Solve the given simultaneous equation

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SUMMARY

This discussion focuses on solving the simultaneous equations involving square roots: ##\sqrt{x} + y = 7## and ##x + \sqrt{y} = 11##. Participants explore various methods, including substituting ##\sqrt{y} = m## to derive the quartic equation ##m^4 - 14m^2 + m + 38 = 0##. They discuss techniques for solving quartic equations, such as polynomial division and the Factor Theorem, ultimately confirming that the polynomial has four real roots. The conversation emphasizes the importance of numerical methods and trial-and-error approaches in finding solutions.

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  • Understanding of simultaneous equations and algebraic manipulation.
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  • Knowledge of polynomial division and the Factor Theorem.
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  • #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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