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Asinense Asinense
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11 years ago
Hi

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

Thanks
second equation is sqrt(y)+x
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#1
11 years ago
This can't be solved.

y = 11 - sqrt(x)

y = 7 - sqrt(x)

11 - sqrt(x) = 7 - sqrt(x)

11 = 7

not true

Well, we sort of solved it without solving a 4th degree equation.
wrote...
#2
11 years ago
sqrt(x)+y=11;
 y+sqrt(x)=7

They are the same equation (ie, they are both y+sqrt(x))
They cannot equal two things at once, therefore I assume you've made a typing error.
wrote...
#3
11 years ago
im assuming the second equation should be x+sqrt(y)=7

So, in order for two numbers added together to be rational, each number must be rational.  thus, sqrt(y) and sqrt(x) must be rational.  some possible addition pairs to get seven are:

1,6
2,5
4,3

whichever we choose for x, if we square it it must be less than 11 due to the first equation.  so, we could have

6+sqrt(1)=7
5+sqrt(4)=7
4+sqrt(9)=11

Using x=4, y=9 is the only of these three which also satisfies the first equation.
wrote...
#4
11 years ago
the given equations can be recast as a+b^2=11 .. and b+a^2=7...[ii] for a,b belonging to Z+.
now b={0,1,2,3,4,5,6,7}
test for those values of b for which a^2 is perfect square in eq ..[ii]
a^2=7-b is perfect iff b=3.
which yields a=2
now the above substitutions hold true in eq
hence x=4,y=9
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