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The number of solutions (x, y, z) to the system of equations `x+2y+4z=9,4yz+2xy=13,xyz=3` Such that at least two of x, y, z are integers is -A. 3B. 5C. 6D. 4 |
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Answer» Correct Answer - B `x+2y+4z=9`, `2xy+8yz+4xz=26` , `(x)(2y)(4z)=24` Say roots of `P^(3)-9P^(2)+26P-24=0` are x, 2y and 4z Here `(P-2)(P^(2)-7+12)=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 "(2,2,(3)/(4))` `x=3,2y=4,4z=2" then "(3,2,(1)/(2))` `x=2,2y=3,4z=4" then "(2,(1)/(2),1)` `x=4,2y=2,4z=3" then"(4,1,(3)/(4))` Hence there are five solutions. |
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