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

Jul 4, 2018

$3 < k < 7$

#### Explanation:

Rewrite as

${x}^{2} + \left(k + 3\right) x + \left(4 k - 3\right) > 0$

Look for the roots of the polynomial:

${x}_{1 , 2} = \setminus \frac{- \left(k + 3\right) \setminus \pm \setminus \sqrt{{\left(k + 3\right)}^{2} - 4 \left(4 k - 3\right)}}{2}$

which simplifies into

${x}_{1 , 2} = \setminus \frac{- \left(k + 3\right) \setminus \pm \setminus \sqrt{{k}^{2} - 10 k + 21}}{2}$

so, if $\setminus \Delta = {k}^{2} - 10 k + 21 < 0$, 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 $\setminus \Delta$:

${k}^{2} - 10 k + 21 = \left(x - 3\right) \left(x - 7\right)$

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