Question

Given positive integers, if x^2+y^2+4y-96=0 , find the value of xy .

Steven Zheng Steven Zheng posted 3 weeks ago


Answer

\because x^2+y^2+4y-96=0

\therefore x^2=100-(y+2)^2>0

If y=1 , 100-(1+2)^2=91 , not a square number.

If y=2 , 100-(2+2)^2=84 , not a square number.

If y=3 , 100-(3+2)^2=75 , not a square number.

If y=4 , 100-(4+2)^2=64 , x=8

If y=5 , 100-(5+2)^2=51 , not a square number.

If y=6 , 100-(6+2)^2=36 , x=6

If y=7 , 100-(7+2)^2=19 , not a square number.

If y=8 , 100-(8+2)^2=0 , x=0

\therefore y=4 or 6 , xy=32 or 36

Steven Zheng Steven Zheng posted 3 weeks ago

Scroll to Top