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Math Help - solving this equation problem

  1. #1
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    solving this equation problem

    so far i have solved the answer to part a).

    but i can't seem to solve the equation for part b) though. to solve this qudractic do i have to use the qaudractic formula?

    any help is appreichated thank you.

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  2. #2
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    Hi Rpatel:

    Strictly speaking, the quadratic formula is nor required in this instance because the trinomial 10x^2-11x-84 is factorable. FYI, any quadratic trinomial, ax^2+bx+c, is factorable whenever its discriminant, b^2-4ac, is a perfect square. In the present case, b^2-4ac = (-11)^2 - 4(10)(-84) = 3481, which is indeed a perfect square (Any integer, k, is a perfect square if sqrt(k) is itself an integer. Because sqrt(3481)=59, and 59 is an integer, it follows that 3481 is by definition a perfect square). Nevertheless, were this my problem, I would use the q.formula just the same, as the coefficients 10 and 84 make for more factor combinations than I care to test.

    Regards,

    Rich B.
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  3. #3
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    Quote Originally Posted by rpatel
    so far i have solved the answer to part a).

    but i can't seem to solve the equation for part b) though. to solve this qudractic do i have to use the qaudractic formula?

    any help is appreichated thank you.

    There is someting wrong.

    Your 10x^2 -11x -84 = 0 is correct per shown figure.
    And that gives x = 3.5cm or -2.4cm

    x=3.5 fails at x-7.
    3.5 -7 = -3.5cm ------cannot be. No negative dimensions.

    x = -2.4m ------cannot be.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Rich B.
    Hi Rpatel:

    Strictly speaking, the quadratic formula is nor required in this instance because the trinomial 10x^2-11x-84 is factorable.


    In what sense? By the Fundamental Theorem of Algebra every
    quadratic has a root, and so is factorisable. It's just that these
    roots may not be real, or rational.

    FYI, any quadratic trinomial, ax^2+bx+c, is factorable whenever its discriminant, b^2-4ac, is a perfect square.
    Discriminant Positive - if we are interested in real factors.

    For instance the discriminant of: x^2-2 is 8, which
    is not a perfect square, but that does not mean we cannot factorise it:

    x^2\ -\ 2\ =\ (x-\sqrt2)(x+\sqrt2).

    When the discriminant is a perfect square it does ensure that
    the roots are rational.

    In the present case, b^2-4ac = (-11)^2 - 4(10)(-84) = 3481, which is indeed a perfect square (Any integer, k, is a perfect square if sqrt(k) is itself an integer. Because sqrt(3481)=59, and 59 is an integer, it follows that 3481 is by definition a perfect square). Nevertheless, were this my problem, I would use the q.formula just the same, as the coefficients 10 and 84 make for more factor combinations than I care to test.

    Regards,

    Rich B.
    By the time you have computed the discriminant and determined it
    is a perfect square you have done almost all the work involved
    in finding the roots using the quadratic formula and may as well
    go ahead and find them that way, and then complete the factorisation,
    or not (in this case we are interested in the roots not the factors).

    RonL
    Last edited by CaptainBlack; December 4th 2005 at 12:33 AM.
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