Need a formula; the product of two consecutive integers is 41 more than their sum. Find the integers.

let the two numbers be x and x+1

then x(x+1) = x + y + 41
expanding and simplifying we get
x^2 - x - 42 = 0
(x-7)(x+6) = 0
x = 7 or x = -6

case 1: the two numbers are 7 and 8
check: product = 56
sum = 15, 56 is greater than 15 by 41

case 2: the two numbers are -6 and -5
check: product is 30
sum is -11
30 is 41 greater than -11

so the two numbers are either 7 and 8 or
-6 and -5

×^-×-42=0

Find such 3 consecutive integers that the sum of the first and third is 22 more than the second one.

Why did the two consecutive integers start bickering? Because one of them thought the other was 41 more than their sum! To find the solution to this puzzle, let's call the smaller integer "x" and the larger integer "x+1". According to the problem, the product of these consecutive integers is 41 more than their sum. So we can write the equation as:

x(x + 1) = (x + x + 1) + 41

Let's simplify that equation and solve it to find the values of x and x+1!

To solve this problem, let's start by assigning variables to the unknown integers. Let's call the first integer "x" and the second integer "x+1" (since they are consecutive integers).

According to the problem statement, the product of the two consecutive integers is 41 more than their sum. So, we can set up the following equation:

x(x + 1) = x + (x + 1) + 41

Now, let's solve this equation step by step:

Step 1: Simplify the equation

x^2 + x = 2x + 42

Step 2: Move all the terms to one side to get a quadratic equation

x^2 + x - 2x - 42 = 0

Simplifying further, we get:

x^2 - x - 42 = 0

Step 3: Factorize the quadratic equation

(x - 7)(x + 6) = 0

Now, we have two possible solutions for x:

x - 7 = 0 OR x + 6 = 0

Solving these equations, we find:

x = 7 OR x = -6

Step 4: Find the consecutive integers

If x = 7, the consecutive integers are 7 and 8.
If x = -6, the consecutive integers are -6 and -5.

So, the possible pairs of consecutive integers that satisfy the given conditions are (7, 8) and (-6, -5).