You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers.
2) Call the two consecutive integers "n" and "n+1", and solve the equation:
n(n+1)=210
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Let the integers by n & n+1
Their product is
n(n+1) = 210
n^2 + n = 210
n^2 + n - 210 = 0
Factor
( n - 14)(n + 15) = 0
n = 14 & n = -15
The consecutive integers being 14 & 15.
Let's denote the two consecutive integers as ( n ) and ( n+1 ). The product of these two integers can be expressed as ( n(n+1) = 210 ). Simplifying, we get the quadratic equation ( n^2 + n - 210 = 0 ). Solving this equation using the quadratic formula, we find that the two integers are 14 and 15.
Yes, the integers are 12 and 13.
Well, darling, if you're looking for the product of two consecutive integers that equals 380, then you're in luck. The two integers are 19 and 20. See, no need to break a sweat over this math riddle. Just keep on strutting your stuff!
Let the two consecutive integers be x and x+1. The product of these two integers is x(x+1). Setting this equal to 132 gives us the equation x(x+1) = 132. By expanding the left side of the equation, we get x^2 + x = 132. Rearranging terms, we have x^2 + x - 132 = 0. This is a quadratic equation that can be factored as (x+12)(x-11) = 0. Therefore, the two consecutive integers are 11 and 12, and their product is 132.
-10 -11
-8 -9