Can You Solve for m and n in x^2+mx+n=0 with Only One Possible Value for x?

[mathdad]

My friend and I love to challenge each other from time to time when we come across a new question in a textbook or online. Here is the question he found. In the equation x^2+mx+n=0, m and n are integers. The only possible value for x is -3. What is the value of m and n?

I am thinking discriminant: b^2 - 4ac, where a = 1, b = m and c = n.

Correct approach?

 

[MarkFL]

My friend and I love to challenge each other from time to time when we come across a new question in a textbook or online. Here is the question he found. In the equation x^2+mx+n=0, m and n are integers. The only possible value for x is -3. What is the value of m and n?

I am thinking discriminant: b^2 - 4ac, where a = 1, b = m and c = n.

Correct approach?

What I would do is from the given information, we know:

$$x^2+mx+n=(x+3)^2$$

Expand the RHS, and then equate coefficients to determine $m$ and $n$.

Now, you could use the discriminant...what must its value be? Can you determine another equation with $m$ and $n$ using the given root?

 

[mathdad]

How about this solution?

Let x = -3

x +3 = 0

(x+3)^2 = x^2 + 6x + 9

m = 6; n = 9

Correct?

 

[MarkFL]

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:

$$(m,n)=(6,9)$$

 

[mathdad]

Interesting question for sure.