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How do you find two consecutive positive integers such that the square of the first decreased by 25 equals three times the second?
Wiki User
∙ 14y ago
If the integers are consecutive, you can call them 'X' and 'X+1' until you know what they are. The square of the first is X2. Decreased by 25, it's X2-25. Three times the second number is 3 times (X+1), or 3X+3. Since these quantites are equal, X2 - 25 = 3X + 3. This can be re-arranged into the standard quadratic-equation form, without changing the value of anything: X2 - 3X - 28 = 0 . Solving this equation will give you 'X', the first integer, and the other integer is greater by 1. Solving a standard quadratic equation is beyond the scope of this discussion, and if you were given this problem in a class, then somebody expects you to know how to solve it, or at least how to factor it, so that doesn't need to be detailed here. The two solutions to it are: X = 7 and X = -4. Since you specified positive integers, we can throw away X = -4. The two integers are 7 and 8. Now check these to make sure they satisfy the given facts: The square of the first is 49. Decreased by 25, it becomes 24. Three times the second integer (8) is also 24. By golly, they work !
Wiki User
∙ 14y ago
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