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Math Help - Function question advanced pre-cal

  1. #1
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    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?
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  2. #2
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    Re: Function question advanced pre-cal

    Hello, steward0099!

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

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


    You can use Algebra to answer the question.

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

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

    Now multiply them . . .

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  3. #3
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    Re: Function question advanced pre-cal

    Thank you!
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  4. #4
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    Re: Function question advanced pre-cal

    Also, if m and n are roots of x^2+ mx+ n= 0 (Strictly speaking, an equation has "roots". You are looking for the zeros of the function.) then 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.
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