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

1 Answer
KillerBunny
Jul 4, 2018

#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.

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