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Math Help - I REALLY need help with this quadratic proof.. Would appriciate it.

  1. #1
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    I REALLY need help with this quadratic proof.. Would appriciate it.

    QUESTION: Suppose that r and s are distinct real solutions of the quadratic equation ax^2 + bx + c = 0. In terms of a,b,c, obtain forumlas for r + s and rs.

    SOLUTION?
    so what i know the answer is r+s= -b/a and rs=c/a but im confused as to why.

    SO i started this...
    suppose r=x and s=y:
    (x+y)= M ( where M is an element of the real #)
    (xy)=T (where M is an element of the real #)

    so then x(M-x)=T
    x^2-Mx+T=0
    ax^2-aMx+aT=0

    assuming aM =b and aT=c we get out original quadratic equation.

    Now im stuck.. i tried solving for r+s(aMx) using what i have but i get something different. Please help =/
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  2. #2
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    If you know the Quadratic Formula, then the two solutions are \displaystyle \frac{-b + \sqrt{b^2 - 4ac}}{2a} and \displaystyle \frac{-b - \sqrt{b^2 - 4ac}}{2a}.

    What do you get when you add them? What do you get when you multiply them?
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  3. #3
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    Quote Originally Posted by kensington View Post
    QUESTION: Suppose that r and s are distinct real solutions of the quadratic equation ax^2 + bx + c = 0. In terms of a,b,c, obtain forumlas for r + s and rs.
    First of all, what are the solutions for \displaystyle ax^2 + bx + c = 0 \implies x = \frac{-b\pm\dots }{2a} ?
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  4. #4
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    Quote Originally Posted by Prove It View Post
    If you know the Quadratic Formula, then the two solutions are \displaystyle \frac{-b + \sqrt{b^2 - 4ac}}{2a} and \displaystyle \frac{-b - \sqrt{b^2 - 4ac}}{2a}.

    What do you get when you add them? What do you get when you multiply them?
    Thank you sooo much!
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  5. #5
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    Quote Originally Posted by kensington View Post
    QUESTION: Suppose that r and s are distinct real solutions of the quadratic equation ax^2 + bx + c = 0. In terms of a,b,c, obtain forumlas for r + s and rs.

    SOLUTION?
    so what i know the answer is r+s= -b/a and rs=c/a but im confused as to why.

    SO i started this...
    suppose r=x and s=y:
    (x+y)= M ( where M is an element of the real #)
    (xy)=T (where M is an element of the real #)

    so then x(M-x)=T
    x^2-Mx+T=0
    ax^2-aMx+aT=0

    assuming aM =b and aT=c we get out original quadratic equation.

    Now im stuck.. i tried solving for r+s(aMx) using what i have but i get something different. Please help =/
    a(x-r)(x-s)=ax^2+bx+c

    Expand the left hand side:

    ax^2-a(r+s)x+a(rs)=ax^2+bx+c

    Now equate coefficients of like powers of $$x.

    CB
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