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How to solve equations of the form (a*x+b)^(1/2)+(m*x+n)^(1/2)=c

  1. Jul 12, 2012 #1
    Hello,
    I was wondering how you would go about solving for x in an equation like [itex]\sqrt{ax+b}[/itex]+[itex]\sqrt{mx+n}[/itex]=C (where a,b,m, and n are constant numbers). The problem is if you square the expression you just end up with some linear terms multiplied by terms to the power of 1/2. If you keep squaring you never get rid of them. So how do you go about solving something like this?

    Thanks for any help
     
  2. jcsd
  3. Jul 12, 2012 #2
    (a*x+b)^(1/2)+(m*x+n)^(1/2)=c
    (a*x+b)+(m*x+n)+2*[(a*x+b)*(m*x+n)]^(1/2)=c²
    2*[(a*x+b)*(m*x+n)]^(1/2)=c² -(a*x+b)-(m*x+n)
    4*(a*x+b)*(m*x+n)=[c² -(a*x+b)-(m*x+n)]²
    Expand and solve (...)*x²+(...)*x+(...)=0
    Then bring back the roots x into the first equation in order to check if each root is valid or not.
     
  4. Jul 12, 2012 #3
    After squaring both sides, move all the terms without square root on one side and leave the square root alone on the other side (of the equal sign). Then square again.
     
  5. Jul 12, 2012 #4
    Thank you. I just kept trying to simplify the left hand after squaring, i didnt even think about moving everything over. I feel kind of silly now.
     
  6. Jul 13, 2012 #5

    HallsofIvy

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    Welcome to the club!
     
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