Consider the quadratic function of $x$ $$x ^ { 2 } + ( 4 a - 6 ) x + 2 a + b + 5 = 0 . \tag{1}$$ We are to find the conditions under which one solution is $-1$ and the other solution satisfies the inequality $$| x + 2 a | < a + 1 . \tag{2}$$ (1) The condition for equation (1) to have the solution $-1$ is $$b = \mathbf { L } a - \mathbf { M N } .$$ Denote the other solution by $\alpha$. When $\alpha$ is expressed in terms of $a$, we have $$\alpha = \mathbf { O P } a + \mathbf { Q } .$$ (2) When $a > \mathbf{RS}$, the inequality (2) has a solution, and its solution is $$\mathbf { TU } a - \mathbf { VV } < x < - a + \mathbf { W } .$$ Hence the conditions are: that $a$ and $b$ satisfy (3) and that $a$ satisfies $$\mathbf { X } < a < \mathbf { Y } .$$
Consider the quadratic function of $x$
$$x ^ { 2 } + ( 4 a - 6 ) x + 2 a + b + 5 = 0 . \tag{1}$$
We are to find the conditions under which one solution is $-1$ and the other solution satisfies the inequality
$$| x + 2 a | < a + 1 . \tag{2}$$
(1) The condition for equation (1) to have the solution $-1$ is
$$b = \mathbf { L } a - \mathbf { M N } .$$
Denote the other solution by $\alpha$. When $\alpha$ is expressed in terms of $a$, we have
$$\alpha = \mathbf { O P } a + \mathbf { Q } .$$
(2) When $a > \mathbf{RS}$, the inequality (2) has a solution, and its solution is
$$\mathbf { TU } a - \mathbf { VV } < x < - a + \mathbf { W } .$$
Hence the conditions are: that $a$ and $b$ satisfy (3) and that $a$ satisfies
$$\mathbf { X } < a < \mathbf { Y } .$$