To solve the quadratic equation of one variable (1) x2 + 3x + 1 = 0 (2) x2-10x + 9 = 0 (3) (2x-1) 2 = (3x + 2) 2 (4) (x-1) (x + 2) = 2 (x + 2)

To solve the quadratic equation of one variable (1) x2 + 3x + 1 = 0 (2) x2-10x + 9 = 0 (3) (2x-1) 2 = (3x + 2) 2 (4) (x-1) (x + 2) = 2 (x + 2)

(1) X2 + 3x + 1 = 0, where a = 1, B = 3, C = 1, ∵ b2-4ac = 9-4 × 1 × 1 = 5 ＞ 0, ∵ x = − 3 ± 52, ∵ X1 = − 3 + 52, X2 = − 3 − 52; (2) the factorization formula is: (x-1) (X-9) = 0, X-1 = 0 or X-9 = 0, and the solution is: X1 = 1 (3) the square root is 2x-1 = ± (3x + 2), that is, 2x-1 = 3x + 2 or 2x-1 = - (3x + 2), х X1 = - 3, X2 = - 15; (4) the factorization is (x + 2) (x-1-2) = 0, then x + 2 = 0 or x-3 = 0, and the solution is X1 = - 2, X2 = 3