the product of 2 consecutive integers is 156. find the integers.
I found them mentally 12 and 13 but, how do I find them while solving?
a = first number
b = a + 1 = socond number
a b = 156
a ( a + 1 ) = 156
a ^ 2 + a = 156 Add 1 / 4 to both sides
a ^ 2 + a + 1 / 4 = 156 + 1 / 4
a ^ 2 + a + 1 / 4 = 624 / 4 + 1 / 4
a ^ 2 + a + 1 / 4 = 625 / 4
________________________________________
Remark :
( x + y ) ^ 2 = x ^ 2 + 2 x y + y ^ 2
( a + 1 / 2 ) ^ 2 = a ^ 2 + 2 a * 1 / 2 + ( 1 / 2 ) ^ 2 =
a ^ 2 + a + 1 / 4
so
a ^ 2 + a + 1 / 4 = ( a + 1 / 2 ) ^ 2
________________________________________
a ^ 2 + a + 1 / 4 = 625 / 4
( a + 1 / 2 ) ^ 2 = 625 / 4 Take the square root of both sides
a + 1 / 2 = + OR - 25 / 2 Subtract 1 / 2 to both sides
a + 1 / 2 - 1 / 2 = + OR - 25 / 2 - 1 / 2
a = + OR - 25 / 2 - 1 / 2
First solution:
a = 25 / 2 - 1 / 2 = 24 / 2 = 12
Second solution:
a = - 25 / 2 - 1 / 2 = - 26 / 2 = - 13
Final solutions :
a = 12
b = a + 1 = 12 + 1 = 13
a * b = 12 * 13 = 156
a = - 13
b = a + 1 = - 13 + 1 = - 12
a * b = - 13 * ( - 12 ) = 156
To find the consecutive integers, let's call the first integer x and the second integer (which follows the first one) x+1.
According to the problem, the product of these two consecutive integers is 156:
x * (x+1) = 156
Now, let's solve this equation to find the values of x and x+1.
Expanding the equation:
x^2 + x = 156
Rearranging the equation:
x^2 + x - 156 = 0
Now, we need to factorize or use the quadratic formula to solve for x. In this case, the equation can be factored easily:
(x - 12)(x + 13) = 0
This means that either (x - 12) = 0 or (x + 13) = 0.
Solving for x in these separate equations:
1) x - 12 = 0
x = 12
2) x + 13 = 0
x = -13
We found two possible values for x.
Since we are looking for consecutive integers, we take x = 12.
So, the consecutive integers are 12 and 13.