Finding the set of all $k\in R$ so that the roots of equation $x^2-(5k+3)x+(k+3)^2$ obey $x_1<4

Question

Stuck on this one. I get inequations with square roots in them, ones which we aren't supposed to know to solve this problem.

Answer 1

Let's denote $f(x)=x^{2}-(5k+3)x+(k+3)^{2}$. Since this is a parabola with its branches directed upwards,you need to call for the $f(4)<0$.

Answer 2

Think to: $$ f(x)= x^2-(5k+3)x+(k+3)^2 $$ as the equation of a parabola. Assuming that the parabola intersects the $x$-axis in two points, i.e.: $$ \Delta = (5k+3)^2-(k+3)^2 = 24\,k(k+1) > 0,\tag{1}$$ since $\lim_{x\to\pm\infty}f(x)=+\infty$, in order to have $x_1<4<x_2$ we just need $f(4)<0$, i.e.: $$ k^2-14k+13 = (k-1)(k-13) < 0,\tag{2} $$ and by putting together $(1)$ and $(2)$ we get: $$ 1 < k < 13.\tag{3}$$