KVPY SA Exam 2012 Question Paper

x + 2y + 4z = 9 ;
2xy + 8yz + 4xz = 26 ;
(x) (2y) (4z) = 24
Say roots of ${\mathit{P}}^3-9{\mathit{P}}^2+26\mathit{P}-24=0$ are x, 2y and 4z
Here $\left(\mathit{P}-2\right)\left({\mathit{P}}^2-7\mathit{P}+12\right)=0$
(P – 2) (P – 3) (P – 4) = 0
now 4z = 4 ; 2y = 2 ; x = 3 then (3, 1, 1)
x = 2 ; 2y = 4 ; 4z = 3 then $\left(2,2,\frac{3}{4}\right)$
x = 3; 2y = 4 ; 4z = 2 then $\left(3,2,\frac{1}{2}\right)$
x = 2; 2y = 3 ; 4z = 4 then $\left(2,\frac{3}{2},1\right)$
x = 4; 2y = 2 ; 4z = 3 then $\left(4,1,\frac{3}{4}\right)$
Hence there are five solutions.