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I am trying to figure out the following inequality with absolute value,

$$\begin{split} |2-x| &>1\\ x^2-4x +3 &> 0 \\ x>3&,\ x<1 \end{split}$$

Am I correct with this process?

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Your method is correct. However there is an easier method. Note that the the first line is equivalent to $$ 2-x>1\quad \text{or}\quad2-x<-1 $$ i.e. $$ x<1\quad \text{or} \quad x>3. $$

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    $\begingroup$ It is one's opinion if a method is "easier" or not. $\endgroup$ – Yash Jain Feb 3 '18 at 16:50
  • $\begingroup$ While the listed formulas are equivalent, IMO it's not clear that the OP's method to arrive at them is correct. It's unclear that the OP even meant equivalences; to wit, an editor and several answers are only considering the forward implications. $\endgroup$ – Hurkyl Feb 4 '18 at 1:18
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your inequality is equivalent to $$(\lambda-1)(\lambda-3)>0$$ so $\lambda>3$ or $\lambda<1$

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Squaring both sides of the first inequality to remove the absolute value, $$(2-x)^2>1$$

$$4-4x+x^2>1$$

$$x^2-4x+3>0$$

$$(x-1)(x-3)>0$$

You are correct.

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The answer is correct. Another way to solve it is to set the absolute value negative and then positive and solve the inequality.

$2 - x > 1$ so $x < 1$

$2 - x < -1$ so $x > 3$

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Yes. Another solution:

$|2 - \lambda| > 1 \Longrightarrow -1 > 2 - \lambda or 2 - \lambda > 1 \Longrightarrow -\lambda < -3$ or $-\lambda>-1 \Longrightarrow 3 < \lambda$ or $1 > \lambda$.

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  • $\begingroup$ Establishing the $\implies$ relationship is only half the problem; you need the $\Longleftarrow$ relationship too. $\endgroup$ – Hurkyl Feb 4 '18 at 1:14
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In words, your inequality says that the distance between $2$ and $x$ is greater than $1$. Does that help?

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Since your work lacks any exposition, it is impossible to know if you are "correct with the process". In particular, your work is consistent with having made two very common mistakes:

  • Forgetting that algebraic manipulations can have complicated interactions with the direction of an inequality
  • Only establishing half of what you are trying to prove.

I, as the reader, cannot tell whether or not you got lucky that this happens to be a problem where squaring the original inequality preserves it, or whether you actually knew what you were doing.

Similarly, I cannot tell if you've even put any thought to the question of whether every $x > 3$ and every $x < 1$ satisfies the original equality.

Finally, while each of the three lines you write are equivalent, since you don't show your work I cannot tell if you reasoned correctly to arrive at the equivalences.

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In general : $|x|$ > $a$
$\Rightarrow x>a $ or $x<-a$
In this case we have $|2-x|>1$ which means $2-x>1$ or $2-x<-1$
Hence : $x<1$ or $x>3$

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