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

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?

• Use the fact that we get two different real roots when $b^2-4ac \gt 0.$ – Mohammad Zuhair Khan Jan 26 at 16:31
• What do you need the first inequality for? – Dr. Mathva Jan 26 at 16:43

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 $$a^2-4(a-1)>0.$$
• Why you want that $f(-\frac{1}{3}) > 0$ ? – StackUser Jan 26 at 21:31
• @StackUser Because we need that roots of the quadratic equation would be less that $-\frac{1}{3}.$ Draw the graph of $f$. – Michael Rozenberg Jan 26 at 21:38
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 $$\frac43. If $$a<0$$, then (1) does not hold.
So if $$\frac43, the equation has two different roots satisfying the inequality.