1. ## Function question advanced pre-cal

The function f(x) = x^2+mx+n
m and n are roots of the function
what is the product of m and n?

I'm stumped on this one. Any help?

2. ## Re: Function question advanced pre-cal

Hello, steward0099!

Given: $\displaystyle f(x) \:=\: x^2+mx+n$
$\displaystyle m$ and $\displaystyle n$ are roots of the function.
What is the product of $\displaystyle m$ and $\displaystyle n$?

There is a theorem which says: the product of the roots is the contant term, $\displaystyle n.$

You can use Algebra to answer the question.

Use the Quadratic Formula to find the two roots: .$\displaystyle x \;=\;\frac{-m \pm \sqrt{m^2 - 4n}}{2}$

The two roots are: .$\displaystyle \frac{-m + \sqrt{m^2-4n}}{2}\,\text{ and }\,\frac{-m - \sqrt{m^2-4n}}{2}$

Now multiply them . . .

Thank you!

4. ## Re: Function question advanced pre-cal

Also, if m and n are roots of $\displaystyle x^2+ mx+ n= 0$ (Strictly speaking, an equation has "roots". You are looking for the zeros of the function.) then $\displaystyle x^2+mx+n= (x-m)(x-n)= x^2- (m+n)x+ mn$. So, as Soroban said, the product mn is -n. Of course, mn= -n is the same as mn+ n= n(m+n)= 0.
Either m= 0 or m= -n. We also have m= -(m+n) so that m= -m- n which yields n= -2m. So if n is not 0, we must have -n= 2m= m which says that m must be 0. That is, either m= 0 or n= 0. In either case mn= 0.