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anonymous

  • 5 years ago

the sum of the squares of two consecutive, positive even integers is 41. find the integers

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  1. anonymous
    • 5 years ago
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    a miracle since if you square an even integer you get an even integer and if you add two even integers together you get another even integer.

  2. anonymous
    • 5 years ago
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    relax the "even" part and the numbers 4 and 5 work

  3. shadowfiend
    • 5 years ago
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    satellite is right. Below is how you would solve the problem if you didn't specifically require even numbers: \[a^2 + (a + 1)^2 = 41\] \[a^2 + a^2 + 2a + 1 = 41\] \[2a^2 + 2a + 1 = 41\] \[2a^2 + 2a - 40 = 0\] Since \(b^2 - 4ac = 4 - (4 \cdot 2 \cdot -40) = 4 + 320 = 324 > 0\), we know this equation has a solution. You can use the quadratic formula to find a. If you want to run it with even integers, you know that two even numbers must be two apart; you do the same as above: \[a^2 + (a + 2)^2 = 41\] \[a^2 + a^2 + 4a + 4 = 41\] \[2a^2 + 4a + 4 = 41\] \[2a^2 + 4a - 37 = 0\] But, running this through the quadratic formula will leave you with a decimal number, not an integer.

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