Question

The quadratic equation 2qx(x+1)-8x^2-1, where q is a constant, has no real roots. Find the range of values of q.

Answer given is -2<q<4

Answer 1

I will assume that you meant the equation to be
2qx(x+1) - 8x^2 - 1 = 0

2qx^2 + 2qx - 8x^2- 1 = 0
x^2(2q - 8) + 2qx - 1 = 0
a quadratic with
a = 2q-8
b=2q
c=-1

to have no real roots, b^2 - 4ac < 0
4q^2 - 4(2q-8)(-1) < 0
4q^2 + 8q - 32 < 0
q^2 + 2q - 8 < 0
(q+4)(q-2) < 0

critical values: q = -4, q = 2

so -4 < q < 2

notice my signs are opposite of the given answer.
I can't find any errors in my solution, perhaps you can if there are any.
Let me know

Answer 2

The equation is 2qx(x+1)=-8x^2 -1

Answer 3

that's not what you had, so make the necessary changes in my solution.

Answer 4

I'm really sorry, but how?