Solve the given simultaneous equation

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The discussion revolves around solving a simultaneous equation involving square roots, specifically ##\sqrt{x} + y = 7## and ##x + \sqrt{y} = 11##. Participants explore various methods, including an alternative approach that leads to a quartic equation, ##m^4 - 14m^2 + m + 38 = 0##, where ##m = \sqrt{y}##. There is a focus on finding real roots and using polynomial division to simplify the problem, with hints about numerical methods and the Factor theorem. The conversation also touches on the nature of polynomials, discussing the likelihood of having real versus complex roots, ultimately leading to a confirmed solution of ##x = 3## and ##y = 2##.
  • #31
Orodruin said:
Since both sets are aleph one, it is possible to construct a bijection between them.
I was thinking about that too. I recalled a problem in an real analysis book of mine which I did not read that long time ago which asked how many algebraic vs. trancendetal numbers (over the rationals) there are. If I remember correctly, there should be countable many (##\aleph_0##) algebraic numbers and uncountable (##\aleph_1##) trancendental numbers... I should revisit my notes at some points
 
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  • #32
Delta2 said:
Hmm, can we say that the probability a random polynomial to have only real roots is 50%? Hard though to imagine how we define this probability as if we try to define it as a fraction, both the numerator and the denominator are uncountable many...
I guess we'd have to choose a distribution to determine probabilities.
 
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  • #33
chwala said:
Homework Statement:: Solve for ##x## and ##y## given;
##\sqrt x+ y=7##
##x+\sqrt y=11##
Relevant Equations:: Simultaneous equations
Presuming this is a Diophantine (integer only) equation , we don't have too many choices:

1 + 6
2 + 5
3 + 4

One number has to be a perfect square. Exclude 2 + 5. If x=1, y=6 and that doesn't work either.
 
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