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Vaibhav Mathur Grade: 12
        

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

8 years ago

Answers : (2)

AskIITians Expert Hari Shankar IITD
17 Points
										

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


 

8 years ago
mycroft holmes
266 Points
										

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


 

8 years ago
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