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Math Help - quadratic equation problem 2

  1. #1
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    quadratic equation problem 2

    6x^2 - x - 2=0
    6x^2 -4x +3x -2=0
    2x(3x-2)1(3x-2)=0
    (2x+1)=0 2x=-1 ,x=-1/2

    3x-2=0,x=-2/3

    I cant understand where I went wrong on this the answer is supposed to be x=-1.33 and x=1
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  2. #2
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    If x = 1, then 6x^2 - x - 2 = 6(1)^2 - (1) - 2 = 6 - 1 - 3 = 3, not zero.

    Either the book has a typo, or maybe you looked at the solution to the wrong exercise...?
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  3. #3
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    Quote Originally Posted by wolfhound View Post
    6x^2 - x - 2=0
    6x^2 -4x +3x -2=0
    2x(3x-2)1(3x-2)=0
    (2x+1)=0 2x=-1 ,x=-1/2

    3x-2=0,x=-2/3

    I cant understand where I went wrong on this the answer is supposed to be x=-1.33 and x=1
    (2x+1)(3x-2)=0

    x=-1/2 , x=2/3

    the answer is wrong .
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  4. #4
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    Quote Originally Posted by stapel View Post
    If x = 1, then 6x^2 - x - 2 = 6(1)^2 - (1) - 2 = 6 - 1 - 3 = 3, not zero.

    Either the book has a typo, or maybe you looked at the solution to the wrong exercise...?
    Sorry I mean x=-1.33 and x=1 as the final answers
    I didnt mean x=1 as the initial value!?
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    (2x+1)(3x-2)=0

    x=-1/2 , x=2/3

    the answer is wrong .
    I know its Wrong!
    That why I have asked for help here????
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  6. #6
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    Quote Originally Posted by wolfhound View Post
    I know its Wrong!
    That why I have asked for help here????
    i mean the answer of the book is wrong , not yours but of course yours is wrong too .
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    i mean the answer of the book is wrong , not yours but of course yours is wrong too .
    If mine is wrong then what is the correct way of doing it?
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  8. #8
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    Quote Originally Posted by wolfhound View Post
    If mine is wrong then what is the correct way of doing it?
    I just posted the correct answers . Well , you are almost correct but you added a negative sign to one of the roots . Refer back to my post .
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  9. #9
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    Quote Originally Posted by mathaddict View Post
    I just posted the correct answers . Well , you are almost correct but you added a negative sign to one of the roots . Refer back to my post .
    could you tell me how change the fractions into decimal numbers please?
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  10. #10
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    Quote Originally Posted by wolfhound View Post
    6x^2 - x - 2=0
    6x^2 -4x +3x -2=0
    2x(3x-2)+(3x-2)=0 <<<<<< here you made a tiny but very effective typo
    (2x+1)=0 2x=-1 ,x=-1/2

    3x-2=0,x=-2/3

    I cant understand where I went wrong on this the answer is supposed to be x=-1.33 and x=1
    6x^2 -4x +3x -2=0~\implies~2x(3x-2)+(3x-2)=0~\implies~(3x-2)(2x+1)=0 and so on ...
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  11. #11
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    Quote Originally Posted by earboth View Post
    6x^2 -4x +3x -2=0~\implies~2x(3x-2)+(3x-2)=0~\implies~(3x-2)(2x+1)=0 and so on ...
    thanks
    could you tell me if x=1/2 or x=2/3 is the correct answer??
    and how could I change these out of fractions please?
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