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Thread: Solve equation for all real numbers a

  1. #1
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    Solve equation for all real numbers a

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    Super Member Quacky's Avatar
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    Re: Solve equation for all real numbers a

    Your first step is to put everything over a common denominator. Factor $\displaystyle x^2-3x$ and see if you can work out what the common denominator is. Then, once you've done that, remember that the only way a fraction can be equal to $\displaystyle 0$ is if the numerator is equal to $\displaystyle 0$. This will give you an equation to solve for $\displaystyle a$.

    If you require further help, please show what you can do and we'll guide you from there.
    Last edited by Quacky; Oct 16th 2011 at 03:47 AM.
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  3. #3
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    Re: Solve equation for all real numbers a

    Okay, so x(x-3) is the common denominator. Which gives me (x^2 + ax-3a + x-12)/ x(x-3) =0 , or (x^2 + x(a+1) - 3(a+4))/ x(x-3).

    If I solve the numerator for x, I get x=3, and x=-a-4.

    Should I try to solve the equation first with x=3, and then x=-a-4?
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  4. #4
    Super Member Quacky's Avatar
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    Re: Solve equation for all real numbers a

    Quote Originally Posted by cristianodoni72 View Post
    Okay, so x(x-3) is the common denominator. Which gives me (x^2 + ax-3a + x-12)/ x(x-3) =0 , or (x^2 + x(a+1) - 3(a+4))/ x(x-3).

    If I solve the numerator for x, I get x=3, and x=-a-4.

    Should I try to solve the equation first with x=3, and then x=-a-4?
    Great work so far! You should notice quickly that x=3 cannot be a solution. Why?
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    Re: Solve equation for all real numbers a

    Thanks! Yes, with x=3, the denominator is 0. So I put x=-a-4, which gives me the numerator 2a-4 and the denominator a^2 +11a+28.
    Do I solve the numerator for a now?
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  6. #6
    Super Member Quacky's Avatar
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    Re: Solve equation for all real numbers a

    Quote Originally Posted by cristianodoni72 View Post
    Thanks! Yes, with x=3, the denominator is 0. So I put x=-a-4, which gives me the numerator 2a-4 and the denominator a^2 +11a+28.
    Do I solve the numerator for a now?
    You're overcomplicating things and getting slightly muddled. We have learned that $\displaystyle x=-a-4$. Therefore, for all real values of $\displaystyle a$,

    $\displaystyle a=-4-x$

    We can't simplify any further.
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    Re: Solve equation for all real numbers a

    Okay! Thanks for guiding me!
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