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Hard simultaneous equations

  1. Jan 28, 2010 #1
    Hi

    How do you solve this equations analyticaly without solving a 4th degree polynomial
    sqrt(x)+y=11; x+sqrt(y)=7

    Thanks
     
  2. jcsd
  3. Jan 28, 2010 #2
    I'll give a general outline on how to solve this:

    First of all, use new variables. a = sqrt(x) and b = sqrt(y). That will remove those ugly square roots for now.

    Then, add the 2 equations and form a new equation a + a² + b + b² = 18. In this equation, 18 can be split up into 2 unknown values c and d to form 2 quadratic equations. Of these 2 values, you only need to know 1 to calculate the other, that is d = 18 - c. The roots of these 2 quadratic equations can be written in terms of these 2 values. Keeping in mind that the solutions can only be positive, that should lead you to the right answer
     
    Last edited: Jan 28, 2010
  4. Jan 28, 2010 #3

    HallsofIvy

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    I may be misunderstanding but you seem to be saying that if you have [itex]a+ a^2+ b+ b^2= 18= c+ d[/itex] then you must have [itex]a+ a^2= c[/itex] and [itex]b+ b^2= d[/itex]. That is not true.
     
    Last edited: Jan 28, 2010
  5. Jan 28, 2010 #4
    Do u want to explain why this is not true? If I define c as being a+a², and d as being b+b², then a + a² + b + b² = c + d = 18

    Edit: nvm, you probably did misunderstand me^^
     
  6. May 8, 2010 #5
    can you plees put full steps for this equation plees
     
  7. Sep 24, 2010 #6
    Hello I'm 14 years old so i don't know what a 4th degree polynomial is... however i agree with the guy named Kyouran. This is how i did it:

    sqrt(x) + y = 11 ...(1)
    sqrt(y) + x = 7 ...(2)

    Let sqrt(x) = a, and sqrt(y) = b.

    The eqs will then be:
    a + b^2 = 11 ...(3)
    b + a^2 = 7 ...(4)

    Add them together and you get:

    a + a^2 + b + b^2 = 18

    The way i solved it is just by doing guess and check. This only takes common sense:

    I first tried a = 2 and b = 3 which when substituted into the equation equals 18:

    2 + 2^2 + 3 + 3^2 = 18

    However when subbing them into eqs 3 and 4, they weren't correct:

    So i then switched the values around saying a = 3, and b = 2 which when subbed into eqs 3 and 4 were correct!

    And so, as said before...

    Since I let sqrt(x) = a, and the sqrt(y) = b:

    sqrt (x) = 3
    and so therefore x = 9

    sqrt (y) = 2
    and so therefore y = 4

    therefore x = 9, y = 4 ;)

    Now you will find, that when subbing these x and y values into your original eqs (1) and (2), they are sure to be correct!
     
    Last edited: Sep 24, 2010
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