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Math Help - Simple factorization/equation problems. Need help!

  1. #1
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    Simple factorization/equation problems. Need help!

    (Posted this on mymathforum.com as well. This is a copy-paste)

    I'm a Swedish student, reading course C in mathematics. I've read course A and B before, but it was more than five years ago, and I've forgot about everything (they predict that everyone knows course A and B).

    I'm having problem with factorization and equations.

    Equations
    :
    I can solve very easy equations like x^2-4x+3 and stuff like that. What I'm having trouble with are √-equations.
    I can solve for example √(2x+5)=x+1 by doing this:
    1. 2x+5 = (x+1)^2
    2. 2x+5 = x^2+2x+1
    3. x^2-4 = 0
    4. x=2

    But then, when I come to this problem, √(4t+1)=3-3t, I can't solve it:
    1. 4t+1 = (3-3t)^2
    2. 4t+1 = 9-18t+9t^2
    3. 9t^2-22t+8 = 0
    4. What here? The answer is (4/9). I know that I somehow should divide everything with 9. Should I first calculate everything and the divide by 9?

    Two more problems: 3√(x+13) = x+9 & √(x+1)-1=x. How do I solve this?

    Factorizations:

    In short: How do I get the answer 7(x-1)(x+2/7) from 7x^2-5x-2? Are there any steps/formulas to get to this answer? How do I know that 7x^2-5x-2 should be written in the form a(b-c)(b+d), and not something like a(a-b-c) etc. When I know that it should be written like a(b-c)(b+d), how do I calculate what should be in "c" and in "d"?

    I hope I've explained my problems well. As said, I haven't done any mathematics in over five years, and the course expect me to know course A and B already, and I feel totally lost.
    Appreciate answer
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  2. #2
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    Response

    1. 4t+1 = (3-3t)^2
    2. 4t+1 = 9-18t+9t^2
    3. 9t^2-22t+8 = 0

    For this one use the quadratic formula:

    <br />
x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}<br />

    So, it would be:
    <br />
x = \frac{-(-22)\pm\sqrt{(-22)^2-4*9*8}}{2*9}<br />

    <br />
x = \frac{22\pm\sqrt{484-288}}{18}<br />

    <br />
x = \frac{22\pm\sqrt{196}}{18}<br />

    <br />
sqrt{196} = 14<br />

    t = (22 + 14) / 18 or t = (22 - 14) / 18
    t = 36/18 or t = 8/18
    t = 2 or t = 4/9

    OR, you could divide the original equation through by 9 to get:
    t^2 - (22/9)t + 8/9 = 0

    Then factor it as:
    (t-2)(t-4/9) = 0
    but, these factors might not have been so obvious. When in doubt, use the quadratic formula to find the values
    Last edited by ajj86; November 11th 2008 at 08:58 AM.
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  3. #3
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    Response

    3√(x+13) = x+9
    Start by squaring both sides to get:
    9(x+13) = (x+9)^2
    9x + 117 = x^2 + 18x + 81
    0 = x^2 + 18x + 81 - 9x - 117
    x^2 + 9x -36 = 0

    This one can be factored of the form: (x-a)(x+b)
    Try some of the factors of 36 to be substituted into a,b, namely 12 and 3

    (x + 12)(x - 3) = x^2 - 3x + 12x -36 = x^2 + 9x - 36 = 0
    So, you would have:
    x = -12 or x = 3

    When in doubt you could always use the quadratic formula to get the solutions, like I did in the last problem.

    For some reason, when I substitute x = -12 back into the original equation, I get 3 = -3. I've checked and can't find the error. Maybe you could go over this one with your teacher or a classmate.
    Last edited by ajj86; November 11th 2008 at 09:07 AM.
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  4. #4
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    Response

    √(x+1)-1=x

    For this one, start by adding 1 to both sides:
    √(x+1) = x+1 Now, square both sides to get:
    x+1 = (x+1)^2
    x+1 = x^2 + 2x + 1
    0 = x^2 + 2x + 1 - x - 1
    x^2 + x = 0

    Factor out an x:
    x(x + 1) = 0

    Solve for x:
    x = 0 or x = -1
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  5. #5
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    Additional

    I can solve very easy equations like x^2-4x+3 and stuff like that. What I'm having trouble with are √-equations.
    I can solve for example √(2x+5)=x+1 by doing this:
    1. 2x+5 = (x+1)^2
    2. 2x+5 = x^2+2x+1
    3. x^2-4 = 0
    4. x=2

    When you get it down to step 3, you need to factor it as:
    (x-2)(x+2) = 0

    Then, you get the answers:
    x = 2 or x = -2.

    Substitute x = -2 back into the original equation in step 1. and you will see that it checks out.
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  6. #6
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    Last part

    How do I get the answer 7(x-1)(x+2/7) from 7x^2-5x-2? Are there any steps/formulas to get to this answer? How do I know that 7x^2-5x-2 should be written in the form a(b-c)(b+d), and not something like a(a-b-c) etc. When I know that it should be written like a(b-c)(b+d), how do I calculate what should be in "c" and in "d"?

    Well, you would start by factoring out a 7 from the equation:
    7(x^2 - (5/7)x - 2/7)
    Then you would factor it as such:
    7(x-1)(x+2/7)

    This sometimes takes guess and check work. If you wanted to, you could use the quadratic formula, find the x values, and then write it of this form.
    I hope all of this helps.
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  7. #7
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    Wow, thank you ajj86 for your help! You have really helped me out
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