用配方法解一元二次方程(1)x2+2x-8=0 (2)x2-4=2x (3)3y2-6y-24=0 (4)4x2-7x+2=0 (5)二分之一x2-2x-9=0
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![用配方法解一元二次方程(1)x2+2x-8=0 (2)x2-4=2x (3)3y2-6y-24=0 (4)4x2-7x+2=0 (5)二分之一x2-2x-9=0](/uploads/image/z/10171750-22-0.jpg?t=%E7%94%A8%E9%85%8D%E6%96%B9%E6%B3%95%E8%A7%A3%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B%281%29x2%2B2x-8%3D0+%282%29x2-4%3D2x+%283%293y2-6y-24%3D0+%284%294x2-7x%2B2%3D0+%285%29%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80x2-2x-9%3D0)
用配方法解一元二次方程(1)x2+2x-8=0 (2)x2-4=2x (3)3y2-6y-24=0 (4)4x2-7x+2=0 (5)二分之一x2-2x-9=0
用配方法解一元二次方程(1)x2+2x-8=0 (2)x2-4=2x (3)3y2-6y-24=0 (4)4x2-7x+2=0 (5)二分之一x2-2x-9=0
用配方法解一元二次方程(1)x2+2x-8=0 (2)x2-4=2x (3)3y2-6y-24=0 (4)4x2-7x+2=0 (5)二分之一x2-2x-9=0
(1)x ² + 2 x - 8 = 0
x ² + 2 x + 1 = 8 + 1
(x + 1)² = 9
x + 1 = ± 3
x = ± 3 - 1
x1 = 3 - 1 = 2
x2 = - 3 - 1 = - 4
(2)x ² - 4 = 2 x
x ² - 2 x - 4 = 0
x ² - 2 x + 1 = 4 + 1
(x - 1)² = 5
x - 1 = ± √5
x = ± √5 + 1
x1 = √5 + 1
x2 = 1 - √5
(3)3 y ² - 6 y - 24 = 0
y ² - 2 y - 8 = 0
y ² - 2 y + 1 = 8 + 1
(y - 1)² = 9
y - 1 = ± 3
y = ± 3 + 1
y1 = 3 + 1 = 4
y2 = - 3 + 1 = - 2
(4)4 x ² - 7 x + 2 = 0
x ² - (7 / 4)x + 1 / 2 = 0
x ² - (7 / 4)x + (7 / 8)² = - 1 / 2 + (7 / 8)²
(x - 7 / 8)² = - 1 / 2 + 49 / 64
(x - 7 / 8)² = 17 / 64
x - 7 / 8 = ±√17 / 8
x = ± √17 / 8 + 7 / 8
x1 = √17 / 8 + 7 / 8 = (√ 17 + 7)/ 8
x2 = - √17 / 8 + 7 / 8 = (7 - √17)/ 8
(5)(1 / 2)x ² - 2 x - 9 = 0
x ² - 4 x - 18 = 0
x ² - 4 x + 4 = 18 + 4
(x - 2)² = 22
x - 2 = ±√22
x = ± √22 + 2
x1 = √22 + 2
x2 = 2 - √22