Find the integer $m$ in $x^{2}+mx+n=0$ if the only solution is $x=-3$

Question

Find the integer $m$ in $x^{2}+mx+n=0$ if the only solution is $x=-3$.

I understand that you get $(x+3)^{2}=x^{2}+6x+9$ and get the value of $m$ this way, but why, when you plug $x=-3$ do you get a family of solutions for $m$?

$(-3)^{2}+m(-3)+n=0$

$3m=9+n$

$m=3+\frac{n}{3}$

This will yield a family of solutions as long as $n$ is a multiple of $3$. How can I constrain $m$ to get the solution $m=6$?

Answer 1

You need also $$m^2-4n=0,$$ which gives $$m^2-4(3m-9)=0$$ or $$(m-6)^2=0$$ or $m=6.$

Answer 2

Well let's try it:

The solutions to $x^2 + (3+\frac n3)x + n=0$ will be $

$x = \frac {-(3 + \frac n3) \pm \sqrt{(3+\frac n3)^2 -4n}}2 = \frac {-(3+ \frac n3) \pm \sqrt{9 - 2n + \frac {n^2}9-}}2 = \frac {-(3 +\frac n3) \pm (3-\frac n3)}{2} = -3, -\frac n3$

Thus $-3$ is always a solution and the other solution is $-\frac n3$. But to have only one solution we must have $-\frac n3 = -3$ or in other words $n = 9$.

Or to do it straight from the beginning to have a quadratic equation with a single root in must be in the form $(x - k)^2$ and not $(x -k_1)(n- k_2)$. So we must have $(x +3)^2 = x^2 + 6x + 9$ and ... well, our choice for $n$ is forced from the beginning.

The upshot is "plugging in $x =-3$" gives you ONE solution but the other solution would be $x = a$ so $(x +3)(x-a) = x^2 + (3-a)x -3a$ with $m= 3-a$ and $n = -3a$. Yes, it is a class of equations $x^2 + (3 + \frac n9) + n$ and solutions $(-3, -\frac n3)$ of which $x^2 + 6x + 9$ and solutions $(-3, -3)$ is just one. But is the only one with a double root.