Question 1167984
<i>1. The square of a positive integer is 98 less than twice the square of the next consecutive integer. Find the integers.</i>
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First integer = x
Second integer = x + 1
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x² = 2(x + 1)² - 98
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x² = 2(x² + 2x + 1) - 98
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x² = 2x² + 4x + 2 - 98
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x² = 2x² + 4x - 96
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x² + 4x - 96 = 0
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(x - 8)(x + 12) = 0
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x = 8 and x = -12
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Since the problem stated the integers were positive, we can eliminate the x = -12 result.  We are left with x = 8.  So...
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First integer = x = <b>8</b>
Second integer = x + 1 = <b>9</b>
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<i>2. Find two integers differ by 4, while their squares differ by 56.</i>
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First integer = x
Second integer = x + 4
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(x + 4)² - x² = 56
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x² + 8x + 16 - x² = 56
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8x + 16 = 56
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8x = 40
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x = 5
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First integer = x = <b>5</b>
Second integer = x + 4 = <b>9</b>