The sum of the squares of two consecutive odd positive integers is 74. Find the integers.

Let first odd number be x

Then that would be [tex]x^2 + (x+2)^2=74[/tex]. We need to solve for x.

[tex]x^2 + (x+2)^2=74\\\ \\x^2 + x^2 + 4x+4 = 74\\\ \\2x^2 + 4x +4=74\\\ \\2x^2+4x-70 = 0\\\ \\2(x^2+2x-35)=0\\\ \\2(x+7)(x-5)=0\\\ \\x=-7\text{ or }5[/tex]

But we need positive integers so we would have [tex]\boxed{x=5}[/tex], so then our integers would be x, x+2 = 5, 7

Check work:

5² + 7² = 25 + 49 = 74.

So our integers would be 5 and 7.

Hope this helps.