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Consecutive Integers
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Consecutive Integers

Consecutive Integers

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Author: Anthony Varela
This lesson will demonstrate how to use multi-step equations to solve for consecutive integer problems (ex. find four consecutive integers that have a sum of 54).
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Show your understanding of this concept by correctly answering 3 questions.

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Questions and Answers

  • Answer 1
    Grace Mariz Gozarin 3 months ago

    find three consecutive integer whicch have 110 as the sum of thier squares

      Anthony Varela answered 3 months ago

      Hi Grace,

      This is a fun question. We'll need to develop an equation, and then solve that equation to give us one of the three consecutive integers that fit the problem. At first, it might seem like our equation is going to have three variables (one for each integer), but we can make it simpler. Since consecutive integers only differ by 1 unit, we can say that the first integer is n, the second integer is (n+1), and the third integer is (n+2). For example, if your problem was finding three consecutive integers that add to 9, our equation would be: n + (n+1) + (n+2) = 9. We'd then combine like terms, and have a linear equation to solve: 3n + 3 = 9. Solving this equation gives us n = 2, which remember is the first consecutive integer in our solution. So the answer to that problem would be 2, 3, and 4 as summing to 9.

      With your question, not only is the sum different (110), but it's also the sum of their squares. We'll still define our first integer as n, our second integer as (n+1), and our third integer as (n+2). However, in our equation, each of these quantities needs to be squared, or raised to the second power.

      So our equation is: n^2 + (n+1)^2 + (n+2)^2 = 110. This is a quadratic equation.

      To solve for n, you'll need to expand/foil each term. n^2 is already expanded for us, but to simplify (n+1)^2, you'll need to use FOIL to expand (n+1)(n+1). Same thing for (n+2)^2, FOIL it to expand (n+2)(n+2). Then combine your like terms to get a quadratic equation in the form an^2+bn+c = 110. This is shown below:

      n^2 = n^2
      (n+1)^2 = (n+1)(n+1) = n^2 + 2n + 1
      (n+2)^2 = (n+2)(n+2) = n^2 + 4n + 4

      So n^2 + (n+1)^2 + (n+2)^2 can be written as: n^2 + (n^2 + 2n + 1) + n^2 + 4n + 4)

      When we combine like terms, we see that there are three n^2 terms, six n-terms, and our constant term is 5. So this simplifies to 3n^2 + 6n + 5. This is set equal to 110.

      So far, we've created a quadratic equation 3n^2 + 6n + 5 = 110. Now we need to solve for n.

      Here are some things to remember when solving the quadratic equation you've constructed:

      We can use the quadratic formula when the quadratic expression is set equal to zero. This means that you'll want to subtract 110 from both sides of the equation. This will not affect 3n^2 or 6n, but our constant term will now be -105. Our equation is now 3n^2 + 6n - 105 = 0

      If all terms share a common factor, we can factor this out. So I'm looking at the coefficients 3, 6, and -105. Each of these is divisible by 3. So I can divide each term by 3, to get: n^2 + 2n - 35 = 0

      Now I'm going to use the coefficients 1, 2, and -35 in the quadratic formula to find n. Often times, there are two solutions to quadratic equations. The solutions are 5 and -7.

      This means that there are actually two sets of three consecutive integers which have 110 as the sum of their squares. The first set is 5, 6, and 7. The second set is -7, -6, and -5.

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