## Radical eqn. system: sqrt(x) + sqrt(y) = 25, x + y = 533

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Luke53
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### Radical eqn. system: sqrt(x) + sqrt(y) = 25, x + y = 533

How to solve this system?
sqrt(x) + sqrt(y) = 25
x + y = 533
Squaring both sides of the first eqn. gives x + y = 625, but given the second eqn. : x + y = 533
625 is not 533!!!
(I know the solution is x = 529 and y = 4 or x = 4 and y = 529.
So, how is this done?
Greetings.

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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sqrt(x) + sqrt(y) = 25
x + y = 533
Squaring both sides of the first eqn. gives x + y = 625
No; the square of (a + b) is not a2 + b2!

Besides reviewing how to multiply polynomials, it might be helpful to start by solving the second equation for one of the variables, and then substitute into the first equation. Then square both sides.

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
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### Re: Radical eqn. system: sqrt(x) + sqrt(y) = 25, x + y = 533

OK thanks, I can see where I went wrong.
Squaring both sides of the first eqn. sqrt(x) + sqrt(y) = 25 gives: x + y + 2 *(sqrt(x) *sqrt(y)) = 625
The second eqn. gives me: x + y = 533
So: 2 * (sqrt(x) *sqrt(y)) = 625 - 533 and so: sqrt(x) * sqrt(y) = 46
Squaring this on both sides gives: x * y = 2116
Out of the system: x * y = 2116 and x + y = 533 I get the quadratic eqn.: -(y^2) + 533 y - 2116 = 0 ; solving it yields: y = 4 or y = 529 ; (this is what I'm looking for).
I've got the same result doing it the way you suggested me:
Out of enq. (2): x = 533 - y and putting this in eqn. (1), yields: sqrt(533 - y) + sqrt(y) = 25
Sqaring both sides gives: 533 - y + 2*(sqrt(533 - y) * sqrt(y)) + y = 625 the y's cancelling out)
2 * qsrt(533 - y) * sqrt(y) = 92 so sqrt(533 -y) * sqrt(y) = 46
Squaring both sides of the last eqn. gives me : (533 - y) * y = 2116 => -(y^2) + 533y - 2116 = 0
I knew the results of that eqn.
Great stuff!