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Math Help - quadratic equation

  1. #1
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    quadratic equation

    hi, i have a quadratic equation and i have done it but im not sure its correct so if someone could check i will be very grateful, heres the question:

    quadratic equation-question2.jpg

    this is what i got:
    quadratic equation-answer.jpg

    is this correct? if not where am i going wrong.
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  2. #2
    MHF Contributor harish21's Avatar
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    correct
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  3. #3
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    This Quadratic factorises nicely...

    \displaystyle 3x^2 - 4x + 1 = 0

    \displaystyle 3x^2 - 3x - x + 1 = 0

    \displaystyle 3x(x - 1) - 1(x - 1)= 0

    \displaystyle (x - 1)(3x - 1) = 0

    \displaystyle x -1 = 0 or \displaystyle 3x - 1 = 0

    \displaystyle x = 1 or \displaystyle x = \frac{1}{3}.
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  4. #4
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    Quote Originally Posted by harish21 View Post
    correct
    Yes it is, but here is an easier way without the formula
    You can do this:

    3x^2 -4x + 1 = 0

    So if you know how to do quadratic equations, you must have a 3 and a 1 in the brackets to achieve the 3x^2

    Therefore your looking at:
    (3x ...)(x ...)
    Obviously the factors of 1, is 1 and itself so you can get:
    (3x .. 1)(x .. 1)
    Now about the signs, you need a + 1, so you need either 2 +'s or 2 -'s, seeing as you have -4 in the middle you can achieve this with 2 -'s

    so:
    (3x - 1)(x - 1) = 0
     x = 1 or x = 1/3

    Regards
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  5. #5
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    Strictly speaking, "0.33" is NOT a correct answer- \frac{1}{3} is.
    0.33 is only approximately equal to 1/3.

    Also, you should understand that you could have checked these by simply substituting you answers into the given equation.

    If x= 1, then 3x^2- 4x+ 1= 3(1)^2- 4(1)+ 1= 3- 4+ 1= 0 and 3\left(\frac{1}{3}\right)^2- 4\left(\frac{1}{3}\right)+ 1= \fra{1}{3}- \frac{4}{3}+ 1= 0.

    And, as others have pointed out, if a polygon can be factored, that is a better way of solving it than using a formula.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    And, as others have pointed out, if a polygon can be factored, that is a better way of solving it than using a formula.
    If a \neq 1 I usually bang out the formula to save time. I don't really think there is a preferred way (unless the question demands it of course)
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