Originally Posted by

**whitecorp** Posting a long list of questions and their answers is not the way to ask for guidance. If you post your workings and show us where you are stuck, others would be more obliged to help. Peace.

Alright, I apologize

For Q1, I will try since there is a hint now

Here are how I attempt the questions:

For Q2, I do not know how to start working

For Q3,

a)i)

3x^2 + 2kx - k =0

(2k)^2 - 4(3)(-k) = 0

4k^2 - 12(-k) = 0

4k^2 + 12k = 0

4k = 0 or k + 3 = 0

k = 0 (rejected) or k = -3

for ii)

4k(k + 3) > or equal to 0

4k > or equal to 0, k + 3 > or equal to 0

k> or equal to 1/-3 <- is that what they mean by reciprocal?

iii)

4k(k + 3) > or equal to 0

4k(k + 3)^2 > or equal to 0

4k > or equal to 0, (k + 3)^2 > or equal to 0

(k^2 + 6k + 9) > or equal to 0 <- from here if I factorise out, it is still the same so I do not know what to do

b)

4(x+1)(x-4) = k

(4x+4)(x-4) = k

4x^2 - 12x - 16 - k = 0

4x^2 - 12x - 16 - k = 0

(-12)^2 - 4(4)(-16-k) > or equal to 0 - 8

144 + 256 + 16k > or equal to 0 - 8

400 + 16k > or equal to 0 - 8

400 < or equal to -16k - 8

408 < or equal to -16k

-25.2 < or equal to K

For Q4

2x^2 + 7x + 9 = 0

roots are A + B

a = 2, b = 7, c = a

A+B = 7/2

AB = a/2

4x^2 + bx + 16 = 0

a = 4, b = b, c = 16

A+B = -b/4

AB = 4

A^2+B^2

= (A+B)^2 - 2AB

= (-b/4)^2 - 2(4)

= b^2/16 - 81

A^2B^2

= (AB)^2

= (4)^2

= 16 <- from this point, I am not sure what to do next

For Q5

3x^2 - squareroot24x - 2 = 0

a = 3, b = -squareroot24, c=-2

m+n = squareroot24/3

mn= -2/3

4/m - 2/n

= 4n - 2m/mn

= (4n)^2 + (2m)^2 - 2mn/mn

= 16n^2 + 4m^2 - 2mn/mn

= 4(4n^2 + m^2) - 2mn/mn <- from this point, I do not know how to factorise 4(4n^2 + m^2) into somthing like (m + n) so that I can sub m + n into it