SOLUTION: two positive integers have a sum of 21 and a product of 108. what are the integers?

Question 320442: two positive integers have a sum of 21 and a product of 108. what are the integers?
Answer by Sunny Day(15) (Show Source): You can put this solution on YOUR website!
Let one integer be x
Then the second integer = 21 - x.
Their product = 108
ie. x(21-x) = 108
21x - x^2 = 108
21x - x^2 - 108 = 0
-x^2 + 21x - 108 = 0
Multiplying the whole with -1, (in order to make the co-efficient of x^2 +ve)
x^2 - 21x + 108 = 0 -------- (1)
[Now we have to find two numbers such that their product is 108 and sum is -21. and these two numbers will eb factors of 108. They are -9 and -12]
So (1) can be written as
x^2 - 9x - 12x + 108 = 0
x(x-9) - 12(x-9) = 0
(x-9) (x-12) = 0
So either x-9 = 0 or x-12 = 0
x = 9 or x = 12
These are the required integers
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