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?

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- Apr 16th 2012, 07:39 PMsteward0099Function 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? - Apr 16th 2012, 08:07 PMSorobanRe: Function question advanced pre-cal
Hello, steward0099!

Quote:

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 . . .

- Apr 16th 2012, 08:13 PMsteward0099Re: Function question advanced pre-cal
Thank you!

- Apr 17th 2012, 09:08 AMHallsofIvyRe: 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.