Let the 2 consecutive odd integers be represented by:
“x” and “(x+2)
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The product of these two consecutive odd integers is:
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→  x*(x + 2); or, write as: x(x + 2)
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The sum
of these two consecutive odd integers is:
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→  x + (x + 2) = (2x + 2)
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The product of 2 conductive integers, “
x(x + 2)” , is 1 less than
4 times their sum, “(2x + 2)”.
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→  Write as: 4*(2x + 2) − 1 = x(x + 2)
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Note the distributive property of multiplication:
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→  a*(b + c) = ab + ac ;
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We have:
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→  4*(2x + 2) − 1 = x(x + 2)
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→ 4*(2x + 2) = (4*2x)  + (4*2) = 8x + 8
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On the “right side of the equation; we have:
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→  x(x + 2) = (x*x) + (x*2) = x² + 2x
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We can rewrite the equation:
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→  4*(2x + 2) − 1 = x(x + 2) ;
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by substituting our obtained “expanded values” for:
“[
4*(2x + 2)]” ; and for: “[x(x + 2)]” ;
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→  4*(2x + 2) − 1 = x(x + 2) =
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→  8x + 8 − 1 = x² + 2x ;
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→ Simplify the “+8 − 1” on the “left-hand side” of the equation to “7”; and subtract “2x” from EACH SIDE of the equation:
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→  8x + 7 − 2x = x² + 2x − 2x ; to get:
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→ 6x + 7 = x² ;
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→To solve for “x”;  Subtract “6x” and subtract “7”; from EACH SIDE of the equation; to get an equation in “quadratic format” ; that is:
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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