Question 1204189: The product of the first and third of three negative consecutive integers is 59 more than 4 times the second integer. Find the integers.
Found 3 solutions by ikleyn, mananth, Edwin McCravy:
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
The integers are (n-1), n, (n+1), and
(n-1)*(n+1) = 4n + 59,
or
n^2 - 1 = 4n + 59,
which implies
n^2 - 4n - 60 = 0,
(n-10)*(n+6) = 0.
So, the middle integer can be either 10 or -6.
And since the integers are negative, the only possibility is that the three numbers are -7, -6, -5.
This and only this triple satisfies the problem's condition.
Solved.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! The product of the first and third of three negative consecutive integers is 59 more than 4 times the second integer. Find the integers.
Let the integers be x-1 , x , x+1
(x-1)(x+1)=59+4x
x^2-1-4x =59
x^2-4x-60=0
x^2-10x +6x-60=0
x(x-10)+6(x-10)=0
(x-10)(x+6)=0
x=-6 or 10
the numbers are negative
so x=-6
the integers are -7,-6,-5 as per given condition
CHECK
(x-1)(x+1)=59+4x
(-7)(-5) = 59 -4*(-6)
35 = 35
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
Let the middle integer be x, then the first one is x-1 and the third one is x+1.
x-1 = smallest
x = middle
x+1 = largest
The product of the first and third of three negative consecutive integers
That's (x-1)(x+1)
is 59 more than 4 times the second integer. Find the integers.
That's 4x + 59
(x-1)(x+1) = 4x + 59
x2 - 1 = 4x + 59
x2 - 4x - 60 = 0
(x + 6)(x - 10) = 0
x + 6 = 0; x - 10 = 0
x = -6; x = 10
We are told that the integers are negative, so we discard x=10
x-1 = smallest = -6-1 = -7
x = middle = -6
x+1 = largest = -6+1 = -5
Checking the words:
The product of the first and third...
(-7)(-5) = 35
...is 59 more than 4 times the second integer.
That's 59 more than 4 times -6, which is -24. Sure enough,'
it's true that (-24) + 59 = 35.
Edwin
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