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 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.
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.
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
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 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.