Respuesta :
Answer:
The 3 positive consecutive integers are 7, 8 and 9.
Step-by-step explanation:
We can find the product of the second integer and the last integer is 72 for 3 positive consecutive integers as follows:
Let the 2nd integer = [tex]x[/tex]
Since all 3 numbers are consecutive integers, then:
- the 1st integer = [tex]x-1[/tex]
- the 3rd integer = [tex]x+1[/tex]
The product of the second integer and the last integer is 72:
[tex]x(x+1)=72[/tex]
[tex]x^2+x-72=0[/tex]
[tex](x-8)(x+9)=0[/tex]
Since [tex]x[/tex] has to be positive, then [tex]\bf x=8[/tex]
- the 1st integer = [tex]x-1[/tex]
= [tex]8-1[/tex]
= [tex]\bf7[/tex]
- the 2nd integer = [tex]x[/tex]
= [tex]\bf8[/tex]
- the 3rd integer = [tex]x+1[/tex]
= [tex]8+1[/tex]
= [tex]\bf9[/tex]
The three positive consecutive integers where the product of the second and the third is 72 are 7, 8, and 9. We found these by setting up an equation with the first integer as n, resulting in the equation (n+1)(n+2) = 72 and solving for n.
We are tasked with finding three positive consecutive integers such that the product of the second and third integers is 72.
Let the first integer be n, which makes the second integer n+1 and the third integer n+2. Given the product of the second and third integers is 72:
(n+1)(n+2) = 7
We solve for n by expanding the equation and subtracting 72 from both sides:
n2 + 3n + 2 - 72 = 0
n2 + 3n - 70 = 0
Factor the quadratic equation:
(n+10)(n-7) = 0
This gives us two possible solutions for n:
n = -10 (ignore since we need positive integers)
n = 7 (valid solution)
Therefore, the integers are 7, 8, and 9.
