Question 1124936


let three positive consecutive integers be {{{n-1}}}, {{{n}}}, and {{{n+1}}}


{{{(n-1)*(n+1) = 3n+17}}}

{{{n^2-1 = 3n+17}}}

{{{n^2 - 3n - 18 = 0}}}...factor


{{{n^2 +3n- 6n - 18 = 0}}}

{{{(n^2 +3n)- (6n + 18) = 0}}}

{{{n(n +3)- 6(n + 3) = 0}}}

{{{(n - 6) (n + 3) = 0}}}

solutions:

if {{{(n - 6) = 0}}}->{{{n=6}}}

if {{{(n + 3) = 0}}}->{{{n=-3}}}

since you need three {{{positive}}} consecutive integers, use {{{n=6}}}

and three numbers are:

{{{n-1=highlight(5)}}}
{{{n=highlight(6)}}}
{{{n+1=highlight(7)}}}


check:

{{{(n-1)*(n+1) = 3n+17}}}

{{{(6-1)*(6+1) = 3*6+17}}}

{{{5*7 = 18+17}}}

{{{35 = 35}}}