Mathematics

Q1: put x=1 into the equation
(a+1)^2-3a+2a-1=0
a^2+2a+1-3a+2a-1=0
a^2+a=0
a(a+1)=0
a=0 OR a= -1


Q2:A)PUT (k,0) and (-k,0) seperately
then, we have 2 equations
k^2+km+n=0------(1)
k^2-km+n=0------(2)
(1)-(2):
2km=0
so,m=0 (as k not equal to 0)

B) BY (A),m=0
then ,the equation become y=x^2+n
put (0, -k) to the equation
then, n= -k

2008-07-25 17:48:30 補充：
C) AS k and -k are roots of the equation y=x^2-k
so product of roots = -k
so -k*k=-k
-k^2=-k
k=1

Sub a = 1
(a+1)^2 - 3a + 2a -1 = 0
a^2 + 2a + 1 - a -1 = 0
a^2 + a = 0
a(a+1) = 0
a = 0 or -1

a)
Sub x= 0
n = -k

when y= 0
k , -k is the root of the equation
y = x^2 + mx + n
y = x^2 + mx - k


product of the roots = -k = -k^2
sum of the roots = -m = 0
m =0 // , k=1