Parameter "m" for which $P(x)=4(m+1)x^3+(m-3)x+1-m$ has a root with multiplicity two...

Question

Can you please help me solve this parametric problem.

So, we have to find all the values of real parameter $m$ for which the following equation has a solution with multiplicity two.$$P(x)=4(m+1)x^3+(m-3)x+1-m$$

What I've done is this:

I used the property: if $P(a)=P'(a)=0$ (where $a$ is the respective solution) then $a$ is a root with multipicity higher than one.

• I've formed a system of equations and got that

$$m={3-12a^2\over {1+12a^2}}$$

• Then I substituted $m$ in the first equation, and after making some calculations I came to the point that:

$$16a^3-12a^2+1=0$$

Now the problem is that this equation has no solutions, so I hope you may help me find my mistake or may be another way to solve this exercise if you have time of course :). $NOTE:$I've tried to solve this exercise four times , and every time I had the same result.

Answer 1

Your idea with $P$ and $P'$ is fine. However, you seem to try to express $m$ in terms of $a$. Instead, you should find $m$ such that there exists $a$ such that ..., so $m$ must be given as an "absolute" expression and then $a$ may depend on $m$. Finally, $16a^3-12a^2+1=0$ surely does have a real solution, as have all odd-degree real polynomials.

So you want to solve the system of equations $$ \tag1 4(m+1)a^3+(m-3)a+1-m=0$$ and $$ \tag2 12(m+1)a^2+m-3=0.$$ Compute $3(1)-a(2)$: $$\tag3 2(m-3)a+3(1-m)=0$$ Any common solution of $(1)$ and $(2)$ must also be a solution of $(3)$, but not vice versa. If $m=3$, then $(3)$ has no solutoin; otherwise we conclude that $a=\frac{3(m-1)}{2(m-3)}$, but when is this $a$ a sulution of $(1)$? Plug it in to arrive after some simplifications at $$ \frac{\text{simple polynomial of degree four}}{(m-3)^3}=0$$ The possible values of $m4 are easily found, but be sure to check them against the original condition!

Answer 2

If the polynomial $P$ has multiple roots, then $P$ and $P'$ have a common factor. Consequently, can be useful to compute the gcd of $P$ and $P'$ (with Euclidean's algorithm) and impose that it has degree at least 1.

Since $P^\prime(x)=12(m+1)x^2+(m-3)$, successive divisions give: $P(x)=\frac x3 P^\prime(x)+\frac 23(m-3)x+(1-m)$ $P^\prime(x)=(18\frac{m+1}{m-3}x+27\frac{m^2-1}{(m-3)^2})(\frac 23(m-3)x+(1-m))+\frac{28m^3-36m^2}{(m-3)^2}$. We must to have $28m^3-36m^2=0$, from which $m=0$ or $m=\frac 97$.

Answer 3

A third degree equation always has a real solution. In this case will be a root of the (2nd degree) derivative.