For what value of a parameter $m\in \mathbb{R}$ is the inequality $$ (m+3)x^2+(m−2)x+m−2>0 $$ satisfied by every real number?
Filip presented his solution to this problem in the class:
(1) Let us first assume that the inequality is linear, i.e., the coefficient of the quadratic term is zero: $$ \begin{gather} m+3=0 \cr m=−3 \end{gather} $$ For $m=−3$, we get the inequality $−5x−5>0$, that is not satisfied by all real $x$.
(2) Now, let's consider that the given inequality is quadratic. The expression on the left side will be positive for all real $x$ if the parabola, that is the graph of the corresponding quadratic function, lies entirely above the $x$-axis. This means that the parabola has no intersections with the $x$-axis, so the discriminant of the quadratic polynomial must be negative: $$ D=(m−2)^2−4(m+3)(m−2)<0 $$ Solving the inequality above, we get: $$ \begin{gather} −3m^2−8m+28<0 \cr m_{1,2}=\frac{8\pm\sqrt{64+336}}{−6} \cr m_1=−\frac{14}{3}\mathrm{~and~} m_2=2 \end{gather} $$ By factoring the quadratic expression for the discriminant, we get: $$ −3\left(m+\frac{14}{3}\right)(m−2)<0 $$ This inequality holds for $m<−\frac{14}{3}$ or $m>2$.
(3) From this, it follows that the initially assigned inequality is satisfied by every real number $x$ if and only if $$m\in\left(−\infty;−\frac{14}{3}\right)\cup (2;+\infty)$$
The teacher asked the students whether they believed Filip's solution was correct. Which of the following comments is right?
Laura: Filip's solution is not quite right. In step (2), he forgot one condition, namely that $$m+3>0$$
Sofia: Filip's solution is quite right. The result is correct, after all.
Petr: Filip made an error in step (2). He should have got the result $m>−\frac{14}{3} \land m<2$. The right solution is $$ m\in \left(−\frac{14}{3};2\right) $$
Libor: Filip made an error in step (3). The right solution is $$ m\in \{-3\} \cap \left[\left(−\infty;−\frac{14}{3}\right)\cup (2;+\infty)\right] $$
If the given inequality is to be quadratic, then it will hold for every real $x$ if two conditions are met: $$ \Delta=(m−2)^2−4(m+3)(m−2)<0 \mathrm{~and~} m+3>0 $$ The first condition $\Delta=(m−2)^2−6(m^2−m−2)<0$ gives the result $m<−\frac{14}{3} \lor m>2$ and guarantees that the parabola has no intersections with the $x$-axis. The second condition $m+3>0$ guarantees that the parabola opens upwards, which, together with the first condition, means that it lies entirely above the $x$-axis. When we solve the inequality $m+3>0$, we get $m>−3$. From this and from the condition $m\in \left(−\infty;−\frac{14}3\right)\cup (2;+\infty)$ it is clear that the inequality holds for all real $x$ if and only if $$m\in(2;+\infty)$$