SOLUTION: Find two consecutive integers whose product is 5 less than the square of the smaller number. How would i write a verbal model and label it to figure out the two numbers

Question 345201: Find two consecutive integers whose product is 5 less than the square of the smaller number. How would i write a verbal model and label it to figure out the two numbers
Found 2 solutions by checkley77, jsmallt9:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
Let x & x+1 be the 2 integers.
x(x+1)-5=x^2
x^2+x-5=x^2
x^2-x^2+x=5
x=5 ans. for the smaller integer.
5+1=6 ans. for the larger integer.
Proof:
5*6-5=5^2
30-5=25
25=25

Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
(Note: The solution provided by another tutor is incorrect. The product of those answers is 30 and the square of the smaller number, 5, is 25. The product is 5 more than the square of the smaller number not 5 less than.)

Let x = the smaller integer.
Then the next integer is x+1.
The product of these two integers is:
The square of the smaller number is:
The equation that says: "The product is 5 less than the square of the smaller number" is:

To solve this we'll subtract from each side:

This is the smaller number. The next integer, x+1, is: -5 + 1 = -4
The two consecutive integers are: -5, -4

Check:
The product of -5 and -4 is 20
The square of the smaller number, -5, is 25.
The product, 20, is indeed 5 less than 25, the square of the smaller number.
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