Question 546011
Consecutive odd integers are two units apart. Example 3 & 5 or 101 & 103. 
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That being the case, if n represents the first positive odd integer, then n + 2 represents the next positive integer. n +2 is the larger odd integer.
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The problem tells you that the first of these two integers (that means n) is to be squared. Squaring n results in {{{n^2}}}
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Next you are told that this quantity is to be equal to "4 more than 5 times the larger." Well then, 5 times the larger is 5*(n + 2) and 4 more than that is 4 + 5*(n + 2). This is to equal n squared. So we can write the equation:
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{{{n^2 = 4 + 5*(n + 2)}}}
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Multiply out the 5*(n + 2) on the right side. It becomes 5n + 10. So this makes the equation:
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{{{n^2 = 4 + 5n + 10}}}
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Add the 4 and the 10 on the right side to get:
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{{{n^2 = 5n + 14}}}
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This is a quadratic relationship. Put it into standard quadratic form by subtracting 5n + 14 from both sides to get:
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{{{n^2 - 5n -14 = 0}}}
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The left side of this equation can be factored as follows:
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{{{(n - 7)*(n + 2) = 0}}}
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and this equation will be true whenever one of the factors equals zero. This is because multiplication by a zero on the left side makes the entire left side equal to the zero on the right side. This means there are two possible solutions. Either:
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{{{n - 7 = 0}}} which means that {{{n = 7}}}
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or
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{{{n + 2 = 0}}} which means that {{{n = -2}}}
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But the problem says that n must be positive and odd. The only answer that satisfies both those conditions is {{{n = 7}}}.
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So if {{{n = 7}}} then the next consecutive odd number is {{{n = 9}}}.
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The problem asks for the sum of these two digits so the answer is {{{7+9 = 16}}}
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Let's just check to make sure that the two digits satisfy the conditions of the problem.
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{{{n^2 = 7*7}}} and that is {{{49}}}
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{{{4+ 5*9 = 4 + 45}}} and that also equals {{{49}}}
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Therefore, we can say that we worked it out correctly. We have the two correct positive odd integers, and their sum (which the problem asked us to find) is 16.
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Hope this helps you to understand how to work this problem and gives you some insight into how to solve similar problems.
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