Let the 2 consecutive odd integers be represented by:

“x” and “(x+2)

_________________

The product of these two consecutive odd integers is:

________________

→ x*(x + 2); or, write as: x(x + 2)

__________________

The sum of these two consecutive odd integers is:

________________________________

→ x + (x + 2) = (2x + 2)

_______________________________

The product of 2 conductive integers, “x(x + 2)” , is 1 less than

4 times their sum, “(2x + 2)”.

______________________________

→ Write as: 4*(2x + 2) − 1 = x(x + 2)

________________________________

Note the distributive property of multiplication:

_______________________________

→ a*(b + c) = ab + ac ;

________________________________

We have:

___________

→ 4*(2x + 2) − 1 = x(x + 2)

_____________________________

→ 4*(2x + 2) = (4*2x) + (4*2) = 8x + 8

____________________________________

On the “right side of the equation; we have:

______________________________________

→ x(x + 2) = (x*x) + (x*2) = x² + 2x

_____________________________________

We can rewrite the equation:

__________________________

→ 4*(2x + 2) − 1 = x(x + 2) ;

___________________________

by substituting our obtained “expanded values” for:

“[4*(2x + 2)]” ; and for: “[x(x + 2)]” ;

______________________________________

→ 4*(2x + 2) − 1 = x(x + 2) =

____________________________

→ 8x + 8 − 1 = x² + 2x ;

__________________________________

→ Simplify the “+8 − 1” on the “left-hand side” of the equation to “7”; and subtract “2x” from EACH SIDE of the equation:

____________________________________

→ 8x + 7 − 2x = x² + 2x − 2x ; to get:

____________________________

→ 6x + 7 = x² ;

________________________________

→To solve for “x”; Subtract “6x” and subtract “7”; from EACH SIDE of the equation; to get an equation in “quadratic format” ; that is:

_____________________________________________

ax

“x” and “(x+2)

_________________

The product of these two consecutive odd integers is:

________________

→ x*(x + 2); or, write as: x(x + 2)

__________________

The sum of these two consecutive odd integers is:

________________________________

→ x + (x + 2) = (2x + 2)

_______________________________

The product of 2 conductive integers, “x(x + 2)” , is 1 less than

4 times their sum, “(2x + 2)”.

______________________________

→ Write as: 4*(2x + 2) − 1 = x(x + 2)

________________________________

Note the distributive property of multiplication:

_______________________________

→ a*(b + c) = ab + ac ;

________________________________

We have:

___________

→ 4*(2x + 2) − 1 = x(x + 2)

_____________________________

→ 4*(2x + 2) = (4*2x) + (4*2) = 8x + 8

____________________________________

On the “right side of the equation; we have:

______________________________________

→ x(x + 2) = (x*x) + (x*2) = x² + 2x

_____________________________________

We can rewrite the equation:

__________________________

→ 4*(2x + 2) − 1 = x(x + 2) ;

___________________________

by substituting our obtained “expanded values” for:

“[4*(2x + 2)]” ; and for: “[x(x + 2)]” ;

______________________________________

→ 4*(2x + 2) − 1 = x(x + 2) =

____________________________

→ 8x + 8 − 1 = x² + 2x ;

__________________________________

→ Simplify the “+8 − 1” on the “left-hand side” of the equation to “7”; and subtract “2x” from EACH SIDE of the equation:

____________________________________

→ 8x + 7 − 2x = x² + 2x − 2x ; to get:

____________________________

→ 6x + 7 = x² ;

________________________________

→To solve for “x”; Subtract “6x” and subtract “7”; from EACH SIDE of the equation; to get an equation in “quadratic format” ; that is:

_____________________________________________

ax

let x and x+2 be the consecutive odd integers.

Their product is x(x+2)

Their sum is x + x+2 or 2x+2

x(x+2)=4(2x+2)-1

Domain is odd integers