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Thread: check my work on quadratic equations?

  1. #1
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    check my work on quadratic equations?

    can someone check this for me?
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  2. #2
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    what about this one?
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  3. #3
    Super Member Bacterius's Avatar
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    No, both are wrong (first one in the answer, second one in the writing)

    $\displaystyle \Rightarrow$ In the first one, you must write both answers :

    $\displaystyle x = \frac{4}{\sqrt{5}}$ and $\displaystyle x = - \frac{4}{\sqrt{5}}$

    $\displaystyle \Rightarrow$ The second one has the correct answer but is not correctly written. Here is an example of how it should be written :

    $\displaystyle x^2 - 8x + 5 = 0$

    $\displaystyle x^2 - 8x = -5$

    $\displaystyle x^2 - 8x + 16 = 11$

    $\displaystyle (x - 4)^2 = 11$

    $\displaystyle x - 4 = \sqrt{11}$ or $\displaystyle x - 4 = - \sqrt{11}$

    $\displaystyle x = \sqrt{11} + 4$ or $\displaystyle x = - \sqrt{11} + 4$

    Thus $\displaystyle x = 4 \pm \sqrt{11}$



    EDIT : talking about the first question in this post ...
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  4. #4
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    $\displaystyle \frac{-4\pm 2\sqrt{10}}{2}= -2\pm \sqrt{10}$
    Last edited by pickslides; Dec 8th 2009 at 06:25 PM. Reason: Bad Latex
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  5. #5
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    You must show all your working

    What if I didn't know what the quadratic formula was?

    You're up to here:

    $\displaystyle \frac {-4 \pm 2\sqrt10}{2} $

    The next bit is wrong. You didn't factor 2 out of the -4.
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  6. #6
    Super Member Bacterius's Avatar
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    The other question you posted is wrong too
    I am not taking account of the quadratic equation written on it because I do not see the link with the following.

    Here is how you do it :

    $\displaystyle \frac{-4 \pm \sqrt{40}}{2} = \frac{-4 \pm \sqrt{4 \times 10}}{2}$

    $\displaystyle = \frac{-4 \pm 2 \sqrt{10}}{2} = \frac{2(-2 \pm \sqrt{10})}{2}$ --> This is where you messed up : you did not factorize the $\displaystyle -4$.

    $\displaystyle = -2 \pm \sqrt{10}$

    Or, more conveniently :

    $\displaystyle = 2 \mp \sqrt{10}$
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  7. #7
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    can someone help me understand the factorization part?
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  8. #8
    Super Member Bacterius's Avatar
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    Remember that $\displaystyle ab + ac = a(b + c)$. Here, you have :

    $\displaystyle \frac{-4 \pm 2 \sqrt{10}}{2}$

    This can be written :

    $\displaystyle \frac{-2 \times 2 \pm 2 \sqrt{10}}{2}$

    The common factor is two, so you factorize it over the addition. Since you are factorizing all the $\displaystyle 2$ (and not $\displaystyle -2$ !), they become $\displaystyle 1$ (some people prefer to remove them directly though) :

    $\displaystyle \frac{2(-2 \times 1 \pm 1 \sqrt{10})}{2}$

    Now you can cancel $\displaystyle 2$ out :

    $\displaystyle -2 \times 1 \pm 1 \sqrt{10}$

    Which is simplified to :

    $\displaystyle -2 \pm \sqrt{10}$

    Or, even more conveniently :

    $\displaystyle 2 \mp \sqrt{10}$
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  9. #9
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    still dont really folllow, sry
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  10. #10
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    alright let's break it down.

    We have

    $\displaystyle -4\pm2\sqrt10 $

    if we had $\displaystyle 2x^2+4x$

    we would factorise it by taking 2x out of the equation.

    So we divide all terms by 2x:

    $\displaystyle x+2$

    So same as $\displaystyle -4\pm2\sqrt10 $

    we can take 2 out of the equation;

    So divide everything by 2
    you will get: $\displaystyle -2\pm\sqrt10 $

    I divided the -4 by 2 (which made -2) and 2 by 2 ( which made 1)

    Is this more clear now?
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  11. #11
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    i think i got it now, thx a bunch
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  12. #12
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    Cool!
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