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)<0`

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`