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 β 2y β156 = 0
y^2 β y β156 = 0
Multiply the first term (i.e y^2) and the last term (β156) together. This gives β156y^2.
Now find two factors of β156y^2 such that when you add the two factors together, it will result to the 2nd term (ie βy) in the equation. These numbers are 12y and β 13y
Now substitute these numbers (i.e 12y and β 13y) in place of βy in the equation above
y^2 β y β156 = 0
y^2 + 12y β 13y β156 = 0
y(y + 12) β 13(y + 12) = 0
(y β13)(y + 12) = 0
y β13 =0 or y +12 =0
y = 13 or y = β12
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
96
144
192
Step-by-step explanation: