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Find all reals:

  1. Sep 7, 2008 #1
    find all real values such that:
    [tex]x+y=1[/tex]
    and [tex]xy-z^2=1[/tex]

    This one was in one of the exams we had.
     
  2. jcsd
  3. Sep 7, 2008 #2

    HallsofIvy

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    Nothing at all difficult about that, except that I presume you mean "find all 'triples' of real numbers, (x, y, z) that satisfy the equations". From x+ y= 1, y= 1- x. Putting that for y in the second equation, x(1- x)- z2= x- x2- z2= 1 which we can rewrite as the quadratic equation x2- x- (z2+ 1)= 0. Now we can solve that for x, as a function of z, using the quadratic formula:
    [tex]x= \frac{-(-1)\pm\sqrt{(-1)^2- 4(1)(-(z^2+ 1)}}{2}= \frac{1\pm \sqrt{8z^2+ 9}}{2}[/tex]
    Now, since 8z2+ 9 is positive for all z, z can be any number, x can be calculated from that formula, and then y can be calculated from y= 1- x.
     
  4. Sep 7, 2008 #3

    gel

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    As x must be positive, that can't be right.

    The problem is easy to see graphically. The set of values for (x,y) is just the intersection of the region xy>=1 with the line x + y = 1, which gives a closed line segment.
     
  5. Sep 7, 2008 #4

    disregardthat

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    [tex]x^2+2xy+y^2=1[/tex]
    [tex]x^2+y^2+2(1+z^2)=1[/tex]
    [tex]x^2+y^2+2z^2=-1[/tex]

    or

    We observe that both y and x must be positive.

    [tex](x+y)^2/4>=xy[/tex], so [tex]xy<=1/4[/tex]. (AM-GM)
    [tex]z^2=xy-1<=-3/4[/tex], so no solutions for z if x and y are real.
     
  6. Sep 7, 2008 #5

    disregardthat

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    How on earth can you deduce that?
     
  7. Sep 7, 2008 #6

    Hurkyl

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    z2 is nonnegative.
     
  8. Sep 7, 2008 #7

    disregardthat

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    Of course.
    Intersection of the region xy>=1 and the line x+y=1 :

    y>=1/x, and y=1-x Suppose 1-x > 1/x, then 1/x+x<1 which is impossible! (as x is positive)
     
  9. Sep 7, 2008 #8

    Hurkyl

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    Right. I find it slightly quicker to visualize the result -- xy >= 1 fills in the top-right and bottom-left parts of the hyperbola xy=1... and x+y=1 passes right between the two branches.
     
  10. Sep 7, 2008 #9

    disregardthat

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    Either way, the closed line segment doesn't exist, right?
     
  11. Sep 7, 2008 #10
    The question has no solution.... Consider the following argument...


    x + y = 1
    xy = 1 + z^2

    Now if x > 1, then
    y = 1 - x => y < 0

    This gives x > 1, y < 0, Thus xy < 0. But since xy = 1 + z^2, xy must be positive.

    Thus x <= 1 and y <= 1

    Now if x < 0, then
    y = 1 - x => y > 1

    In this case x < 0 and y > 1. Therefore xy < 0.
    Thus x >= 0 and y >= 0.

    Taking in both arguments
    0 <= x <= 1
    and 0 <= y<= 1.

    In this case xy <= 1. However since xy = z^2 + 1, xy must be greater than or equal to 1. The only possibility is xy = 1

    Solve for

    x+y = 1
    xy = 1

    Since no real value satisfies this equation, no solution exists
     
    Last edited: Sep 7, 2008
  12. Sep 7, 2008 #11
    Well, u made a mistake. x(1-x) - z2 = 1 does not reduce to x^2 - x - (z^2 - 1) = 0. Infact, the equation is [tex]x^2 - x + z^2 + 1 = 0[/tex].

    [tex]D = (-1)^2 - 4(1)(z^2 + 1) = 1 - 4 - 4z^2 = -3 - 4 z^2[/tex] which is negative for all values of z. Thus, no real solution exists
     
  13. Sep 7, 2008 #12

    arildno

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    Thanks for clearing that up,prahamirta.
    Another way of seeing this is by rewriting the correct quadratic as
    [tex](x-\frac{1}{2})^{2}+z^{2}=-\frac{3}{4}[/tex]
     
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