## Task B-2.2.

Determine $a \in \mathbb{R}$ so that the solutions of the equation $a\left(x^{2}-1\right)+4 x=5$ are distinct real numbers greater than 1.
## First Solution.

The given equation is equivalent to the equation $a x^{2}+4 x-a-5=0$.

For the given equation to have two distinct real solutions, the following must hold:

$D=16-4 a(-a-5)=16+20 a+4 a^{2}>0$ and $a \neq 0$, that is, $a \in\langle-\infty,-4\rangle \cup\langle-1,0\rangle \cup\langle 0,+\infty\rangle$.

According to the condition of the problem, for the solutions $x_{1}$ and $x_{2}$, it must hold that $x_{1}>1$ and $x_{2}>1$, or equivalently $t_{1}=x_{1}-1>0$ and $t_{2}=x_{2}-1>0$.

Two real numbers are positive if both their product and sum are positive numbers.

1. condition: $t_{1} \cdot t_{2}>0$

$\left(x_{1}-1\right) \cdot\left(x_{2}-1\right)>0$

$x_{1} \cdot x_{2}-x_{1}-x_{2}+1>0$

$\frac{c}{a}-\left(-\frac{b}{a}\right)+1>0$

$\frac{-a-5}{a}+\frac{4}{a}+1=\frac{-a-5+4+a}{a}=\frac{-1}{a}>0$

Solving the inequality yields the solution of the first condition:

$a \in\langle-\infty, 0\rangle$.

2. condition: $t_{1}+t_{2}>0$

$x_{1}-1+x_{2}-1>0$

$x_{1}+x_{2}-2=-\frac{b}{a}-2=-\frac{4}{a}-2=\frac{-4-2 a}{a}>0$

Solving the inequality yields the solution of the second condition:

$a \in\langle-2,0\rangle$.

The final solution is the intersection of all conditions: $a \in\langle-1,0\rangle$.