1 f.4 Maths question

a. 4y2 + 4y - 3 = 0
(2y - 1)(2y + 3) = 0
2y - 1 = 0 or 2y + 3 = 0
y = 1/2 or y = -3/2
b. 4(5x - 1/2)2 + 4(5x - 1/2) - 3 = 0
Let u = 5x - 1/2
then 4(5x - 1/2)2 + 4(5x - 1/2) - 3 = 0
=> 4u2 + 4u - 3 = 0
=> u = 1/2 or u = -3/2
u = 1/2 => 5x - 1/2 = 1/2
=> 5x = 1
=> x = 0.2
u = -3/2 => 5x - 1/2 = -3/2
=> 5x = -1
=> x = -0.2

圖片參考：http://l.yimg.com/f/i/tw/ugc/rte/smiley_1.gif
a.
4y^2 + 4y -3 =0
(2y)^2 + 2( 2y) -3 = 0
Let a= 2y,
a^2 +2a -3= 0
(a-1)(a+3) =0
a-1=0 or a+3 =0
So, a=1 or a= -3
--> 2y=1 or 2y = -3
==> y = 1/2 or y= -3/2

b. since y = 1/2 or y = -3/2,
y= (5x-1/2) or y = (5x-1/2)
1/2= 5x-1/2 or -3/2 = 5x-1/2
x= 1/5 or x=-1/5
I hope u understand....
圖片參考：http://l.yimg.com/f/i/tw/ugc/rte/smiley_1.gif

a. y=1/2 or y=-3/2

b. 代y,
4y^2 + 4y - 3 = 0
y=1/2 or y=-3/2
1/2=5x - 1/2 3/2=5x - 1/2
x=1/5 x= 2/5