Question

Find the range of values of k for which `" "x^2 + (k+3)x +4k >3` for all real values of `x`?

Answer 1

`3 < k < 7`

Explanation:

Rewrite as

`x^2 + (k+3)x + (4k-3) > 0`

Look for the roots of the polynomial:

`x_{1,2} = \frac{-(k+3) \pm \sqrt((k+3)^2 - 4(4k-3))}{2}`

which simplifies into

`x_{1,2} = \frac{-(k+3) \pm \sqrt(k^2-10k+21)}{2}`

so, if `\Delta=k^2-10k+21<0`, the quadratic equation has no solutions, which means that it is always greater than zero.

So, this time, we look for the roots of the quadratic equation for `\Delta`:

`k^2-10k+21 = (x-3)(x-7)`

So, this is a parabola, concave up, with roots `3` and `7`. This means that for every `3 < k < 7`, the expression `k^2-10k+21` is negative, which in turn means that `x^2 + (k+3)x + (4k-3)` has no solutions.