F4 math problem~!!

1. a,b are roots of the equation x^2 - 3x - 5 = 0
a + b = 3
ab = -5
(a) If the roots are a + 1/a and b + 1/b
Sum of roots = a + 1/a + b + 1/b
= (a+b) + (1/a+1/b)
= (a+b) + (a+b)/(ab)
= 3 - 3/5
= 12/5
Product of roots = (a + 1/a)(b + 1/b)
= ab + a/b + b/a + 1/(ab)
= -5 - 1/5 + (a²+b²)/(ab)
= -26/5 - (a²+2ab+b²-2ab)/5
= -26/5 - [(a+b)²-2ab]/5
= -26/5 - [3²+2*5]/5
= -9
So the new equation becomes
x² - (sum of roots)x + (product of roots) = 0
x² - 12x/5 - 9 = 0
5x² - 12x - 45 = 0

(b) If the roots are b² + a² and b² + a²
Each root is b² + a² = (b + a)² - 2ab
= 3² + 2*5
= 19
So the equation becomes,
(x - 19)² = 0

2. Let m, n be the roots of x² - 2x - 5 = 0,
then the roots of 2x² - bx + c = 0 are 3m and 3n.
From the 1st equation,
mn = -5
From the 2nd equation,
(3m)(3n) = 9mn = c/2

So 9mn = 9(-5) = c/2
c/2 = -45
c = -90
So the answer is D.

1.a)
sum of roots : a + b = 3
product of roots : ab = -5

(x - a - 1/a)(x - b - 1/b) = 0
x^2 - bx - x/b - ax + ab + a/b - x/a + b/a + 1/ab = 0
x^2 - [a + b + 1/a + 1/b]x + ab +a/b + b/a + 1/ab = 0
x^2 - [(a + b) + (a + b)/ab]x + ab + 1/ab + (a^2 + b^2)/ab = 0
x^2 - [(a + b) + (a + b)/ab]x + ab + 1/ab + [(a + b)^2 - 2ab)/ab = 0
x^2 - (3 - 3/5)x - 5 - 1/5 + (3^2 + 10)/(-5) = 0
x^2 - 12x/5 - 9 = 0
5x^2 - 12x - 45 = 0

b)
b^2 + a^2
= (b + a)^2 - 2ab
= (-3)^2 - (-5)
=14

(x - 14)^2 = 0
x^2 - 28x + 196 = 0


2.
let h, k be the roots of 2x^2 - bx + c = 0
c = hk

then the roots of x^2 - 2x - 5 = 0 would be h/3, k/3
-5 = hk/9
hk = -45
c = -45

2006-12-17 10:53:39 補充：
1. b) 應該係b^2 a^2= (b a)^2 - 2ab= (-3)^2 - 2(-5)=19(x - 19)^2 = 0x^2 - 38x 361 = 0

2006-12-17 10:55:13 補充：
2.let h, k be the roots of 2x^2 - bx c = 0c/2 = hkthen the roots of x^2 - 2x - 5 = 0 would be h/3, k/3-5 = hk/9hk = -45c/2 = -45c = -90