SOLUTION: Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 4 times the third.

Question 148935: Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 4 times the third.
Found 2 solutions by ankor@dixie-net.com, checkley77:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 4 times the third.
:
Let the three consecutive numbers be x, (x+1), (x+2)
:
Write an equation for this statement:
"product of the first and third, minus the second,is 1 more than 4 times the third."
x(x+2) - (x+1) = 4(x+2) + 1
:
x^2 + 2x - x - 1 = 4x + 8 + 1
:
x^2 + x -1 = 4x + 9
:
Arrange as a quadratic equation on the left
x^2 + x - 4x - 1 - 9 = 0
:
x^2 - 3x - 10 = 0
Factors to:
(x-5)(x+2) = 0
Positive solution wanted
x = +5, 6, 7 are the 3 numbers
;
:
Check solution in the statement
"product of the first and third, minus the second,is 1 more than 4 times the third."
5*7 - 6 = 4(7) + 1
35 - 6 = 28 + 1
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
LET X BE THE FIRST NTEGER, (X+1) BE THE SECOND & 9X+2) BE THE THIRD.
X(X+2)-(X+1)=4(X+2)+1
X^2+2X-X-1=4X+8+1
X^2+2X-X-4X-1-8-1=0
X^2-3X-10=0
(X-5)(X+2)=0
X-5=0
X=5 ANSWER FOR THE FIRST INTEGER.
5+1=6 THE SECOND.
5+2=7 THE THIRD.
PROOF:
5*7-6=4*7+1
35-6=28+1
29=29

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