# Math Help - Solve the Quadratic Congruence

1. ## Solve the Quadratic Congruence

3x2 + 2x ≡ 4 mod 17

2. Originally Posted by mndi1105

$3x^2 + 2x \equiv 4 \mod 17$
since 6 and 17 are coprime, the above equation is equivalent to: $18x^2 + 12 x \equiv 24 \mod 17,$ which is equivalent to: $x^2 + 12 x - 7 \equiv 0 \mod 17.$

now we have: $x^2 + 12x - 7=(x+6)^2 - 43 \equiv (x+6)^2 - 9 \mod 17.$ but $(x+6)^2 - 9 = (x+3)(x+9).$ so the solutions to your equation are:

$x \equiv -3 \equiv 14 \mod 17$ and $x \equiv -9 \equiv 8 \mod 17. \ \ \ \Box$