# The number of Integer solutions of the equations x *y1/2+y *x1/2=20 & x *x1/2 +y *y^1/2=65 isa.1  b.2 c.none d. 3

17 Points
14 years ago

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

mycroft holmes
272 Points
14 years ago

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