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

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

Postby Luke53 on Wed Apr 13, 2011 2:24 pm

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.
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Postby stapel_eliz on Wed Apr 13, 2011 9:49 pm

Luke53 wrote: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! :shock:

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

Postby Luke53 on Fri Apr 15, 2011 10:55 am

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!
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