there are three consecutive positive integers such that the sum of the square of the first and product of other two is 191 what are the integers
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let the three consecutive positive numbers be x+1, x+2, x+3 say
according to the problem
(x+1)^2+(x+2)(x+3)=191
x^2+2x+1+x^2+3x+2x+6=191
2x^2+7x+7-191=0
2x^2+7x-184=0
2x^2+23x-16x-184=0
x(2x+23)-8(2x+23)=0
(2x+23)(x-8)=0
x=-23/2
or
x=8
according to the problem
(x+1)^2+(x+2)(x+3)=191
x^2+2x+1+x^2+3x+2x+6=191
2x^2+7x+7-191=0
2x^2+7x-184=0
2x^2+23x-16x-184=0
x(2x+23)-8(2x+23)=0
(2x+23)(x-8)=0
x=-23/2
or
x=8
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0
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