因式分解解一元二次方程例题

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因式分解解一元二次方程例题

例题 1

方程:x^2 5x = 0

解析:提取公因式x,得到x(x 5) = 0,则x = 0或x 5 = 0,解得x_1 = 0,x_2 = 5

例题 2

方程:x^2 + 6x + 8 = 0

解析:因式分解为(x + 2)(x + 4) = 0,则x + 2 = 0或x + 4 =

0,解得x_1 = 2,x_2 = 4

例题 3

方程:x^2 7x + 10 = 0

解析:因式分解为(x 2)(x 5) = 0,则x 2 = 0或x 5 = 0,解得x_1 = 2,x_2 = 5

例题 4

方程:2x^2 5x 3 = 0

解析:因式分解为(2x + 1)(x 3) = 0,则2x + 1 = 0或x 3 =

0,解得x_1 = \frac{1}{2},x_2 = 3

例题 5

方程:3x^2 + 2x 1 = 0

解析:因式分解为(3x 1)(x + 1) = 0,则3x 1 = 0或x + 1 =

0,解得x_1 = \frac{1}{3},x_2 = 1

例题 6

方程:4x^2 12x + 9 = 0 解析:因式分解为(2x 3)^2 = 0,则2x 3 = 0,解得x_1 =

x_2 = \frac{3}{2}

例题 7

方程:x^2 8x + 16 = 0

解析:因式分解为(x 4)^2 = 0,则x 4 = 0,解得x_1 = x_2

= 4

例题 8

方程:5x^2 10x + 5 = 0

解析:提取公因式5得5(x^2 2x + 1) = 0,再因式分解为5(x

1)^2 = 0,则x 1 = 0,解得x_1 = x_2 = 1

例题 9

方程:6x^2 + 7x + 1 = 0

解析:因式分解为(6x + 1)(x + 1) = 0,则6x + 1 = 0或x + 1

= 0,解得x_1 = \frac{1}{6},x_2 = 1

例题 10

方程:x^2 10x + 25 = 0

解析:因式分解为(x 5)^2 = 0,则x 5 = 0,解得x_1 = x_2

= 5

例题 11

方程:2x^2 8 = 0

解析:提取公因式2得2(x^2 4) = 0,再因式分解为2(x +

2)(x 2) = 0,则x + 2 = 0或x 2 = 0,解得x_1 = 2,x_2 = 2

例题 12

方程:3x^2 12 = 0 解析:提取公因式3得3(x^2 4) = 0,再因式分解为3(x +

2)(x 2) = 0,则x + 2 = 0或x 2 = 0,解得x_1 = 2,x_2 = 2

例题 13

方程:4x^2 + 8x = 0

解析:提取公因式4x得4x(x + 2) = 0,则4x = 0或x + 2 =

0,解得x_1 = 0,x_2 = 2

例题 14

方程:5x^2 25x = 0

解析:提取公因式5x得5x(x 5) = 0,则5x = 0或x 5 = 0,解得x_1 = 0,x_2 = 5

例题 15

方程:x^2 + 4x 21 = 0

解析:因式分解为(x + 7)(x 3) = 0,则x + 7 = 0或x 3 =

0,解得x_1 = 7,x_2 = 3

例题 16

方程:2x^2 + 5x 3 = 0

解析:因式分解为(2x 1)(x + 3) = 0,则2x 1 = 0或x + 3 =

0,解得x_1 = \frac{1}{2},x_2 = 3

例题 17

方程:3x^2 11x 4 = 0

解析:因式分解为(3x + 1)(x 4) = 0,则3x + 1 = 0或x 4 =

0,解得x_1 = \frac{1}{3},x_2 = 4

例题 18

方程:4x^2 + 7x 2 = 0 解析:因式分解为(4x 1)(x + 2) = 0,则4x 1 = 0或x + 2 =

0,解得x_1 = \frac{1}{4},x_2 = 2

例题 19

方程:5x^2 13x + 6 = 0

解析:因式分解为(5x 3)(x 2) = 0,则5x 3 = 0或x 2 =

0,解得x_1 = \frac{3}{5},x_2 = 2

例题 20

方程:6x^2 11x + 3 = 0

解析:因式分解为(2x 3)(3x 1) = 0,则2x 3 = 0或3x 1 =

0,解得x_1 = \frac{3}{2},x_2 = \frac{1}{3}

例题 21

方程:x^2 6x 16 = 0

解析:因式分解为(x 8)(x + 2) = 0,则x 8 = 0或x + 2 =

0,解得x_1 = 8,x_2 = 2

例题 22

方程:2x^2 7x 4 = 0

解析:因式分解为(2x + 1)(x 4) = 0,则2x + 1 = 0或x 4 =

0,解得x_1 = \frac{1}{2},x_2 = 4

例题 23

方程:3x^2 8x 3 = 0

解析:因式分解为(3x + 1)(x 3) = 0,则3x + 1 = 0或x 3 =

0,解得x_1 = \frac{1}{3},x_2 = 3

例题 24

方程:4x^2 9x + 2 = 0 解析:因式分解为(4x 1)(x 2) = 0,则4x 1 = 0或x 2 =

0,解得x_1 = \frac{1}{4},x_2 = 2

例题 25

方程:5x^2 16x + 3 = 0

解析:因式分解为(5x 1)(x 3) = 0,则5x 1 = 0或x 3 =

0,解得x_1 = \frac{1}{5},x_2 = 3

例题 26

方程:6x^2 17x + 5 = 0

解析:因式分解为(2x 5)(3x 1) = 0,则2x 5 = 0或3x 1 =

0,解得x_1 = \frac{5}{2},x_2 = \frac{1}{3}

例题 27

方程:x^2 18x + 81 = 0

解析:因式分解为(x 9)^2 = 0,则x 9 = 0,解得x_1 = x_2

= 9

例题 28

方程:2x^2 10x + 8 = 0

解析:提取公因式2得2(x^2 5x + 4) = 0,再因式分解为2(x

1)(x 4) = 0,则x 1 = 0或x 4 = 0,解得x_1 = 1,x_2 = 4

例题 29

方程:3x^2 15x + 12 = 0

解析:提取公因式3得3(x^2 5x + 4) = 0,再因式分解为3(x

1)(x 4) = 0,则x 1 = 0或x 4 = 0,解得x_1 = 1,x_2 = 4

例题 30

方程:4x^2 16x + 12 = 0 解析:提取公因式4得4(x^2 4x + 3) = 0,再因式分解为4(x

1)(x 3) = 0,则x 1 = 0或x 3 = 0,解得x_1 = 1,x_2 = 3