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One positive integer is 2 less than twice another. The sum of their squares is 628. Find the integers.

User Kizoso
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let one of the integers be x,

other one = 2x - 2

now, according to question :


  • {x}^(2) + (2x - 2) {}^(2) = 628


  • {x}^(2) + 4 {x}^(2) + 4 - 8 x = 628


  • 5 {x}^(2) - 8x = 628 - 4


  • 5 {x}^(2) - 8x = 624


  • 5 {x}^(2) - 8x - 624 = 0


  • 5 {x}^(2) - 60 x+ 52 x- 624 = 0


  • 5x(x - 12) + 52(x - 12) = 0


  • (5x + 52)(x - 12) = 0

now, since the Integers are positive so the value obtained from (5x + 52) = 0 can't hold true.

so, x - 12 = 0


  • x = 12

the first number is :


  • x = 12

second number is :


  • 2x - 2


  • (2 * 12) - 2


  • 24 - 2


  • 22
User Patrick Michaelsen
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