Question

Find three consecutive odd integers such that the square of the first increased the product of the other two is 28.

Answer 1

Since these numbers are odds, they are of the form:

$2x+1$

Let's take this as the first.

The second one must be:

$2x+3$

And the last one:

$2x+5$

The square of the first one increased the product of the other two is 28:

$(2x+1)^2+(2x+3)(2x+5)=28$

If we take k=2x+1, we get:

$k^2+(k+2)(k+4)=28$$k^2+k^2+6k+8=28$$2k^2+6k+8=28$$2k^2+6k+8-28=0$$2k^2+6k-20=0$$k^2+3k-10=0$$(k+5)(k-2)=0$

K could be -5 or 2, but because has to be a odd number k=-5.

Lets check:

$(-5)^2+(-3)(-1)$$25+3$$28$

The consecutive odd numbers are -5 , -3 and -1