(a) Solve (x^2)-x-2.
(b) Hence, solve [(x^2)-2x]^2-(x^2)+2x-2=0. (Leave your answers in surd form.)
15分1條~ F.4 quadratic equation
2009-10-04 9:52 pm
回答 (2)
2009-10-04 10:32 pm
✔ 最佳答案
Here is the solution.圖片參考：http://www.acm313.com/knowledge/1.gif
圖片參考：http://www.acm313.com/knowledge/2.gif
參考: me
2009-10-04 10:30 pm
a. x2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = 2 or -1
b. [x2 - 2x]2 - x2 + 2x - 2 = 0
(x2 - 2x)2 - (x2 - 2x) - 2 = 0
Put y = x2 - 2x
So, it becomes y2 - y - 2 = 0
From part a, y = -1 or 2
So, x2 - 2x = -1 or x2 - 2x = 2
x2 - 2x + 1 = 0 or x2 - 2x - 2 = 0
(x - 1)2 = 0 or x = {-(-2) +- sqrt[(-2)2 - 4(1)(-2)]}/2
x = 1 (repeated) or x = 1 +- sqrt3
(x + 1)(x - 2) = 0
x = 2 or -1
b. [x2 - 2x]2 - x2 + 2x - 2 = 0
(x2 - 2x)2 - (x2 - 2x) - 2 = 0
Put y = x2 - 2x
So, it becomes y2 - y - 2 = 0
From part a, y = -1 or 2
So, x2 - 2x = -1 or x2 - 2x = 2
x2 - 2x + 1 = 0 or x2 - 2x - 2 = 0
(x - 1)2 = 0 or x = {-(-2) +- sqrt[(-2)2 - 4(1)(-2)]}/2
x = 1 (repeated) or x = 1 +- sqrt3
參考: Physics king
收錄日期: 2021-04-13 16:53:00
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