maths

Q1:
Sorry I did find it.

Q2:

(x+a)(x+a-1)=(x+a-1)

x+a=1

X=(1-a)

ans:X=1-a

Q3:
(x-2)^2-8(x-2)^2+16=0

should→(x-2)^2-8(x-2)+16=0

[(x-2)-4]^2=0

ans:X=6(root)

2012-11-05 20:47:08 補充：
My English is too bad Sorry~~

Q1. y = a + c/x where a and c are constants.
This is the graph of y = 1/x ( a rectangular hyperbola) but enlarged/contracted c times along the y - direction and then translated upward/downward by a units.
Q2.
(x + a)(x + a - 1) = (x + a - 1)
(x + a)(x + a - 1) - (x + a - 1) = 0
(x + a - 1)[(x + a) - 1] = 0
(x + a - 1)^2 = 0
so x = 1 - a.
Q3.
(x - 2)^2 - 8(x - 1) + 16 = 0 [assume 8(x - 2)^2 is a typing mistake.]
Put x - 2= y
so y^2 - 8y + 16 = 0
(y - 4)^2 = 0
so y = 4
x = 2 + y = 6

Q1. y = a + c/x where a and c are constants.
This is the graph of y = 1/x ( a rectangular hyperbola) but enlarged/contracted c times along the y - direction and then translated upward/downward by a units.
Q2.
(x + a)(x + a - 1) = (x + a - 1)
(x + a)(x + a - 1) - (x + a - 1) = 0
(x + a - 1)[(x + a) - 1] = 0
(x + a - 1)^2 = 0
so x = 1 - a.
Q3.
(x - 1)^2 - 8(x - 1) + 16 = 0 [assume 8(x - 1)^2 is a typing mistake.]
Put x - 1 = y
so y^2 - 8y + 16 = 0
(y - 4)^2 = 0
so y = 4
x = 1 + y = 5 (two equal real roots).

2012-11-06 20:58:27 補充：
Correction to Q3: Sorry that it should be (x - 2), not (x - 1), so x = 4 + 2 = 6.

Wrong Question?