anonymous
  • anonymous
Find four consecutive intergers such that the product of the 1st interger and the 3rd interger is 25 greater than the product of -13 and the fourth integer Answers: -14,-13,-12,-11
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Let’s say that your sequence of four numbers starts at some number n. Then the four numbers are n, n+1, n+2, and n+3. You’re asked to find that number n so that n(n+2) = 25 + (-13)(n+3). One way to solve this is to notice that this is a quadratic equation in n. You’re guaranteed that an integer solution exists, so you’d expect to be able to factor the quadratic expression. If you try, you’ll simplify it down to n^2 + 15n + 14 = 0. And this is (n + 14)(n + 1) = 0. So n can be either -14 or -1. If n = -14, then you get the sequence -14, -13, -12, -11. If you let n = -1, then you get the sequence -1, 0, 1, 2. You can check that both sequences satisfy our condition on the numbers.

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