Can u help plz ........

[Unregistered]

1) Find the value of q such that the equation:

qx^2+3x=q-3 has a repeated root.

2) Solve

2x^2+3k^2=5kx

3) Factorise -x^3+5x^2-7x+3
State the range of values of x for which f(x) is less than or equal to 0

[CoolerShaker]

1.) For a repeated root, the quadratic of the equation's discriminant is equal to zero, hence:

qx^2 + 3x + (3-q) = 0
discriminant b^2 - 4ac = 0
---> 3^2 - 4(q)(3-q) = 0
---> 9 - 4q(3-q) = 0
---> 9 - 12q + 4q^2 = 0
---> q = {12(+/-)Sq.root(144-4(4)(9)]}/8
---> q = [12(+/-)0]/8

---> Solution: q = 3/2

2.) 2x^2+3k^2=5kx
---> 2x^2 - 5kx + 3k^2 = 0
---> x = {5k(+/-)sq.root([5k]^2-4(2)(3k^2)}/4
---> x = {5k(+/-)sq.root(25k^2 - 24k^2)}/4
---> x = {5k(+/-)sq.root(k^2)}/4
---> x = 5k (+/-)k/4
---> x = (5k + k)/4 = 6k/4
---> Solution: x = 3k/2
Or: ---> x = (5k - k)/4 = 4k/4
---> Solution: x = k

3.) Let f(x) = -x^3 + 5x^2 - 7x + 3
f(1) = -(1)^3 + 5(1)^2 - 7(1) + 3 = -1 + 5 - 7 + 3 = 4 - 4 = 0
As f(1) = 0 ----> (x-1) is a factor of f(x)
Hence: (x-1)(ax^2 + bx + c) = -x^3 + 5x^2 - 7x + 3
By comparing co-efficients:
---> a = -1
---> -a + b = 5 ---> 1 + b = 5 ---> b = 4
---> -c = 3 ---> c = -3

Hence: f(x) = (x - 1)(-x^2 + 4x - 3) = -x^3 + 5x^2 - 7x + 3
---> -x^3 + 5x^2 - 7x + 3 = (x - 1)(x - 3)(-x + 1)

When f(x) = 0 ---> x = 1 or x = 3
When x < 1 ---> f(x) = (-ve)(-ve)(+ve) = +ve
When x lies between 1 and 3 ---> f(x) = (+ve)(-ve)(-ve) = + ve
When x > 3 ---> f(x) = (+ve)(+ve)(-ve) = -ve

Hence: f(x) is smaller than or equal to zero when x > 3