I can't solve $x-2<\frac{1}{x-1}$

Question

We have to solve this inequation.

$x-2<\frac{1}{x-1}$.

So we try $(x-1)(x-2)<1$ what's wrong ?

Please, give me a hint.

Answer 1

Yes, it's wrong because $x-1$ can be negative.

We need $$x-2-\frac{1}{x-1}<0$$ or $$\frac{x^2-3x+1}{x-1}<0$$ or $$\frac{\left(x-\frac{3+\sqrt5}{2}\right)\left(x-\frac{3-\sqrt5}{2}\right)}{x-1}<0,$$ which gives the answer.

I got $$\left(-\infty,\frac{3-\sqrt5}{2}\right)\cup\left(1,\frac{3+\sqrt5}{2}\right).$$

Answer 2

hint: solve $$x^2-3x+2<1 , \hspace{10pt} x>1 $$ $$x^2-3x+1<0 , \hspace{10pt} x>1 $$

and

$$x^2-3x+2>1 , \hspace{10pt} x>1 $$ $$x^2-3x+1>0 , \hspace{10pt} x<1 $$

Answer 3

If you have $x<y$ and multiply it by positive number $a$, then it is still true that $ax < ay$. If you multiply it by negative number $b$, then the inequality changes direction, so $bx > by$ is true instead.

For example, we know that $2<3$ and if we multiply it by $2$, we get $4<6$, which is still true, but if we multiply it by $-2$, it is not true that $-4<-6$, but on the contrary, $-4>-6$.

What does it have to do with solving inequalities?

You have $$x-2 < \frac 1{x-1}$$ and want to multiplity it by $x-1$. So, what is true: $(x-2)(x-1)<1$ or $(x-2)(x-1)>1$? According to the above discussion, if $x-1>0$, the former is true, and if $x-1<0$ the latter is true. But, the problem is that we don't know whether $x-1$ is positive or negative, so we would entangle ourselves into some casework.

However, there is a common trick to avoid this, multiply by $(x-1)^2$ instead of $x-1$! We do know that $(x-1)^2>0$, so the inequality becomes $$(x-2)(x-1)^2<x-1.$$ Can you finish from here?

Answer 4

$$x-2<\frac { 1 }{ x-1 } \\ x-2-\frac { 1 }{ x-1 } <0\\ \frac { { x }^{ 2 }-3x+1 }{ x-1 } <0\\ \frac { \left( x-\frac { 3-\sqrt { 5 } }{ 2 } \right) \left( x-\frac { 3+\sqrt { 5 } }{ 2 } \right) \left( x-1 \right) }{ { \left( x-1 \right) }^{ 2 } } <0\\ \left( x-\frac { 3-\sqrt { 5 } }{ 2 } \right) \left( x-\frac { 3+\sqrt { 5 } }{ 2 } \right) \left( x-1 \right) <0\\ x\in \left( -\infty ,\frac { 3-\sqrt { 5 } }{ 2 } \right) \cup \left( 1,\frac { 3+\sqrt { 5 } }{ 2 } \right) \\ $$

Answer 5

$x-1<1/(x-1) +1;$

Set $a:= x-1$, $a \not =0$.

$a < 1/a +1;$

1) Let $a >0$; then

$a^2-a -1 <0.$

$(a-1/2)^2 \lt 5/4;$

$|a-1/2| \lt (1/2)√5;$

We get for $a >0$: $a < 1/2 +(1/2)√5$.;

2) $a < 0$:

$a^2-a-1 >0;$

$(a-1/2)^2 >5/4.$

$|a-1/2| >(1/2)√5$;

We get for a <0: $1/2-a >(1/2)√5$, or

$a < 1/2-(1/2)√5$.

Combining:

$a \in (-\infty, \dfrac{1-√5}{2})\cup (0,\dfrac{1+√5}{2}).$

Now revert to $x$.