Question

The product of 2 consecutive odd integers is 1 less than 4 times their sum. A quadratic function models this situation. What is the domain of the situation?

A) all integers
B) all odd integers
C) all odd numbers less than zero
D) all odd integers greater than zero

Answer 1

b; all odd integers

Explanation:

since the problem is built around any odd integer than the domain would be the set of all odd integers--both positive and negative.

Answer 2

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)".
______________________________
→ Write as: 4*(2x + 2) − 1 = x(x + 2)
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Note the distributive property of multiplication:
_______________________________
→ 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:
__________________________
→ 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 ;
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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:
____________________________
→ 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:
_____________________________________________
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