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

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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

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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.