If the value of two integers differs by 3, and the product of the two integers is 108, what is the sum of the two integers?
A.21
B.19
C.25 D.23

Answer:

A

Step-by-step explanation:

Let the smallest number be n.

If the next integer differs by 3 then:

n + 3 is our second integer.

If the product is 108 then we would multiply them together.

n * (n + 3) = 108    Now solve for n.

(n*n) + (n*3) = 108

[tex]n^{2} +3n = 108[/tex]     Now we move 108 to the other side by subtracting so we can factor.

[tex]n^{2} + 3n - 108 = 0[/tex]

(n       )(n       ) = 0    

Find all the multiples of 108

1 x 108

2 x 54

3 x 36

4 x 27

6 x 18

9 x 12

Which one can we add or subtract together to get 3?

9 x 12    because 12 - 9 = 3

[tex]n^{2} + 3n - 108 = 0[/tex]

(n    12)(n     9) = 0    Now we figure out the signs. Since we have -108 one must be negative, and the 3 in the 2nd term (3n) is positive so:

(n + 12)(n - 9) = 0   Is our factored equation.

Our smallest number is 9, so n = 9

Our second number is n + 3 and 9 + 3 = 12

So our 2 integers are 9, 12, there sum would be 9 + 12 = 21

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