Question

The number of solution (x, y, z) to the system of equations
x + 2y + 4z = 9
4yz + 2xz + xy = 13
xyz = 3
Such that at least two of x, y, z are integers is
A. 3
B. 5
C. 6
D. 4

Answer 1

Hint: Here we will first we let $\alpha = x,\beta = 2y,\gamma = 4z$ and using this we will use the given three equations to find an equation whose roots are $\alpha ,\beta {\text{ and }}\gamma $ and then find the roots of that equation. Finally, we will take the various combinations of roots to find the required values of x, y and z.

Complete step-by-step answer:
The given three equations are:
x + 2y + 4z = 9 (1)
4yz + 2xz + xy = 13 (2)
xyz = 3 (3)
First we let $\alpha = x,\beta = 2y,\gamma = 4z$
So, equation 1 can be written as:
$\alpha + \beta + \gamma = x + 2y + 4z = 9$ (4)
Now, on calculating the value of $\alpha \beta + \beta \gamma + \gamma \alpha $, we have:
$\alpha \beta + \beta \gamma + \gamma \alpha = 2xy + 8yz + 4xz$

So, equation 2 can be written as:
$\alpha \beta + \beta \gamma + \gamma \alpha = 2xy + 8yz + 4xz = 2\left( {xy + 4yz + 2xz} \right) = 2 \times 13 = 26$. (5)
Now, on finding product of $\alpha ,\beta {\text{ and }}\gamma $, we have:
$\alpha \beta \gamma = 8xyz = 8 \times 3 = 24$ (6)
As we know that any cubic polynomial in ‘P’ with $\alpha ,\beta \& \gamma $ as its roots can be written as
${p^3} - \left( {\alpha + \beta + \gamma p} \right){p^2} + \left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)p - \alpha \beta \gamma = 0$
Using equations 4, 5 and 6 , we can write the cubic equation as:
$ \Rightarrow {p^3} - 9{p^2} + 26p - 24 = 0$
The cubic polynomial in ‘P’ can be solved by a hit and trial method.
$p(2) = {2^3} - 9 \times {2^2} + 26 \times 2 - 24 = 60 - 60 = 0$
$p(3) = {3^3} - 9 \times {3^2} + 26 \times 3 - 24 = 105 - 105 = 0$
$p(4) = {4^3} - 9 \times {4^2} + 26 \times 4 - 24 = 168 - 168 = 0$
This cubic polynomial is satisfied when we put p = 2 or p = 3 or p = 4.
This means that these are the roots of the cubic polynomial.
Now, the different cases formed are:
$1.\left( {\alpha ,\beta ,\gamma } \right) = \left( {2,3,4} \right) \Rightarrow \left( {x,2y,4z} \right) = \left( {2,3,4} \right) \Rightarrow \left( {x,y,z} \right) = \left( {2,\dfrac{3}{2},1} \right))$$2.\left( {\alpha ,\beta ,\gamma } \right) = \left( {2,4,3} \right) \Rightarrow \left( {x,2y,4z} \right) = \left( {2,4,3} \right) \Rightarrow \left( {x,y,z} \right) = \left( {2,2,\dfrac{3}{4}} \right)$ $3.\left( {\alpha ,\beta ,\gamma } \right) = \left( {3,2,4} \right) \Rightarrow \left( {x,2y,4z} \right) = \left( {3,2,4} \right) \Rightarrow \left( {x,y,z} \right) = \left( {3,1,1} \right)$ $4.\left( {\alpha ,\beta ,\gamma } \right) = \left( {3,4,2} \right) \Rightarrow \left( {x,2y,4z} \right) = \left( {3,4,2} \right) \Rightarrow \left( {x,y,z} \right) = \left( {3,2,\dfrac{1}{2}} \right)$ $5.\left( {\alpha ,\beta ,\gamma } \right) = \left( {4,2,3} \right) \Rightarrow \left( {x,2y,4z} \right) = \left( {4,2,3} \right) \Rightarrow \left( {x,y,z} \right) = \left( {4,1,\dfrac{3}{4}} \right)$ $6.\left( {\alpha ,\beta ,\gamma } \right) = \left( {4,3,2} \right) \Rightarrow \left( {x,2y,4z} \right) = \left( {4,3,2} \right) \Rightarrow \left( {x,y,z} \right) = \left( {4,\dfrac{3}{2},\dfrac{1}{2}} \right)$
Clearly the possible solutions (x, y, z) to the system of equations for which at least two of x, y, z are integers are (1), (2), (3), (4), (5).
So total 5 numbers of solutions (x, y, z) to the system of equations are possible for which at least two of x, y, z are integers.

So, the correct answer is “Option B”.

Note: These types of problems are usually solved by converting the system of equations into a polynomial with the help of the sum of roots, product of roots and sum of product of more than one root at a time. The number of roots of a polynomial is equal to the degree of that polynomial.