In order for both roots of the equation $x^2 + ( m + 1 ) x + 2m - 1 = 0$ to not be real, it must be ... In order for both roots of the equation $x^2 + ( m + 1 ) x + 2m - 1 = 0 $ to not be real, it must be the case that ...
A. $m > 1$
B. $1 < m < 5$
C. $1 \leq m \leq 5$
D. $m < 1$ or $m > 5$
E. $m \leq 1$ or $m \geq 5$
 A: The roots of a quadratic function (with real coefficients) are non-real when the discriminant $D=b^2-4ac$ is negative.  In this case,
\begin{equation}
D= (m+1)^2-4(2m-1) = m^2-6m + 5 = (m-1)(m-5),
\end{equation}
so you must determine when this discriminant is negative.
A: Notice, for roots of the given quadratic equation: $x^2+(m+1)x+2m-1=0$ to be non real (i.e. imaginary), we have determinant $$\Delta =B^2-4AC<0$$
$$\implies (m+1)^2-4(1)(2m-1)<0$$
$$m^2+2m+1-8m+4<0$$
$$m^2-6m+5<0$$
$$(m-5)(m-1)<0$$
$$\color{red}{1<m<5}$$
A: In case you forget the discriminant like me: The equation
$$
x^2 + (m+1) x + 2m - 1 = 0
$$
leads to the condition
$$
\left(x + \frac{m+1}{2}\right)^2 =  \left(\frac{m+1}{2}\right)^2 - 2m + 1 < 0
$$
because then the square root of a negative number prevents real solutions $x$ to exist.
That right hand side is a quadratic function
$$
f(m) = \frac{1}{4} \left(m^2-6m+5\right)
$$
with positive factor for $m^2$, so its graph is an upwards open parabola.
The roots are
$$
m = 3 \pm \sqrt{4} = 3\pm 2\in \{ 1, 5\}
$$
so $f(m)$ is negative on the intervall $(1, 5)$.
