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The number of Integer solutions of the equations x *y1/2+y *x1/2=20 & x *x1/2 +y *y^1/2=65 is a.1 b.2 c.none d. 3
The number of Integer solutions of the equations x *y1/2+y *x1/2=20 & x *x1/2 +y *y^1/2=65 is
a.1 b.2 c.none d. 3
Hi Let p=root(x) and q=root(y). Then, the given equations become p2q+pq2=20 and p3+q3=65 or, pq(p+q)=20 and p3+q3=65. Use the identity p3+q3=(p+q)3-3pq(p+q) in the second equation. From the first equation, we already know that pq(p+q)=20. So the second equation becomes (p+q)3-3(20)=65 (Since pq(p+q)=20 ) So, (p+q)3= 60+65=125 Hence, p+q = 125^(1/3) = 5. Putting this back into pq(p+q)=20, we get pq=4. Or, root(xy) = 4. So, xy=16. Hence, the only possible integer solutions are (1,16),(2,8),(4,4),(8,2),(16,1) Out of these, only (1,16) and (16,1) satisfy the equations. So we have 2 solutions. Option b
Hi
Let p=root(x) and q=root(y).
Then, the given equations become
p2q+pq2=20
and p3+q3=65
or,
pq(p+q)=20 and p3+q3=65.
Use the identity p3+q3=(p+q)3-3pq(p+q) in the second equation.
From the first equation, we already know that pq(p+q)=20.
So the second equation becomes
(p+q)3-3(20)=65 (Since pq(p+q)=20 )
So, (p+q)3= 60+65=125
Hence, p+q = 125^(1/3) = 5.
Putting this back into pq(p+q)=20, we get pq=4.
Or, root(xy) = 4. So, xy=16.
Hence, the only possible integer solutions are (1,16),(2,8),(4,4),(8,2),(16,1)
Out of these, only (1,16) and (16,1) satisfy the equations. So we have 2 solutions. Option b
Squaring the equations and subtracting, we get (x-y)2 (x+y) = 85 X 45 = 17 X 52 X 32 Hence (x-y)2 = 25 and x+y = 153 or (x-y)2 = 9 and x+y = 425 or (x-y)2 = 152 and x+y = 17 Its easy to verify that x=16, y = 1 and x=1 and y = 16 are the only solutions
Squaring the equations and subtracting, we get (x-y)2 (x+y) = 85 X 45 = 17 X 52 X 32
Hence (x-y)2 = 25 and x+y = 153 or
(x-y)2 = 9 and x+y = 425 or
(x-y)2 = 152 and x+y = 17
Its easy to verify that x=16, y = 1 and x=1 and y = 16 are the only solutions
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