急... S4. Maths (20 pts)

Q1a.
Sum of roots = a + b = - 3.........(1)
Product of roots = ab = - 8..........(2)
For the new equation,
sum of roots = a^2 + b^2 + 2 = (a + b)^2 - 2ab + 2 = (-3)^2 - 2(-8) + 2
= 9 + 16 + 2 = 27 (note: that means a^2 + b^2 = 25)
product of roots = (a^2 - 1)(b^2 - 1) = (ab)^2 + 1 - (a^2 + b^2)
= (-8)^2 + 1 - 25 = 64 + 1 - 25 = 40
so the equation is x^2 - 27x + 40 = 0.
Q1b
sum of roots = (a^2 + b^2 ) - (a + b) = 25 - (- 3) = 28
product of roots = (a^2 - b)(b^2 - a) = (ab)^2 - (a^3 + b^3) + ab
since (a + b)^3 = a^3 + 3a^b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b)
that is (- 3)^3 = a^3 + b^3 + 3(-8)(-3)
-27 = a^3 + b^3 + 72
so a^3 + b^3 = - 99
so product of roots = (-8)^2 - (-99) + (-8) = 64 + 99 - 8 = 155
so the equation is x^2 - 28x + 155 = 0.
Q2.
Let annual interest rate = r% = r/100
so amount after 1 year = 50000(1 + r/100)
Interest rate of 2nd year = (1 + r)/100
Amount after 2nd year = 50000(1 + r/100)[1 + (1 + r)/100]
So interest received = 50000(1 + r/100)[1 + (1 + r)/100] - 50000 = 3150
50000(1 + r/100)[1 + (1 + r)/100] = 53150
50000[(100 + r)/100][(101 + r)/100] = 53150
5(100 + r)(101 + r) = 53150
(100 + r)(101 + r) = 10630
10100 + 201r + r^2 - 10630 = 0
r^2 + 201r - 530 = 0
r = (206.2 - 201)/2 = 5.2/2 = 2.6
so 1st year interest rate is 2.6%
Remark: Assumed $3150 is the total interest received in the 2 years. If it is just in the 2nd year, then interest received in the 1st year must be deducted first and adjustment in the equation must be made.




2009-10-05 08:10:29 補充：
Interest received in the 1st year = 50000(r/100) = 500r. Interest received in the 2nd year = 3150 - 500r. The equation becomes r^2 + 301r - 530 = 0. r becomes 1.75%.

2009-10-05 08:16:09 補充：
Sorry, it should be total interest received in 2 years = 3150 + 500r, the equation becomes r^2 + 101r - 530 = 0, (r - 5)(r + 106) = 0, so r = 5%. (Please disregard r = 1.75%)

2009-10-05 08:18:46 補充：
In the remark, it should be .....' added first'..., not.....'deducted first'..... Please correct.

2009-10-05 08:26:05 補充：
Sorry, in Q1a, product of roots should be (ab)^2 + 1 + (a^2 + b^2) = 64 + 1 + 25 = 90, so equation should be x^2 - 27x + 90 = 0.