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One positive integer is 1 less than twice another. The sum of their squares is 157. Find the integers.

User Excellent
by
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1 Answer

2 votes

Answer:

  • 11 and 6

Explanation:

Let the integers are x and y.

We have:

  • x = 2y - 1
  • x² + y² = 157

Substitute x into the second equation and solve for y:

  • (2y - 1)² + y² = 157
  • 4y² - 4y + 1 + y² = 157
  • 5y² - 4y - 156 = 0
  • D = (-4)² - 4*5*(-156) = 3136
  • y = (4 ± √3136)/10 = (4 ± 56)/10
  • y = 6 is the only integer solution

Find x:

  • x = 2*6 - 1 = 11
User Tanysha
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