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Math Help - Fixed Real Numbers...Part 2

  1. #1
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    Fixed Real Numbers...Part 2

    Suppose a, b, and c are fixed real numbers such that
    b^2 - 4ac ≥ 0. Let r and s be the solutions of

    ax^2 + bx + x = 0.

    (a) Use the quadratic formula to show that r + s = -b/a and
    rs = c/a.

    (b) Use part (a) to verify that ax^2 + bx + c =
    a(x - r)(x - s).
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  2. #2
    Senior Member I-Think's Avatar
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    ax^2 + bx + x = 0. Solutions: r,s
    I think you mean
    ax^2 + bx + c = 0.

    x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}
    Let r=\frac{-b+{\sqrt{b^2-4ac}}}{2a}
    And s=\frac{-b-{\sqrt{b^2-4ac}}}{2a}<br />
    Hence
    (x-r)(x-s)=x^2+\frac{b}{a}x+\frac{c}{a}=0
    x^2-(r+s)x+rs=x^2+\frac{b}{a}x+\frac{c}{a}=0

    Hence r+s=\frac{-b}{a} and rs=\frac{c}{a}

    2nd part should fall into place now
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  3. #3
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    well...

    Quote Originally Posted by I-Think View Post
    ax^2 + bx + x = 0. Solutions: r,s
    I think you mean
    ax^2 + bx + c = 0.

    x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}
    Let r=\frac{-b+{\sqrt{b^2-4ac}}}{2a}
    And s=\frac{-b-{\sqrt{b^2-4ac}}}{2a}<br />
    Hence
    (x-r)(x-s)=x^2+\frac{b}{a}x+\frac{c}{a}=0
    x^2-(r+s)x+rs=x^2+\frac{b}{a}x+\frac{c}{a}=0

    Hence r+s=\frac{-b}{a} and rs=\frac{c}{a}

    2nd part should fall into place now
    Part 2 does not fall into place but I will play with it later today.
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