11) x2 - (y+4)x + (y2-4y+16) = 0
since x is real : D >= 0
(y+4)2 >= 4(y2-4y+16)
(y-4)2 <= 0
=> y = 4
so only solution is (4,4)
(1) find all real valuse of x and y that satisfy the equation
x2-xy+y2-4x-4y+16=0
(2) all psitive Integral solution of y2+6xy-16x=0
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6 Answers
1
341The first equation is equivalent to:
(x-y)^2+(x-4)^2+(y-4)^2 = 0.
Its easy to see that x=y=4 is the unique solution.
341Its obvious that y is even. So let y=2z.
Then we have z2=x(4-3z).
Since LHS>0 and x>0, we must have 4-3z>0. This happens only for z=1.
From this we get x=1.
Hence x=1, y=2 is the only solution among +ve integers