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Math Help - Please Check My Work on Quadratic Equations

  1. #1
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    Please Check My Work on Quadratic Equations

    This week, we are learning Quadratic Equations and I am having a bit of trouble with them as I am not sure which solution to try, or which one may be easier. Could someone please check the following problems to make sure I am on the right track.












    Thank you in advance for taking your time to look over these.
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  2. #2
    Senior Member eumyang's Avatar
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    1) Why did you change from \pm 9 to \pm 9i? The answers should be \pm 9.

    2-6) ok

    7) ok, but I would add this step: y - 2 = -4 between "y = 6" and "y = 2-4" so that there is no confusion.

    8-10) ok

    11) you forgot the "twice the width" part. If you let the width = x, then the length would be 2x + 2. So the equation would be
    x(2x + 2) = 56.

    EDIT: Added #11
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  3. #3
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    Hi eumyang, thank you very much for your time. For question 1, that was a typo, I need to slow down a bit. These are so confusing to me. Is there on solution that is better than the other (e.g. is quadratic formula easier than factoring, or perhaps completeting the sqare is better). Any thoughts would be greatly appreciated.
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  4. #4
    Senior Member eumyang's Avatar
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    The methods you used for each problem were okay. After a while you kind of get the feel of picking the best method.

    You didn't mention this, but there is also taking the square roots as a method. Use it when there is no x-term, or when you have a binomial squared. Examples:
    5 x^2 - 125 = 0
    x^2 = 333
    (x + 5)^2 = 61
    (2x - 3)^2 - 28 = 0

    For factoring, you'll going to have to just try it. If you check the discriminant (b^2 - 4ac) and it's a perfect square, then the quadratic is factorable.
    For example, 3x^2 + 13x + 12 = 0 is factorable because the discriminant is a perfect square: 13^2 - 4(3)(12) = 169 - 144 = 25. The quadratic factors into (3x + 4)(x + 3)
    2x^2 + 5x - 1 is not factorable: the discriminant is 5^2 - 4(2)(-1) = 25 + 8 = 33

    For completing the square, I use this method the least. If the quadratic is set up so that the x-squared term's coefficient is 1, the x-term's coefficient is even, AND both of these terms are on one side and the constant term is on the other, then maybe I'll try it:
    Examples:
    x^2 + 12x = -4
    6 = x^2 - 8x

    Just don't use quadratic formula all the time, okay? I've seen students who will do that. Yuck!
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  5. #5
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    Wonderful, thank you so much! You have been a tremendous help to me. I will definitely have to do some more practice on these types of problems and hopefully it will soon become second nature. Thank you again, you have no idea how beneficial you have been. Have a good evening :-)
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  6. #6
    Senior Member yeKciM's Avatar
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    one quick quetstion for "dclary"
    is this true ?

     \displaystyle \sqrt{x^2} = x

    yes ? why , no ? why
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  7. #7
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    Hi yeKciM, okay, as I said, I am not very good at these but I would say that it is true because it has 2 real solutions being x=0, x=-1
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  8. #8
    Super Member Bacterius's Avatar
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    \sqrt{x^2} = |x| (absolute value of x)
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  9. #9
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by dclary View Post
    Hi yeKciM, okay, as I said, I am not very good at these but I would say that it is true because it has 2 real solutions being x=0, x=-1
     |x| = \left\{\bar{}\begin{matrix}<br />
x & x>0 \\ <br />
0& x=0 \\ <br />
-x & x<0<br />
\end{matrix}\right.


    P.S. don't know why is there "-" with that zero hm.... i check it twice and in the code i didn't put it but i see it really don't know why, but if u see it to that should not be there
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