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Math Help - [SOLVED] Quadratic Equation

  1. #1
    Member looi76's Avatar
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    [SOLVED] Quadratic Equation

    Question:
    Find the value of x when y = 0
    y = \frac{2x - 1}{x + 3}

    Attempt:

    (2x - 1)(x + 3)^{-1} = 0

    (2x - 1)(x^{-1} + \frac{1}{3}) = 0

    2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0

    \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0

    \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0

    \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0

    I think the quadratic equation that i got is wrong ... Where did I go wrong?
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  2. #2
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    Hello,

    Quote Originally Posted by looi76 View Post
    Question:
    Find the value of x when y = 0
    y = \frac{2x - 1}{x + 3}

    Attempt:

    (2x - 1)(x + 3)^{-1} = 0

    {\color{red}(2x - 1)(x^{-1} + \frac{1}{3}) = 0}

    2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0

    \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0

    \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0

    \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0

    I think the quadratic equation that i got is wrong ... Where did I go wrong?
    It's the step in red which is incorrect...

    (a+b)^n \not = a^n+b^n


    What you have to do is to say that x+3 has to be different of 0.
    Then solve :

    0=\frac{2x-1}{x+3}

    Multiplying both sides by x+3 (which is possible because x+3 is not = 0) :

    \boxed{0=2x-1}



    But I don't see why there is a quadratic equation here... ^^
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  3. #3
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    Quote Originally Posted by looi76 View Post
    Question:
    Find the value of x when y = 0
    y = \frac{2x - 1}{x + 3}

    Attempt:

    (2x - 1)(x + 3)^{-1} = 0

    (2x - 1)(x^{-1} + \frac{1}{3}) = 0

    2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0

    \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0

    \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0

    \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0

    I think the quadratic equation that i got is wrong ... Where did I go wrong?
    You've made very heavy weather of this. Simply solve 2x - 1 - 0 (why?).

    And it's good form to check that the solution is not a solution to x + 3 = 0 (why?)
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  4. #4
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    Quote Originally Posted by looi76 View Post
    Question:
    Find the value of x when y = 0
    y = \frac{2x - 1}{x + 3}
    this is what i did (dont know if it is right or not though)
    y = \frac{2x - 1}{x + 3}

    <br />
{2x - 1}= 0<br />

     <br />
{2x} = 1<br />

     <br />
{x} = 1/2<br />

    Zac
    Last edited by zp3929; May 2nd 2008 at 06:08 AM.
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  5. #5
    Moo
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    Quote Originally Posted by zp3929 View Post
    this is what i did (dont know if it is right or not though)
    y = \frac{2x - 1}{x + 3}

     <br />
{x + 3} = {2x - 1}<br />

     <br />
{-x + 4} = 0<br />

     <br />
{-x} = -4<br />

     <br />
{x} = 4<br />

    Zac
    What you did was for y=1
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  6. #6
    Moo
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    Quote Originally Posted by zp3929 View Post
    this is what i did (dont know if it is right or not though)
    y = \frac{2x - 1}{x + 3}

     <br />
{-x - 3} = {2x - 1}<br />

     <br />
{-3x -2} = 0<br />

     <br />
{-3x} = 2<br />

     <br />
{x} = -2/3<br />

    Zac
    Now, this is for y=-1
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  7. #7
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    Quote Originally Posted by Moo View Post
    Now, this is for y=-1
    i suck at this today
    ok 1 last try
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  8. #8
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    Quote Originally Posted by zp3929 View Post
    this is what i did (dont know if it is right or not though)
    y = \frac{2x - 1}{x + 3}

    <br />
{2x - 1}= 0<br />

     <br />
{2x} = 1<br />

     <br />
{x} = 1/2<br />

    Zac
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  9. #9
    Member looi76's Avatar
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    y = \frac{2x - 1}{x + 3}

    y = 0

    \frac{2x - 1}{x + 3} = 0

    2x - 1 = x + 3
    2x - 1 - x - 3 = 0
    x - 4 = 0
    x = 4

    The answer I have is x = \frac{1}{2} ?!
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  10. #10
    Member looi76's Avatar
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    Quote Originally Posted by zp3929 View Post
    this is what i did (dont know if it is right or not though)
    y = \frac{2x - 1}{x + 3}

    <br />
{2x - 1}= 0<br />

     <br />
{2x} = 1<br />

     <br />
{x} = 1/2<br />

    Zac
    Quote Originally Posted by mr fantastic View Post
    Why do we get rid of x + 3 ?
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  11. #11
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    Quote Originally Posted by looi76 View Post
    y = \frac{2x - 1}{x + 3}

    y = 0

    \frac{2x - 1}{x + 3} = 0

    2x - 1 = x + 3
    2x - 1 - x - 3 = 0
    x - 4 = 0
    x = 4

    The answer I have is x = \frac{1}{2} ?!
    yer it is x = \frac{1}{2}
    i edited my post (several times)

    we get rid of the x = 3 because x = 3/0 = 0 which leaves us with 2x - 1 = 0
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  12. #12
    Member looi76's Avatar
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    Quote Originally Posted by zp3929 View Post
    yer it is x = \frac{1}{2}
    i edited my post (several times)
    Thanks zp3929! I couldn't understand the reason for you removing x + 3 from the equation?
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  13. #13
    Moo
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    Quote Originally Posted by looi76 View Post
    Thanks zp3929! I couldn't understand the reason for you removing x + 3 from the equation?
    Told you, it's because you multiply each side by x+3 ^^
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