find all possible solutions to the equation x^2 + [3]x + [2] = [0] in Z47?

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Find all possible solutions to the equation x^2 + [3]x + [2] = [0] in Z49

Answer 1

x^2 + 3x + 2 = 0

Factor the equation:

(x + 2)(x + 1) = 0
x = -2, -1

Answer 2

That means x^2 + 3x + 2 ≡ 0 (mod 49), or
(x + 1)(x + 2) ≡ 0 (mod 49), the latter fulfilled when
x + 1 ≡ 0 (mod 49) or x + 2 ≡ 0 (mod 49) only,
/x + 1 ≡ 7 (mod 49) and simultaneously x + 2 ≡ 7 (mod 49) is impossible, no other alternatives present for 49 = 7^2/, so
x ≡ -1 ≡ 48 (mod 49) or x ≡ -2 ≡ 47 (mod 49), or in notations You use we have 2 solutions over Z_{sub49}: x = [47] and x = [48]

P.S. (Edit) I haven't noticed Z_{sub47} in the original question, but it is treated the same way producing
x ≡ -1 ≡ 46 (mod 47) or x ≡ -2 ≡ 45 (mod 47), or
x = [45] and x = [46]

Answer 3

Z_47 is an integral domain, so the zero divisor law holds.
So we can factor your equation:
(x+1)(x+2) = 0(mod 47)
So x = -1 or -2, i.e. 45 and 46(mod 47)