• Apr 16th 2012, 07:39 PM
steward0099
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 PM
Soroban
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 PM
steward0099
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.