F4. Quadratic Equations

1.
(x + 1)² + 3(x + 1) - 4 = 0
[(x + 1) + 4] [(x + 1) - 1] = 0
(x + 5)x = 0
x = -5 or x = 0

Alternative method :
Let u = x + 1
(x + 1)² + 3(x + 1) - 4 = 0
u² + 3u - 4 = 0
(u + 4)(u - 1) = 0
u = -4 or u = 1
x + 1 = -4 or x + 1 = 1
x = -5 or x = 0



2.
(4 - x)² - 5(4 - x) - 14 = 0
[(4 - x) - 7] [(4 - x) + 2] = 0
4 - x = 7 or 4 - x = -2
x = -3 or x = 6

Alternative method :
Let u = 4 - x
(4 - x)² - 5(4 - x) - 14 = 0
u² - 5u - 14 = 0
(u - 7)(u + 2) = 0
u = 7 or u = -2
4 - x = 7 or 4 - x = -2
x = -3 or x = 6

2013-12-30 17:21:49 補充：
kit******@ymail.com 的第二題，9 和 0 不是正確的答案。

x² - 3x - 18 = 0
(x - 6)(x + 3) = 0
x = 6 or x = -3

2014-01-01 22:21:01 補充：
1.
y² + 3y - 4 = 0
(y + 4)(y - 1) = 0

所以：()² + 3() - 4 = 0
[() + 4] [() - 1] = 0

將 (x + 1) 看成 () 便可。

若還是看不懂，可參看 alternative method。

(x+1)^2+3(x+1)-4=0

(x+1)^2+3(x+1)-4
=x^2+2x+1+3x+3-4
=x^2+5x
=x(x+5)

x(x+5)=0
x=0 or x+5=0

so x=0 or -5


(4-x)^2-5(4-x)-14
=16-8x+x^2-20+5x-14
=x^2-3x-18

x^2-3x-18=0
x=(-(-3)±√((-3)^2-4(1)(-18)))/2(1)
x=[9±√(9+72)]/2
x=(9+9)/2 or (9-9)/2
x=9 or 0