1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Use the fact that we get two different real roots when $b^2-4ac \gt 0.$ $\endgroup$ – Mohammad Zuhair Khan Jan 26 at 16:31
  • $\begingroup$ What do you need the first inequality for? $\endgroup$ – Dr. Mathva Jan 26 at 16:43
1
$\begingroup$

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 $$a^2-4(a-1)>0.$$

$\endgroup$
  • $\begingroup$ Why you want that $f(-\frac{1}{3}) > 0 $ ? $\endgroup$ – StackUser Jan 26 at 21:31
  • $\begingroup$ @StackUser Because we need that roots of the quadratic equation would be less that $-\frac{1}{3}.$ Draw the graph of $f$. $\endgroup$ – Michael Rozenberg Jan 26 at 21:38
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.