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Math Help - Equations Help

  1. #1
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    Equations Help

    Hey, I'm having trouble with a few of these.

    1. x^2 - 6x + 3 = 0

    2. x^2 + 5x + 1 = 0

    3. 2x^2 - 50 = 0

    4. x = 3/x + 7
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by carlito1 View Post
    x^2 - 6x + 3 = 0
    Is the problem which method to use or using the methods? In general, if you can't figure out how to solve a quadratic, there's always the quadratic formula to fall back on. (This is going to look terrible without LaTeX, sorry!)

    ax^2 + bx + c = 0 implies

    x = -b/(2a) (+/-) [sqrt(b^2 - 4ac)]/(2a) (Where the "+/-" is "plus or minus.")

    The method of "completing the square" always works as well. Then there's the methods you can use to factor, but at a glance I don't think any of these factor.

    I'll do this one by completing the square, if you need help with the quadratic formula, just let me know.

    x^2 - 6x + 3 = 0

    x^2 - 6x = -3

    We want to add a number to both sides such that the LHS is a perfect square. The format for this is the identity:
    (x + a)^2 = x^2 + (2a)x + a^2.

    So, looking at the form (with the "a" in it) and your problem the coefficient of the x term is - 6 = 2a. Thus a = -3. So we want to add (-3)^2 to both sides of the equation:
    x^2 - 6x + 9 = -3 + 9

    (x - 3)^2 = 6

    x - 3 = (+/-)sqrt(6)

    x = 3 (+/-) sqrt(6).

    -Dan
    Last edited by topsquark; September 22nd 2006 at 11:42 AM. Reason: I made it prettier. :)
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by carlito1 View Post
    x^2 + 5x + 1 = 0
    We can do this one using completing the square as well.

    x^2 + 5x + 1 = 0

    x^2 + 5x = -1

    The coefficient of x is 5 = 2a, so a = 5/2. Thus we need to add (5/2)^2 to both sides:

    x^2 + 5x + (5/2)^2 = -1 + (5/2)^2

    (x + 5/2)^2 = -1 + 25/4 = -4/4 + 25/4 = 21/4

    x + 5/2 = (+/-)sqrt(21/4) = (+/-)sqrt(21)/2

    x = -5/2 (+/-) sqrt(21)/2.

    I'll do this with the quadratic formula as well:
    x^2 + 5x + 1 = 0

    Here a = 1, b = 5, c = 1.

    x = -b/(2a) (+/-) [sqrt(b^2 - 4ac)]/(2a) = -5/(2*1) (+/-) [sqrt(5^2 - 4*1*1)]/(2*1)

    x = -5/2 (+/-) [sqrt(25 - 4)]/2 = -5/2 (+/-) sqrt(21)/2.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by carlito1 View Post
    3. 2x^2 - 50 = 0

    4. x = 3/x + 7
    Mostly I'd like you to try these again based on the work in the last two posts, but I'll get you started.

    3. 2x^2 - 50 = 0
    We really don't need the completing the squares technique here. I lied in my first post, this one factors. You would start the same way in either case:
    2x^2 - 50 = 0

    x^2 - 25 = 0

    This is the difference between two perfect squares. You should have a formula to factor this in your book. (Or complete the square, but in this case a = 0, which is a little silly. However moving the constant term to the other side of the equation gives: x^2 = 25. You don't need to fiddle with the LHS because it's already a perfect square! So just take the square root. Remember the "+/-"!)

    4. x = 3/x + 7
    The only "hidden" difficulty here is that we can't have a solution where x = 0. (We don't in any case, but it's always good to check this.)

    I'm going to multiply both sides of this equation by x:
    x^2 = 3 + 7x

    x^2 - 7x - 3 = 0.

    This one doesn't factor, but can you take it from here?

    -Dan
    Last edited by topsquark; September 22nd 2006 at 11:40 AM. Reason: Addendum
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  5. #5
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    I get x=9.25 or x= 4.71 for number 4 - can someone check this
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  6. #6
    Senior Member OReilly's Avatar
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    Quote Originally Posted by carlito1 View Post
    I get x=9.25 or x= 4.71 for number 4 - can someone check this

    I get x = 7.4 or x = -0.4
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  7. #7
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Carlito1 View Post
    I get x=9.25 or x= 4.71 for number 4 - can someone check this
    are you sure your using the quadratic formula?

    Here it is:
    Attached Thumbnails Attached Thumbnails Equations Help-quadratic-formula.jpg  
    Last edited by Quick; September 24th 2006 at 12:03 PM.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by carlito1 View Post
    x = 3/x + 7
    Multiply both sides by x:
    x^2 = 3 + 7x

    x^2 - 7x = 3

    To complete the square, 2a = -7 so a = -7/2. Thus we add a^2 = 49/4 to both sides.

    x^2 - 7x + (-7/2)^2 = 3 + (-7/2)^2

    (x - 7/2)^2 = 3 + 49/4 = 12/4 + 49/4 = 61/4

    x - 7/2 = (+/-)sqrt(61/4) = (+/-)[sqrt(61)]/2

    x = 7/2 (+/-) [sqrt(61)]/2

    Numerically this is x = 7.405124838 or x = -0.405124838

    -Dan
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