Question

Help me solve this please

Answer 1

1) A quadratic will have no real roots when its discriminant is negative

$0 = ax^2 + bx + c = x^2 + 2k x + (4k-3)$

$d = b^2 - 4ac = (2k)^2 - 4(1)(4k-3) = 4k^2 - 16k + 12 < 0$

Dividing by 4,

$k^2 - 4 k + 3 < 0 \quad\checkmark$

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.

$(k-3)(k-1)$

That's true for

$1 < k < 3$

2) We look for the meet of the line and the parabola:

$2x + k + 2 = 2x^2 + (k+2)x + 8$

$0 = 2x^2 + kx + (6-k)$

For two intersections we need a positive discriminant:

$k^2 - 4(6-k)(2) > 0$

$k^2 + 8k - 48 > 0$

$(k+12)(k-4) > 0$

That's means two negative or two positive, so

$k < 12$ or $k > 4$