SOLUTION: Find 2 consecutive even integers whose product is 48.

Question 126650: Find 2 consecutive even integers whose product is 48.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Consecutive even numbers can be represented by x for the first and the next number has to be
2 more than that. So the next even number is x + 2. (Think of 6 and 8 as being consecutive
even numbers ... and note that they are 2 units apart.)
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You are now told that the product of these two consecutive even numbers is 48. So multiply the
two numbers together and set the result equal to 48 as follows:
.

.
Multiply out the left side and this equation becomes:
.

.
Get this into standard quadratic form by subtracting 48 from both sides and you have:
.

.
The left side of this equation can be factored as follows:
.

.
Notice this equation will be true if either of the two factors on the left side equals zero
because a multiplication by zero will make the entire left side become zero and therefore
equal to the right side.
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Set the two factors equal to zero (one at a time) as follows:
.

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Solve for x by subtracting 8 from both sides to get:
.

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then set the other factor equal to zero to get:
.

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Solve by adding 6 to both sides:
.

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So there are two possible answers to this problem because x (the first of the consecutive
even integers) can either be -8 or + 6.
This means that in the first possible set of consecutive even integers you have -8 and as the
first integer and therefore the next consecutive integer is 2 greater than that or -6. These
numbers are consecutive and even and their product is +48.
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For the second answer you have that the first integer (x) is +6 and therefore the next
consecutive integer is +8. This pair (+6 and +8) are consecutive and even and their product
is also +48.
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So the two answers are either -8 and -6 or +6 and +8.
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Hope this helps you to see how this problem can be worked.
.
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