Answer:
numbers are 13 and 14
Step-by-step explanation:
Let the numbers be y and y+1
The products of the numbers is written as:
y(y + 1)
The sum of the numbers is written as:
y + (y + 1) = 2y + 1
From the question,
The products of the two number is great than the sum by 155. This can be written as:
y(y + 1) = 2y + 1 + 155
y^2 + y = 2y + 156
y^2 + y — 2y —156 = 0
y^2 — y —156 = 0
Multiply the first term (i.e y^2) and the last term (—156) together. This gives —156y^2.
Now find two factors of —156y^2 such that when you add the two factors together, it will result to the 2nd term (ie —y) in the equation. These numbers are 12y and — 13y
Now substitute these numbers (i.e 12y and — 13y) in place of —y in the equation above
y^2 — y —156 = 0
y^2 + 12y — 13y —156 = 0
y(y + 12) — 13(y + 12) = 0
(y —13)(y + 12) = 0
y —13 =0 or y +12 =0
y = 13 or y = —12
Since the numbers are positive numbers, y =13
The first number = y = 13
The 2nd number = y + 1 = 13 +1 = 14
Therefore, the consecutive numbers are 13 and 14
