f.4 maths ( 10 PTS )

Q1:
x-2y = 1
x = 1+2y .......(1)

2^2-xy-y^2 = 0 .......(2)

Subsitution (1) into (2)
2(1+2y)^2-(1+2y)y-y^2 = 0
2(1+4y+4y^2)-(y+2y^2)-y^2 = 0
2+8y+8y^2-y-2y^2-y^2 = 0
5y^2+7y+2 = 0
(5y+2)(y+1) = 0
5y+2 = 0 or y+1 = 0
5y = -2 y = -1
y = -2/5

From (1), when y = -2/5, x = 1+2y
x = 1+2(-2/5)
x = 1- 4/5
x = 1/5

From (1), When y = -1, x = 1+2(-1)
x = 1-2
x = -1
So, when x = 1/5, y = -2/5 and when x = -1, y =-1.


Q2:
P(x) = x^3-(k+2)x^2+(3k+3)x-(2k+6)
(x+2) is a factor of P(x), mean P(-2) = 0

P(x) = 0
(-2)^3-(k+2)(-2)^2+(3k+3)(-2)-(2k+6) = 0
-8 - 4(k+2) - 2(3k+3) -2k-6 = 0
-8-4k-8-6k-6-2k-6 = 0
-28-12k = 0
-12k = 28
k = -28/12
k = -7/3

1) x-2y=1.........(1)
2x^2 -xy- y^2=0...........(2)
from (1), x=2y+1............(3)
sub. (3) into (2),
2(2y+1)^2 -(2y+1)y -y^2 =0
8y^2 +8y+2 -2y^2 -y -y^2 =0
5y^2 +7y+2 =0
(5y+2)(y+1)=0
y= -2/5 or y= -1
x= 1/5 x= -1



2) P(x) = x^3 -(k+2)x^2 +(3k+3)x - (2k+6)
P(-2)= 0
(-2)^3 -(k+2)(-2)^2 +(3k+3)(-2) - (2k+6)=0
-8-4k-8-6k-6-2k-6 =0
-28-12k=0
k= -7/3