Question 846601: can you break down find three consecutive even integers such that the product of the first and third is 96?
Answer by josh_jordan(263)   (Show Source):
You can put this solution on YOUR website! To find each of these integers, we need to set up each integer algebraically to represent each integer. We know that each integer is even. So we can represent the first integer with the letter x. If x is an even number, what must we add that number to give us the next consecutive even integer? 2. So, our second integer can be represented by x + 2. What must we add our original integer to to produce the next consecutive integer? 4. So, our third even integer can be represented by x + 4. So, we have:
First integer = x
Second integer = x + 2
Third integer = x + 4
Now, the word problem states that the product of the first and third integer is 96. In other words:
x(x + 4) = 96
Now we have our equation. Let's first distribute x, by multiplying it by both x and 4, giving us:
x^2 + 4x = 96
Subtracting 96 from both sides of the equation will give us a quadratic equation in standard form that we can factor:
x^2 + 4x - 96 = 0
We need to determine what factors of 96 can we multiply together to give us -96 and add together to give us 4. To shorten this solution, I will give you the factors: -8 and 12. So, we have:
(x - 8)(x + 12) = 0
Setting each of these factors equal to zero gives us two possible values of x: 8 and -12. Replacing our first integer with each value of x, will give us our other two integers:
x = 8, so 8 + 2 = 10 and 8 + 4 = 12. So, 8, 10, and 12 are 3 integers that work.
Now, let's replace x with -12:
x = -12, so -12 + 2 = -10, and -12 + 4 = -8
Notice that -8 x -12 = 96 AND 8 x 12 = 96. So, we have two sets of solutions:
-8, -10, -12 AND 8, 10, 12
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