Problem 10

Explanation

Given the inequality:

Rearrange it into the standard quadratic form by moving $-4$ to the left side:

For the quadratic function to be positive for every real $x$ (positive definite), its graph must lie entirely above the $x$-axis. The conditions are:

1. The coefficient of $x^2$ must be positive ($a > 0$).
2. The discriminant must be negative ($D < 0$).

Condition 1: $a > 0$

Condition 2: $D < 0$

Recall that $D = b^2 - 4ac$. With $a = 2m + 1$, $b = -4(m + 1)$, and $c = m + 4$:

Divide both sides by $4$ to simplify:

The roots are $m = 0$ and $m = \frac{1}{2}$. Since the inequality is "<", the solution lies between the roots:

Conclusion

We find the intersection of conditions $(1)$ and $(2)$:

• Condition $(1)$: $m > -\frac{1}{2}$
• Condition $(2)$: $0 < m < \frac{1}{2}$

The intersection of both regions is: