I have to solve the equation x^2 + 8x – 2 = 0 using both
(1) The quadratic formula
(2) Completing the square
Please help solve for both equations.?
2008-12-18 1:35 pm
回答 (4)
2008-12-18 1:43 pm
✔ 最佳答案
x=-4 + - (16+2)^1/2= -4+(18)^1/2 & x=-4-(18)^1/2
2008-12-18 2:31 pm
x1,x2 = [-8 ±√(64+8)] / 2 = -4 ± 3√2
x^2 + 8x – 2 = 0
x^2 + 8x +16-18=0
(x+4)^2=18
x+4= ± 3√2
x=-4 ± 3√2
x^2 + 8x – 2 = 0
x^2 + 8x +16-18=0
(x+4)^2=18
x+4= ± 3√2
x=-4 ± 3√2
2008-12-18 2:09 pm
1)
x^2 + 8x - 2 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 1
b = 8
c = -2
x = [-8 ±√(64 + 8)]/2
x = [-8 ±√72]/2
x = [-8 ±√(2^2 * 3^2 * 2)]/2
x = [-8 ±(2)(3)√2]/2
x = [-8 ±6√2]/2
x = -4 ±3√2
∴ x = -4 ±3√2
= = = = = = = =
2)
x^2 + 8x - 2 = 0
x^2 + 8x = 2
x^2 + 4x + 4x = 2
x^2 + 4x + 4x + 16 = 2 + 16
(x^2 + 4x) + (4x + 16) = 18
x(x + 4) + 4(x + 4) = 18
(x + 4)^2 = 18
x + 4 = ±√18
x = ±√(3^2 * 2) - 4
x = ±3√2 - 4
∴ x = ±3√2 - 4
x^2 + 8x - 2 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 1
b = 8
c = -2
x = [-8 ±√(64 + 8)]/2
x = [-8 ±√72]/2
x = [-8 ±√(2^2 * 3^2 * 2)]/2
x = [-8 ±(2)(3)√2]/2
x = [-8 ±6√2]/2
x = -4 ±3√2
∴ x = -4 ±3√2
= = = = = = = =
2)
x^2 + 8x - 2 = 0
x^2 + 8x = 2
x^2 + 4x + 4x = 2
x^2 + 4x + 4x + 16 = 2 + 16
(x^2 + 4x) + (4x + 16) = 18
x(x + 4) + 4(x + 4) = 18
(x + 4)^2 = 18
x + 4 = ±√18
x = ±√(3^2 * 2) - 4
x = ±3√2 - 4
∴ x = ±3√2 - 4
2008-12-18 1:42 pm
x^2 + 8x – 2 = 0
x^2 + 8x +16-18=0
(x+4)^2=18
x+4= ± 3√2
x=-4 ± 3√2
x_1,x_2 = (-8 ±√(64+8))/2=-4 ± 3√2
x^2 + 8x +16-18=0
(x+4)^2=18
x+4= ± 3√2
x=-4 ± 3√2
x_1,x_2 = (-8 ±√(64+8))/2=-4 ± 3√2
收錄日期: 2021-05-01 11:38:33
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