, we proceed to eliminate $\displaystyle x^3$ and $\displaystyle x^2$ to be left with x,
because this x is the double root, corresponding to f(x)=0 and f'(x)=0.
$\displaystyle 3x^2+6px+3q=0$ for the double root
hence we now have a linear expression for $\displaystyle x^2$ at the double root .....you could use $\displaystyle u=x^2$
$\displaystyle 3x^2=-6px-3q\ \Rightarrow\ x^2=-2px-q$
Therefore we can rewrite f(x)=0 for the double root
$\displaystyle f(x)=(x)x^2+(3p)x^2+3qx+r=0$
becomes
$\displaystyle f(x)=x(-2px-q)+3p(-2px-q)+3qx+r=0$
We want x and f(x) still contains $\displaystyle x^2$ but $\displaystyle x^2=-2px-q$ since the value of x is the x co-ordinate of the double root.
hence
$\displaystyle f(x)=(-2p)x^2-qx-6p^2x-3pq+3qx+r=0$
is $\displaystyle f(x)=(-2p)(-2px-q)-qx-6p^2x-3pq+3qx+r=0$
for the x-value of the double root.
We just replace all $\displaystyle x^2$ terms at any stage by the value of $\displaystyle x^2$ 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.