Question 156843: The product of two consecutive odd integer is 1 less than 4 times their sum find the two integers Found 2 solutions by checkley77, ankor@dixie-net.com:Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! x(x+2)=4(x+x+2)-1
x^2+2x=4(2x+2)-1
x^2+2x=8x+8-1
x^2+2x-8x-8+1=0
x^2-6x-7=0
(x-7)(x+1)=0
x-7=0
x=7 for the smaller integer.
7=2=9 for the larger integer.
Proof:
7*9=4(7+9)-1
63=4*16-1
63=64-1
63=63
You can put this solution on YOUR website! The product of two consecutive odd integer is 1 less than 4 times their sum find the two integers
:
The two consecutive odd integers: x, (x+2)
:
write an equation for what it says:
x(x+2) = 4[x + (x+2)] -1
:
x^2 + 2x = 4(2x+2) - 1
:
x^2 + 2x = 8x + 8 - 1
;
x^2 + 2x = 8x + 7
Arrange as a quadratic equation
x^2 + 2x - 8x - 7 = 0
:
x^2 - 6x - 7 = 0
Factor
(x-7)(x+1) = 0
:
x = +7
and
x = -1
:
The integers: 7, 9 and -1,+1
;
;
Check solution in the statement:
"The product of two consecutive odd integer is 1 less than 4 times their sum
x=7: 7*9 = 4(16) - 1
x=-1: -1*1 = 4(-1+1) - 1
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