Find the values of k for which the equation $x^3-9x^2 +24x +k =0$ may have multiple roots and solve the equation in each case.

Question

This question is from Theory of Equations. Help please.

Answer 1

A polynomial $\;f(x)\;$ over a field has a multiple root $\;\alpha\;$ iff $\;f(\alpha)=f'(\alpha)=0\;$. Now, here we have

$$f(x)=x^3-9x^2+24x+k\implies f'(x)=3x^2-18x+24=0\iff$$

$$x^2-6x+8=(x-2)(x-4)=0\iff x=2,4$$

So it must either $\;x=2,\,4\;$ is a multiple root. Now check for what values of $\;k\;$ this happens.

Answer 2

Hint: What does a cubic equation with positive $x^3$ coefficient look like? Where do the turning points occur? What happens if both turning points are above the $x$-axis? What if they both occur below? What about one above and one below?

Hope this helps you to make some head way! :)

Answer 3

A cubic will have multiple roots when its discriminant is equal to $0$, the cubic discriminant is equal to:

$$\Delta = b^2c^2 -4ac^3-4b^3d-27a^2d^2+18abcd$$

(Wikipedia)

with your $a=1, b=-9, c=24$ and $d=k$, input these in and you get a quadratic that you need to solve wrt to $k$.

Once solved you should get that the two values of $k$ are $-20$ and $-16$

For $-20$ using Cardano's solution to the cubic it has roots equal to $2, 2$ and $5$

and for $-16$ it has roots $1, 4$ and $4$

Sorry there's a lack of equations as well I'm a bit of a noob still :)

Answer 4

One way to solve the problem is to find the roots of the derivative of the polynomial $p(x)$, because if $p(x)$ has multiple roots, then $p'(x)$ will have the same roots. To see this, say $a$ is a root of $p(x)$ with multiplicity $n$, then we can write $p(x)=(x-a)^ng(x)$ where $g(a)\not=0.$ Then $p'(x)=n(x-a)^{n-1}g(x)+(x-a)^ng'(x)$, i.e. $a$ is also a root of $p'(x)$.

So to solve your problem, you can take the derivative of your polynomial and then find the roots and then put those roots back to the original polynomial to calculate the values of $k.$