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Math Help - Another qaudratic equation

  1. #1
    Junior Member Mist's Avatar
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    Another qaudratic equation

    I'm finding it difficult to solve this question
    It's solving quadratic eqaution by completing the square

    x^2+12x+3=0

    This is what I have done so far ...

    x^2+12x+3=0
    x^2 +12x=-3
    (x+12)^2-144=-3
    (x+12)^2=141
    x+12=√141

    Then, I don't know what to do
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Mist View Post
    I'm finding it difficult to solve this question
    It's solving quadratic eqaution by completing the square

    x^2+12x+3=0

    This is what I have done so far ...

    x^2+12x+3=0
    x^2 +12x=-3 .. its correct till here .
    (x+12)^2-144=-3 it should be halved
    (x+12)^2=141
    x+12=√141

    Then, I don't know what to do
    HI

    x^2+12x+3=0

    x^2 +12x=-3

    (x+6)^2=-3+6^2

    (x+6)^2=33

     <br />
x=\pm\sqrt{33}-6<br />
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by Mist View Post
    I'm finding it difficult to solve this question
    It's solving quadratic eqaution by completing the square

    x^2+12x+3=0

    This is what I have done so far ...

    x^2+12x+3=0
    x^2 +12x=-3
    (x+12)^2-144=-3
    (x+12)^2=141
    x+12=√141

    Then, I don't know what to do
    You have the right idea. However, it should be (x+6)^2, not (x+12)^2, you always take half of the coefficient of x.

    Also, when you square root your equation, you end up with two terms on the right hand side. For example \sqrt{4} would give you 4 and -4.

    You seem to be struggling with some of the basic mathematical operations, I suggest that you take some time to read through your notes/books on the subject.

    Hope this helps
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  4. #4
    Junior Member Mist's Avatar
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    Have I solved this question correctly?

    x^2-3x-5=0
    (x-1.5)^2-2.25=5
    x-1.5=√7.25
    x=1.5√7.25

    The answer seems wrong
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by Mist View Post
    Have I solved this question correctly?

    x^2-3x-5=0
    (x-1.5)^2-2.25=5
    x-1.5=√7.25
    x=1.5√7.25
    The answer seems wrong
    No.

    You started off right, however you went wrong from the highlighted point above.

    Because you are only considering the positive square root you are missing out a solution!

    I have x = 1.5 \pm\sqrt{7.25}

    You are not far off, just keep practising, and take your time, dont rush things when you are unsure about them.

    Hope this helps
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  6. #6
    MHF Contributor
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    Quote Originally Posted by Mist View Post
    Have I solved this question correctly?

    x^2-3x-5=0
    (x-1.5)^2-2.25=5
    x-1.5=√7.25 -- you missed the +/- in front of the square root .
    x=1.5√7.25 -- its + 1.5

    The answer seems wrong
    The rest is correct .
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  7. #7
    Super Member craig's Avatar
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    Quote Originally Posted by Mist View Post
    x=1.5√7.25
    Also here, do you mean 1.5 + \sqrt{7.25} or 1.5 \times \sqrt{7.25} ?
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  8. #8
    Junior Member Mist's Avatar
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    Stuck on another question

    How would I solve this?

    2x^2-7=4x

    This is what I have done:
    2x^2-7=4x
    2x^2-4x-7=0
    2(x-2)^2+4=7
    2(x-2)^2=11


    I don't know what to do after that
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  9. #9
    Super Member craig's Avatar
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    Quote Originally Posted by Mist View Post
    How would I solve this?

    2x^2-7=4x

    This is what I have done:
    2x^2-7=4x
    2x^2-4x-7=0
    2(x-2)^2+4=7
    2(x-2)^2=11


    I don't know what to do after that
    If there is a coefficient other that 1 in front of the x^2 term you need to remove it before completing the square.

    In the case of your equation, before you start rearranging the equation divide everything by 2, then proceed as normal.
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  10. #10
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    What I get

    x^2+12x+3=0

    (x+6)^2-33=0

    (x+6)^2=33

    x+6= rt33

    x= 6+- rt33

    p.s is there a way to get all the maths signs? I don't really wanna have to write rt lol.
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  11. #11
    RRH
    RRH is offline
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    you forgot to change the 6 to a negative number when you moved it to the other side of the equal sign

    <br /> <br />
x = -6 \pm\sqrt{33}<br />

    In addition, I need to learn how to properly format text
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  12. #12
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    Quote Originally Posted by RRH View Post
    you forgot to change the 6 to a negative number when you moved it to the other side of the equal sign

    <br /> <br />
x = -6 \pm\sqrt{33}<br />

    In addition, I need to learn how to properly format text
    Oh yehh darn , shouldn't have made that mistake.
    Last edited by mr fantastic; October 15th 2009 at 12:51 AM. Reason: m --> r
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  13. #13
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    Quote Originally Posted by alibond07 View Post
    ...

    p.s is there a way to get all the maths signs? I don't really wanna have to write rt lol.
    Go to the forum LaTex Help and have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html
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