Since these numbers are odds, they are of the form:
[tex]2x+1[/tex]
Let's take this as the first.
The second one must be:
[tex]2x+3[/tex]
And the last one:
[tex]2x+5[/tex]
The square of the first one increased the product of the other two is 28:
[tex](2x+1)^2+(2x+3)(2x+5)=28[/tex]
If we take k=2x+1, we get:
[tex]k^2+(k+2)(k+4)=28[/tex][tex]k^2+k^2+6k+8=28[/tex][tex]2k^2+6k+8=28[/tex][tex]2k^2+6k+8-28=0[/tex][tex]2k^2+6k-20=0[/tex][tex]k^2+3k-10=0[/tex][tex](k+5)(k-2)=0[/tex]
K could be -5 or 2, but because has to be a odd number k=-5.
Lets check:
[tex](-5)^2+(-3)(-1)[/tex][tex]25+3[/tex][tex]28[/tex]
The consecutive odd numbers are -5 , -3 and -1
