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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