amaths

1. (kx - 1)2 = k

k2x2 – 2kx + 1 = k

k2x2 – 2kx + (1 - k) = 0

Product of roots, (1 - k)/k2 = -3/16

i.e. 16(1 - k) = -3k2

3k2 – 16k + 16 = 0

k = 4 or 4/3

2.a. Let b and –b be the roots

x2 + a(3a - 5)x = 2(x + 4a)

x2 + [a(3a - 5) – 2]x – 8a = 0

Sum of roots, b + (-b) = -[a(3a - 5) - 2]

So, 0 = -[a(3a - 5) - 2]

3a2 – 5a – 2 = 0

a = 2 or -1/3

b. Because a > 0, so a = 2

The equation becomes:

x2 – 16 = 0

x = 4 or -4

1)
Firstly re-write the equation...
(kx-1)^2 = k
k^2x^2 - 2kx + 1 - k = 0
k^2x^2 - 2kx + (1-k) = 0
Thus, a=k^2, b=-2k, c=1-k

Let m and n be the roots of the equation...
m*n = c/a = (1-k)/k^2

As the question mentioned, product of roots is -3/16
Thus,
(1-k)/k^2 = -3/16
16(1-k) = -3k^2
16 - 16k = -3k^2
3k^2 - 16k + 16 = 0
(3k-4)(k-4) = 0
k = 4/3 or k = 4

2a)
Firstly re-write the equation...
x^2 + a(3a-5)x = 2(x+4a)
x^2 + 3a^2x - 5ax = 2x + 8a
x^2 + 3a^2x - 5ax - 2x - 8a = 0
x^2 + (3a^2 - 5a - 2) - 8a = 0

Let m be one of the roots of the equation, then another root must be -m
Sum of roots = -B/A
Product of roots = C/A
Thus,
m+(-m) = -(3a^2 - 5a - 2)/1
0 = 3a^2 - 5a - 2
(3a+1)(a-2) = 0
a = -1/3 or a = 2

2b)
If a>0, a must be equal to 2
Thus, put a=2 into the equation x^2 + (3a^2 - 5a - 2) - 8a = 0
x^2 + [ 3(2^2) - 5*2 - 2 ] - 8(2) = 0
x^2 - 16 = 0
x^2 - (4)^2 = 0
(x+4)(x-4) = 0
x = 4 or x = -4