Question

一元二次方程 因式法

Answer 1

1) 21 = 2x2 + 19x 2x2 + 19x﹣21 = 0 (2x + 21)(x﹣1) = 0 x = -21 / 2或x = 1 2)x(x + 5) = 24 x2 + 5x﹣24 = 0 (x﹣3)(x + 8) = 0 x = 3或x = -8 3)(x﹣3)(x + 7) = -9 x2 + 7x﹣3x﹣21 = -9 x2 + 4x﹣12 = 0 (x﹣2)(x + 6) = 0 x = 2或x = -6 4)(x﹣2)(2x + 4) = 5(x﹣1) (x﹣2) [2(x + 2)] = 5x﹣5 2(x﹣2)(x + 2) = 5x﹣5 2(x2﹣4) = 5x﹣5 2x2﹣5x﹣3 = 0 (2x + 1)(x﹣3) = 0 x = -1 / 2或x = 3 5)x2 = 3x x2﹣3x = 0 x(x﹣3) = 0 x = 0或x = 3 6)2x(x + 1) = 3(x + 1) 2x(x + 1)﹣3(x + 1) = 0 (x + 1)(2x﹣3) = 0 x = -1或x = 3 / 2 7)x(2x + 7) = 3(2x + 7) x(2x + 7)﹣3(2x + 7) = 0 (2x + 7)(x﹣3) = 0 x = -7 / 2或x = 3 8)3(x﹣2)2 = x(x﹣2) 3(x﹣2)2﹣x(x﹣2) = 0 (x﹣2)[3(x﹣2)﹣x] = 0 (x﹣2)(3x﹣6﹣x) = 0 (x﹣2)(2x﹣6) = 0 x = 2或x = 3 9)(x﹣1)(2x + 3) = (x﹣1)(x﹣4) (x﹣1)(2x + 3)﹣(x﹣1)(x﹣4) = 0 (x﹣1)[(2x + 3)﹣(x﹣4)] = 0 (x﹣1)(2x + 3﹣x + 4) = 0 (x﹣1)(x + 7) = 0 x = 1或x = -7