Question

The number of integral solutions (x,y) of the equations \[x\sqrt y + y\sqrt x = 20\] and \[x\sqrt x + y\sqrt y = 65\] is

Answer 1

Hint: To find the number of integral solutions (x,y) of the given equations we first change the variables “x and y” into another variable to avoid the fraction power, then solve it by comparison of the equations which are newly formed.

Complete step-by-step answer:
The given equations are,
\[x\sqrt y + y\sqrt x = 20\] …… (1)
\[x\sqrt x + y\sqrt y = 65\]…… (2)
We need to find the number of integral solutions (x,y) of above equations.
Let, \[\sqrt x = a\& \sqrt y = b\]
On squaring on both side the variables we get,
\(x = {a^2}\& y = {b^2}\)
By changing the variables the equation (1) becomes,
\[{a^2}b + {b^2}a = 20\]
Let us take “\[ab\] ” in common we get,
\[ab(a + b) = 20\] …… (3)
Let the above equation be (3)
Now by changing the variable equation (2) becomes,
\[{a^2}a + {b^2}b = 65\]
On multiplying terms in L.H.S we get,
\[{a^3} + {b^3} = 65\]
We use the following algebraic identity
\[{a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)\], in the above equation and get,
\[{(a + b)^3} - 3ab(a + b) = 65\]
Let us substitute the value of \[ab(a + b)\] from (3) in the above equation, we get,
\[{(a + b)^3} - 3 \times 20 = 65\]
By solving the above equation,
\[{(a + b)^3} = 65 + 60\]
\[{(a + b)^3} = 125\]
Let us take cube root on both sides of the equation we get,
\[(a + b) = \sqrt[3]{{125}}\]
\[a + b = 5\]
Let us substitute \[a + b = 5\] in equation (3) we get,
\[ab = \dfrac{{20}}{5} = 4\]
Let us now substitute the values of a and b we get,
\[\sqrt x + \sqrt y = 5\]…… (4)
\[\sqrt x \sqrt y = 4\] …… (5)
Let us substitute the equation 4 in the 5 we get,
\[\sqrt x (5 - \sqrt x ) = 4\]
By multiplying terms in Left hand side we get,
\[5\sqrt x - {(\sqrt x )^2} = 4\]
Let us solve further to get the following quadratic equation,
\[{(\sqrt x )^2} - 5\sqrt x + 4 = 0\]
Let us solve the quadratic equation by middle term factor,
\[{(\sqrt x )^2} - 4\sqrt x - \sqrt x + 4 = 0\]
Let us take \[\sqrt x \] as a common term from first two terms and \[ - 1\] from last two terms we get,
\[\sqrt x (\sqrt x - 4) - 1(\sqrt x - 4) = 0\]
By taking \[\sqrt x - 4\] as common in the above equation we get,
\[(\sqrt x - 4)(\sqrt x - 1) = 0\]
Equating both the terms in the product to zero we get,
\[\sqrt x = 4{\rm{ or }}\sqrt x = 1\]
By squaring the above terms we get,
\[x = 16{\rm{ or }}x = 1\]
Let us substitute \[\sqrt x = 4\] in (5) we get,
\[\sqrt y = \dfrac{4}{4} = 1\]
By squaring the above equation we get,
\[y = 1\]
Hence,\[\;\left( {x,y} \right) = \left( {16,1} \right)\] is one of the integral solutions of the given equations.
Let us substitute \[\sqrt x = 1\] in equation (5) we get,
\[\sqrt y = \dfrac{4}{1} = 4\]
By squaring the above equation we get,
\[y = 16\]
Hence, \[\left( {x,y} \right) = \left( {1,16} \right)\] is the other integral solution.
Thus, The number of integral solutions (x,y) of the equations \[x\sqrt y + y\sqrt x = 20\] and \[x\sqrt x + y\sqrt y = 65\] is 2.

Note: We can solve any quadratic equation by middle term factor.
In Quadratic Factorization using Splitting of Middle Term which is x term is the sum of two factors and product equal to last term.
While changing the variables to “a” and “b” we should choose wisely so that the power of the given function reduces correctly. If we take \(x = a,y = b\) then the equation remains unchanged hence we should choose the change of variable correctly.