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Math Help - Quadratic Equations

  1. #1
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    Quadratic Equations

    Any advise much appreciated

    I have the following equation;

    x^2 - 4x - 3 = 0

    I tried the quadratic formula but could not get any answers to = 0

    So I tried to solve the equation by factorisation. Here are some attempts that I have tried, where I don't want to believe that the equation cannot be factored, rather I think I may just be missing something?

    x^2 - 4x - 3 = 0

    (x + 3)(x - 1) = x^2 - 1 + 3x - 3
    (x - 1)(x + 3) = x^2 + 3x - x - 4
    (x + 4)(x - 1) = x^2 -x + 4x - 4
    (x - 1)(x + 4) = x^2 +4x - x - 4
    (x + 7)(x - 4) = x^2 -4x + 7x - 28
    (x + 2)(x - 2) = x^2 -2x + 2x -4 = x^2 - 4x -4

    My problem is that I can't re-generate the -3

    David

    (x + 3)(x - 1) = x^2 -x + 3x - 3 = x^2 - 4x - 3 = 0

    Sorry
    Last edited by David Green; November 17th 2011 at 03:36 AM. Reason: Very sorry I didn't see it until after posting
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  2. #2
    Super Member Quacky's Avatar
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    Re: Quadratic Equations

    You have made a few illogical attempts. The only things you need to try are the factors of -3.
    These are:
    (1,-3)
    (3,-1)

    You will soon notice, having tried these, that neither works: indeed, it doesn't factor. If you've struggled with the formula, why not give completing the square a jab?

    We have:
    x^2-4x-3=0

    So you want something of the form:

    (x-2)^2-k=0 where k a number you need to work out.

    Edit: Please doublecheck your factors: there's a sign error in your calculations. This quadratic doesn't factor.
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  3. #3
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    Re: Quadratic Equations

    Quadratic formula works nicely: \frac{4\pm\sqrt{4^2-(4)(-3)}}{2}= \frac{4\pm\sqrt{16+ 12}}{2} = \frac{4\pm\sqrt{28}}{2}= \frac{4\pm\sqrt{(4)(7)}}{2} = \frac{4\pm 2\sqrt{7}}{2}= 2\pm\sqrt{7}.

    I wonder if you didn't "lose" the sign on -3 and use \sqrt{4^2- 4(3)}= \sqrt{4}= 2 by mistake.
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  4. #4
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    Re: Quadratic Equations

    Quote Originally Posted by HallsofIvy View Post
    Quadratic formula works nicely: \frac{4\pm\sqrt{4^2-(4)(-3)}}{2}= \frac{4\pm\sqrt{16+ 12}}{2} = \frac{4\pm\sqrt{28}}{2}= \frac{4\pm\sqrt{(4)(7)}}{2} = \frac{4\pm 2\sqrt{7}}{2}= 2\pm\sqrt{7}.

    I wonder if you didn't "lose" the sign on -3 and use \sqrt{4^2- 4(3)}= \sqrt{4}= 2 by mistake.
    I spotted my other mistake in the equation, thanks for pointing this out, however I understand Hallsofivy thinks I made a mistake with the signs?

    I understand there is no value present for "a", and if "a" is either ignored or used as "1", it should not make a difference, however keeping with the formula, there should be two routes, and only one route seems to work.

    - 4 + or - Sq root 28 / 2 = -4 + or - 5.29 / 2

    9.29 / 2 = 4.65

    Equation;

    x^2 - 4x - 3 = 0

    4.65^2 - 4(4.65) - 3 = 0.0225 (I'll take it this route is zero)

    the next route does not work?

    So am I still incorrect or if one of the two routes are OK is that acceptable?
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  5. #5
    Super Member Quacky's Avatar
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    Re: Quadratic Equations

    Both roots pointed out by HallsofIvy are fine. Leave things in exact form to prevent yourself from inducing rounding errors and to keep things simple.

    (2-\sqrt{7})^2-4(2-\sqrt{7})-3

    =4-4\sqrt{7}+7-8+4\sqrt{7}-3

    =0
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  6. #6
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    Re: Quadratic Equations

    of course the pros make it look so easy.

    for a quadratic x^2 + bx + c = 0 the quadratic formula gives us the roots:

    x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}, which is the "method of last resort".

    there is a compromise, of sorts, when your polynomial has integer coefficients, and the x^2 coefficient is 1:

    calculate b^2 - 4c and see if this is a perfect square. if it is, the roots will be "nice" (integers) so it will "factor easy".

    if not, you'll wind up using the quadratic formula anyway, and you have half of it computed already

    (if it's negative, depending on your point of view, you either have "no roots" or "complex roots", and if you aren't considering anything but real roots, you can just stop, and say "no solutions").
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  7. #7
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    Re: Quadratic Equations

    Quote Originally Posted by Quacky View Post
    Both roots pointed out by HallsofIvy are fine. Leave things in exact form to prevent yourself from inducing rounding errors and to keep things simple.

    (2-\sqrt{7})^2-4(2-\sqrt{7})-3

    =4-4\sqrt{7}+7-8+4\sqrt{7}-3

    =0
    Thanks I got it.

    I was getting the signs wrong between x^2 and -4x

    I get both routes = 0
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