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Thread: [SOLVED] Quadratic Equation

  1. #1
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    [SOLVED] Quadratic Equation

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

    Attempt:

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

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

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

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

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

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

    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 $\displaystyle x$ when $\displaystyle y = 0$
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    Attempt:

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

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

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

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

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

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

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

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


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

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

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

    $\displaystyle \boxed{0=2x-1}$



    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 $\displaystyle x$ when $\displaystyle y = 0$
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    Attempt:

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

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

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

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

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

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

    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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    Quote Originally Posted by looi76 View Post
    Question:
    Find the value of $\displaystyle x$ when $\displaystyle y = 0$
    $\displaystyle y = \frac{2x - 1}{x + 3}$
    this is what i did (dont know if it is right or not though)
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    $\displaystyle
    {2x - 1}= 0
    $

    $\displaystyle
    {2x} = 1
    $

    $\displaystyle
    {x} = 1/2
    $

    Zac
    Last edited by zp3929; May 2nd 2008 at 05:08 AM.
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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)
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    $\displaystyle
    {x + 3} = {2x - 1}
    $

    $\displaystyle
    {-x + 4} = 0
    $

    $\displaystyle
    {-x} = -4
    $

    $\displaystyle
    {x} = 4
    $

    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)
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    $\displaystyle
    {-x - 3} = {2x - 1}
    $

    $\displaystyle
    {-3x -2} = 0
    $

    $\displaystyle
    {-3x} = 2
    $

    $\displaystyle
    {x} = -2/3
    $

    Zac
    Now, this is for y=-1
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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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    Quote Originally Posted by zp3929 View Post
    this is what i did (dont know if it is right or not though)
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    $\displaystyle
    {2x - 1}= 0
    $

    $\displaystyle
    {2x} = 1
    $

    $\displaystyle
    {x} = 1/2
    $

    Zac
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  9. #9
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    $\displaystyle y = \frac{2x - 1}{x + 3}$

    $\displaystyle y = 0$

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

    $\displaystyle 2x - 1 = x + 3$
    $\displaystyle 2x - 1 - x - 3 = 0$
    $\displaystyle x - 4 = 0$
    $\displaystyle x = 4$

    The answer I have is $\displaystyle x = \frac{1}{2}$ ?!
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  10. #10
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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)
    $\displaystyle y = \frac{2x - 1}{x + 3}$

    $\displaystyle
    {2x - 1}= 0
    $

    $\displaystyle
    {2x} = 1
    $

    $\displaystyle
    {x} = 1/2
    $

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

    $\displaystyle y = 0$

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

    $\displaystyle 2x - 1 = x + 3$
    $\displaystyle 2x - 1 - x - 3 = 0$
    $\displaystyle x - 4 = 0$
    $\displaystyle x = 4$

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

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