Find 2 consecutive integers with a Product of 20

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We need two consecutive integers (n) and (n + 1) who have

a product = 20

Setup relational equation:

We need to find two integers, n and n + 1 who have a product of 20
n * (n + 1) = 20

Multiplying through, we get n² + n = 20

Rearranging the equation we get n² + n - 20 = 0

Now that it is in Quadratic Format, determine a, b, and c:

a = 1, b = 1, and c = -20

Solution 1 = ½(-b + √b² - 4ac)

Solution 1 = ½(-1 + √1² - 4 * 1 * -20)

Solution 1 = ½(-1 + √1 - -80)

Solution 1 = ½(-1 + √81)

Solution 1 = ½(-1 + 9)

Solution 1 = ½(8)

Solution 1 = 4

Determine Answers:

Solution 2 = Solution 1 + 1

Solution 2 = 4 + 1

Solution 2 = 5

Also, since the product of 2 negative #'s is positive, another solution is:

Solution 3 = (-1 * 4) * (-1 * 5)

Solution 3 = -1 * 4

Solution 3 = -4

Solution 4 = -1 * 5

Solution 4 = -5

Final Answers:

You have 2 free calculationss remaining

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Tags:

• consecutive integer word problems
• consecutive integers
• integer
• product
• sum
• word problem