Question 538267: find two consecutive positive integers such that the square of the second added to 4 times the first equals 113
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let n represent the first positive number. Since the two numbers are consecutive, the next number is n + 1.
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The square of the second number is (n + 1)^2. It squares out to n^2 + 2n + 1.
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4 times the first is 4n
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When the square of the second number is added to 4 times the first as below:
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n^2 + 2n + 1 + 4n
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the sum equals 113 as follows:
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n^2 + 2n + 1 + 4n = 113
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Simplify by adding the 2n and the 4n to get:
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n^2 + 6n + 1 = 113
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Subtract 113 from both sides to get this equation into the standard form of a quadratic equation:
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n^2 + 6n - 112 = 0
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Factor the left side:
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(n - 8)*(n + 14) = 0
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This equation will be true if either of the factors on the left side is equal to zero. (A multiplication by zero on the left side makes the entire left side equal to the zero on the right side.) Set each factor equal to zero and solve for n to see what values of n will make each of the two factors equal to zero:
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n - 8 = 0
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n = 8
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and
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n + 14 = 0
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n = -14
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The answer is that the first number is +8. The next consecutive number is +9. You can ignore the negative 14 because the problem requires that the answer be positive integers.
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So the answer is + 8 and +9.
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Check by squaring the second (9 squared is 81) and add that to 4 times 8 (32).
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81 + 32 = 113
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This is the result that the problem established, so the answer is correct.
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Hope this helps you to understand how to work this problem.
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