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Math Help - Solve an equation

  1. #1
    Senior Member Mukilab's Avatar
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    Solve an equation

    Solve
    \ \frac{3}{(y-1)^2}+\frac{3}{2(y-1)^2}=2

    I tried most things but get stuck at \frac{9}{4}=(y-1)^2 pretty much every time. Any help?
    Last edited by Mukilab; June 5th 2010 at 12:53 PM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    Solve
    \frac{3}{(y-1)^2}+\frac{3}{2(y-1)^2}=2

    I tried most things but get stuck at \frac{9}{4}=(y-1)^2 pretty much every time. Any help?
    You're fine up to there, you can complete the square to find y. Remember that \frac{9}{4} = \left(\frac{3}{2}\right)^2
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  3. #3
    Senior Member Mukilab's Avatar
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    How can I complete the square with that!?

    I thought completing the square is along the lines of

    2x^2+4x+13=(2x^2+4x+4)-11

    o_o

    What can I do with {y-1}^2?

    y^2-2y+1

    completing the square

    (y^2-2y+1)+1-1=9/4
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    How can I complete the square with that!?

    I thought completing the square is along the lines of

    2x^2+4x+13=(2x^2+4x+4)-11

    o_o

    What can I do with {y-1}^2?

    y^2-2y+1

    completing the square

    (y^2-2y+1)+1-1=9/4
    (y-1)^2 is already in completed square form so you can take the square root. In other words you can take the square root of both sides and then add 1.
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  5. #5
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    (y-1)^2 is already in completed square form so you can take the square root. In other words you can take the square root of both sides and then add 1.
    I've done that already but I got 2 and 4 which don't work if you put them into the equation
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  6. #6
    Senior Member Mukilab's Avatar
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    OOhhhhh


    Thanks, I forgot when I did it you have to take the square root on both sides! heh.


    Don't I need an exact answer though? Not allowed a calculate

    final:
    y=\sqrt{\frac{9}{4}}+1
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  7. #7
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Remember that \frac{9}{4} = \left(\frac{3}{2}\right)^2
    Quote Originally Posted by Mukilab View Post
    OOhhhhh


    Thanks, I forgot when I did it you have to take the square root on both sides! heh.


    Don't I need an exact answer though? Not allowed a calculate

    final:
    y=\sqrt{\frac{9}{4}}+1
    See post 2 (quoted at the top of this post) for how to eliminate the square root. Also you still need the \pm sign.

    edit: that answer is correct, it just needs simplifying and also the negative solution
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  8. #8
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    See post 2 (quoted at the top of this post) for how to eliminate the square root. Also you still need the \pm sign.

    edit: that answer is correct, it just needs simplifying and also the negative solution
    <br />
y=\pm\sqrt{\frac{13}{4}} ????

    sorry
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  9. #9
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    Quote Originally Posted by Mukilab View Post
    y=\sqrt{\frac{9}{4}}+1
    That's 3/2 + 1 = 5/2
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  10. #10
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    <br />
y=\pm\sqrt{\frac{13}{4}} ????

    sorry
    Don't ever be sorry for asking for help

    \pm \sqrt{\frac{9}{4}} = \pm \sqrt{\left(\frac{3}{2}\right)^2}

    As \pm \sqrt{a^2} = \pm a that square root can be reduced

    You should get y = \pm \frac{3}{2} + 1 hence y = \frac{5}{2} \text{ or  } y = -\frac{1}{2}


    We can't add the 1 under the square root since \sqrt{a} + \sqrt{b} \neq \sqrt{a+b}
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  11. #11
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Don't ever be sorry for asking for help

    \pm \sqrt{\frac{9}{4}} = \pm \sqrt{\left(\frac{3}{2}\right)^2}

    As \pm \sqrt{a^2} = \pm a that square root can be reduced

    You should get y = \pm \frac{3}{2} + 1 hence y = \frac{5}{2} \text{ or  } y = -\frac{1}{2}


    We can't add the 1 under the square root since \sqrt{a} + \sqrt{b} \neq \sqrt{a+b}
    Thanks for your perserveerance
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