The first step--I'd say--to most math word problems is to turn the words into algebra.
Define your variables.
Let a = your smallest positive integer
Usually you want to define the smallest number of variables possible. We could have defined the middle integer and the largest integer in our sequence, but that would have been unnecessary. Once you pick where you start your sequence, the other numbers fall into place automatically. We can refer to them in terms of "a". The middle term is a + 2, and the largest term is (a + 2) + 2 or a + 4.
Can we write an equation?
"square of the smallest" becomes a2
"twice the largest" becomes 2(a + 4)
"exceeds . . . by 7" means we have to add 7 somewhere in our equation.
a2 = 2(a + 4) + 7
Now we can solve for a.
This is a quadratic equation.
Subtract 2(a + 4) + 7 from both sides to set it equal to 0.
a2 - 2(a + 4) - 7 = 0
Distribute that 2.
a2 - 2a - 8 - 7 = 0
Simplify.
a2 - 2a - 15 = 0
Factor. (Find two numbers that multiply to get -15 and add to get -2)
(a + 3)(a - 5) = 0
Apply zero-product property.
a = {-3, 5}
a = -3 is clearly an extraneous solution as the problem asked for positive integers.
Our sequence is 5, 7, 9.
Check.
52 = 2(9) + 7
25 = 18 + 7
25 = 25
