RTCNTC said:
How about this solution?
Let x = -3
x +3 = 0
(x+3)^2 = x^2 + 6x + 9
m = 6; n = 9
Correct?
Yep. :D
Let's look at using the discriminant as you initially posted.
If a quadratic has only one root, then its discriminant must be zero, and so we obtain:
$$m^2-4n=0$$
Using the given root $x=-3$, we also have:
$$9-3m+n=0$$
If we multiply this by 4, we get:
$$36-12m+4n=0$$
The first equation (regarding the discriminant) implies $4n=m^2$, and we have:
$$m^2-12m+36=(m-6)^2=0\implies m=6\implies n=9$$
Now, we could also reason that a quadratic having only one root, will have its axis of symmetry at that root. Recall the axis of symmetry for the general quadratic $ax^2+bx+c$ is given by:
$$x=-\frac{b}{2a}$$
So we would have:
$$-\frac{m}{2}=-3\implies m=6$$
And then using either the discriminant, or the value of the quadratic at the root, we can easily find $n=9$.
We could also write the quadratic in vertex form:
$$x^2+mx+n=\left(x^2+mx+\frac{m^2}{4}\right)+n-\frac{m^2}{4}=\left(x+\frac{m}{2}\right)^2+\frac{4n-m^2}{4}$$
We know the vertex must be at $(-3,0)$, and so this gives us:
$$\frac{m}{2}=3$$
$$\frac{4n-m^2}{4}=0$$
And from this we easily find:
$$(m,n)=(6,9)$$
