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Math Help - Simultaneous Equations with square root

  1. #1
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    Simultaneous Equations with square root

    Hi, I'm trying to find the solution to \sqrt{x}=x-6 but i can't for the life of me figure out how to do it, it's the square root that has me stumped. I've managed to get it into a form of  x^2 +x+-36 and found the roots of approximately 5.5 and -6.5 which i then substituted into the y=mx+c form equations but still couldnt get the right answer of 4,2.
    So I've returned again seeking help, anyone care to take a stab?
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  2. #2
    Moo
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    Quote Originally Posted by mattty View Post
    Hi, I'm trying to find the solution to \sqrt{x}=x-6 but i can't for the life of me figure out how to do it, it's the square root that has me stumped. I've managed to get it into a form of  x^2 +x+-36 and found the roots of approximately 5.5 and -6.5 which i then substituted into the y=mx+c form equations but still couldnt get the right answer of 4,2.
    So I've returned again seeking help, anyone care to take a stab?
    x=(\sqrt{x})^2
    So the equation is now y=y^2-6, where y=\sqrt{x}

    now can you solve it ? don't forget that y is positive since the square root of a real number is always positive.
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    to be honest... no, i still can't. I suppose I should have put in more information in my original post, but that was one of the steps i went through to get to x^2 +x-36 I suppose the real problem I'm having is when i get x+\sqrt{x}=-6 because i can't figure out how to bring it down to a single x term.
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  4. #4
    Moo
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    Quote Originally Posted by mattty View Post
    to be honest... no, i still can't. I suppose I should have put in more information in my original post, but that was one of the steps i went through to get to x^2 +x-36 I suppose the real problem I'm having is when i get x+\sqrt{x}=-6 because i can't figure out how to bring it down to a single x term.
    Following what I have said, put them all on the same side :
    you get : y^2-y-6=0 (I definitely don't know how you get yours)

    Now this has two roots (which you get by using whatever method you want) : 3 and -2
    Since y>0, y=3
    So x=+ or - sqrt(3)
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    hmmm, that still doesn't satisfy the solution of 4,2 (  x=4, y=2) that I gave in my original post. According to the graph on my calculator, 4,2 is the point where the 2 equations meet. Was I asking for the wrong thing perhaps?
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  6. #6
    Moo
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    Quote Originally Posted by Moo View Post
    Following what I have said, put them all on the same side :
    you get : y^2-y-6=0 (I definitely don't know how you get yours)

    Now this has two roots (which you get by using whatever method you want) : 3 and -2
    Since y>0, y=3
    So x=+ or - sqrt(3)
    I made a very stupid mistake.........................

    So we have y^2-y-6=0
    This means that y=3 or y=-2
    But x=y (and not the contrary)
    So x=9 or 4.


    If you substitute in \sqrt{x}=x-6, 9 will work, but not 4, because \sqrt{4} \neq 4-6
    So 9 is a solution


    I didn't understand this was for finding the intersection of two curves... So you want to find the intersection between y=sqrt(x) and y=x-6 ?
    9,3 is the correct answer. And I checked it on a graph...

    It would have been nice to put the whole problem ^^' to avoid confusion. And I'm sorry for the one I created.
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  7. #7
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    Edit: i'll start a new thread tomorrow, thanks for the help, misguided as it was, Moo
    Last edited by mattty; February 21st 2009 at 05:57 AM. Reason: Tired
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