algebra problem

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If you have the sume of squares of the digits of a positive two-digit number is 20, the tens digit is 2 more than the units digit.

answer is 2^2 and 4^2 (?)

  • algebra problem -

    Yes, that looks right.

  • algebra problem -

    a b
    a^2+b^2 = 20
    a = b+2

    (b+2)^2 + b^2 = 20
    b^2 + 4 b + 4 + b^2 = 20
    2 b^2 + 4 b -16 = 0
    b^2 + 2 b - 8 = 0
    (b-2)(b+4 = 0
    b = 2
    a = 4
    so number is indeed
    42

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