For which values of $a$ we will get two different roots?

Question

In given the following system of equations:
$$ |x-1| > 2x+2 $$ $$ x^2 + ax + a -1 = 0 $$
For which values of $a$ we will get two different roots?

Answer 1

The first gives $$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-\frac{1}{3}.$$ Now, let $f(x)=x^2+ax+a-1$ and solve the following system: $$f\left(-\frac{1}{3}\right)>0$$ $$-\frac{a}{2}<-\frac{1}{3}$$ and $$a^2-4(a-1)>0.$$

Answer 2

The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1\neq x_2$ if $a\neq 2$.

Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes $$ |a|>4-2a. \tag{1}$$ If $a\ge2$, (1) holds. If $0\le a<2$, then the solution of (1) is $\frac43<a<2$. If $a<0$, then (1) does not hold.

So if $\frac43<a<2$, the equation has two different roots satisfying the inequality.