1) A quadratic will have no real roots when its discriminant is negative
[tex]0 = ax^2 + bx + c = x^2 + 2k x + (4k-3)[/tex]
[tex]d = b^2 - 4ac = (2k)^2 - 4(1)(4k-3) = 4k^2 - 16k + 12 < 0 [/tex]
Dividing by 4,
[tex] k^2 - 4 k + 3 < 0 \quad\checkmark[/tex]
We have a positive coefficient on k^2 so this parabola is a CUP (concave up positive) so has a minimum at the vertex. If the vertex y value is less than zero, the inequality will be true in the range between the zeros.
[tex](k-3)(k-1)<0[/tex]
That's true for
[tex]1 < k < 3[/tex]
2) We look for the meet of the line and the parabola:
[tex]2x + k + 2 = 2x^2 + (k+2)x + 8[/tex]
[tex]0 = 2x^2 + kx + (6-k)[/tex]
For two intersections we need a positive discriminant:
[tex]k^2 - 4(6-k)(2) > 0[/tex]
[tex]k^2 + 8k - 48 > 0[/tex]
[tex](k+12)(k-4) > 0[/tex]
That's means two negative or two positive, so
[tex]k < 12[/tex] or [tex]k > 4[/tex]
