The difference of 2 positive integers is 3 and the sum of their squares is 117. find the numbers
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x-y=3
x=3+y
x^2+y^2=117
(3+y)^2+y^2=117
y^2+3y-54=0
(y+9)(y-6)=0
y=-9,or y=6
x=-6,orx=9
since both iteger are positive
so
x=9&y=6
x=3+y
x^2+y^2=117
(3+y)^2+y^2=117
y^2+3y-54=0
(y+9)(y-6)=0
y=-9,or y=6
x=-6,orx=9
since both iteger are positive
so
x=9&y=6
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