If x and y are real numbers, then solve this system of equations:
√x + √y =3,
√(x+5) + √(y+3) = 5
Let √x = a, let √y = b, then b=(3-a)
√(a^2+5) + √((3-a)^2+3) = 5
√(a^2+5)+√(a^2-6a+12) = 5
All sqrt terms will be collected after squaring, so let
(√(a^2+5) + √((3-a)^2+3)) = 25
2a^2+2sqrt((3-a)^2+3)sqrt(a^2+5)-6a+17 = 25
2sqrt((3-a)^2+3)sqrt(a^2+5) = 25+6a-17-2a^2
Square again to eliminate collected square roots:
4a^4-24a^3+68a^2-120a+240 = 4a^4-24a^3+4a^2+96a+64
68a^2-120a+240 = 4a^2+96a+64 Some nice cancelling
64a^2-216a+176 = 0
8 (a-2) (8a -11) = 0 Factor
Solutions a=2, a=11/8
But x=a^2, and y=(3-a)^2
So {x,y}= {4,1}, {121/64,169/64} and these are the only real solutions.
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Posted by broll
on 2024-07-13 04:26:50 |
Possible Solution