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Math Help - Fixed Real Numbers...Part 3

  1. #1
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    Fixed Real Numbers...Part 3

    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.

    (b) Use part (a) to verify that ax^2 + bx + c =
    a(x - r)(x - s)

    Use part (b) to factor x^2 - 2x - 1 and 5x^2 + 8x + 2.
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  2. #2
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    Quote Originally Posted by sologuitar View Post
    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.

    (b) Use part (a) to verify that ax^2 + bx + c =
    a(x - r)(x - s)

    Use part (b) to factor x^2 - 2x - 1 and 5x^2 + 8x + 2.
    ??? What was "part a"? The quadratic formula?
    Perhaps:
    b) The fact that r and s satisfy ax^2+ bx+ c= 0 means that (x- r) and (x- s) are factors. Since just multiplying them together gives x^2- (r+s)x+ rs, to get the "a" multiplying x^2, we must multiply that product by a: a(x-r)(x-s)= ax^2- a(r+s)x+ ars= ax^2+ bx+ c.

    c) Use the quadratic formula (or completing the square) to find the roots of each equation. Then use the fact, from (b) above, that the polynomial can be factored as a(x-r)(x-s) where a is the leading coefficient (1 in the first example, 5 in the second) and r and s are the solutions to the equations with the polynomial set to 0.
    Last edited by HallsofIvy; December 9th 2009 at 04:46 AM.
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  3. #3
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    yes..

    Quote Originally Posted by HallsofIvy View Post
    ??? What was "part a"? The quadratic formula?
    Perhaps:
    b) The fact that r and s satisfy ax^2+ bx+ c= 0 means that (x- r) and (x- s) are factors. Since just multiplying them together gives [tex]x^2- (r+s)x+ rs, to get the "a" multiplying x^2, we must multiply that product by a: a(x-r)(x-s)= ax^2- a(r+s)x+ ars= ax^2+ bx+ c.

    c) Use the quadratic formula (or completing the square) to find the roots of each equation. Then use the fact, from (b) above, that the polynomial can be factored as a(x-r)(x-s) where a is the leading coefficient (1 in the first example, 5 in the second) and r and s are the solutions to the equations with the polynomial set to 0.
    Yes, part a was the quadratic formula. Thanks for the extra tips.
    I'll work on this a little later today.
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