SOLUTION: 1. The square of a positive integer is 98 less than twice the square of the next consecutive integer. Find the integers. Show your solution. 2. Find two integers differ by 4, wh

Algebra ->  Formulas -> SOLUTION: 1. The square of a positive integer is 98 less than twice the square of the next consecutive integer. Find the integers. Show your solution. 2. Find two integers differ by 4, wh      Log On


   

Question 1167984: 1. The square of a positive integer is 98 less than twice the square of the next consecutive integer. Find the integers. Show your solution.
2. Find two integers differ by 4, while their squares differ by 56. Show your solution.
Answer by VFBundy(438) About Me  (Show Source):
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1. The square of a positive integer is 98 less than twice the square of the next consecutive integer. Find the integers.

First integer = x
Second integer = x + 1

x² = 2(x + 1)² - 98

x² = 2(x² + 2x + 1) - 98

x² = 2x² + 4x + 2 - 98

x² = 2x² + 4x - 96

x² + 4x - 96 = 0

(x - 8)(x + 12) = 0

x = 8 and x = -12

Since the problem stated the integers were positive, we can eliminate the x = -12 result. We are left with x = 8. So...

First integer = x = 8
Second integer = x + 1 = 9

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2. Find two integers differ by 4, while their squares differ by 56.

First integer = x
Second integer = x + 4

(x + 4)² - x² = 56

x² + 8x + 16 - x² = 56

8x + 16 = 56

8x = 40

x = 5

First integer = x = 5
Second integer = x + 4 = 9