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anonymous

  • one year ago

I need help solving this equation please. The problem say that i need to find the total numbers of pairs of integers which satisfy the equation:

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    First bring them to perfect squares form: => (x^2 - 4xy + 4y^2) + (y^2 + 2y + 1) - 5 = 0 => (x-2y)^2 + (y+1)^2 = 5 Only 2 perfect squares can make 5 are 4+1, so there are 4 cases: y+1 = 1; y=0 => (x-2y)^2 =4 => x-2y =+/- 2 but y=0 so, x=+/-2 Solutions: (-2,0) and (2,0) similarly, y+1 = -1; y = -2 => x-2y = x+4 = +/-2; x = -2 or -6 Solutions: (-2,-2) and (-6,-2 y+1 = 2; y = 1 => x-2y = x-2 = +/-1; x = 1 or 3 Solutions: (1,1) and (3,1) y+1 = -2; y = -3 => x-2y = x+6 = +/-1; x = -5 or -7 Solutions: (-5, -3) and (-7, -3) So, 8 pairs of integers that can satisfy the given eqn.

  3. anonymous
    • one year ago
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    Thank you man! , now I understand! :D

  4. anonymous
    • one year ago
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    Nice explanation

  5. UsukiDoll
    • one year ago
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    I wish I had that explanation last semester. I got stuck with those find all values of x and y that produces a zero.

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