7
$\begingroup$

Simplify: $x^2 > 1$.

My solution: Taking square root on both sides:

$±x > ±1$

So my results are:

  1. $x > 1$
  2. $x > -1$
  3. $-x > 1$ $\implies$ $(-1 > x)$
  4. $-x > -1$ $\implies$ $(1 > x) $

But I strongly feel this is wrong. What is wrong here?

A step-by-step explanation will help me.

$\endgroup$
  • $\begingroup$ Why do you feel this is wrong ? $\endgroup$ – servabat Feb 19 '15 at 8:45
  • 8
    $\begingroup$ $a^2 > b^2$ does not imply $\pm a > \pm b$, as "taking both positive and negative square roots" on both sides is not a well defined, monotone increasing function. $\endgroup$ – Macavity Feb 19 '15 at 9:16
  • 2
    $\begingroup$ @servabat: probably because it is wrong ;-) For example, $x = 0$ satisfies (2) but doesn't satisfy the original inequality. So the questioner's working hasn't produced results equivalent to the original, as hoped. $\endgroup$ – Steve Jessop Feb 19 '15 at 14:39
  • $\begingroup$ @SteveJessop : $x = 0$ doesn't satisfy (1), (2), (3) and (4) (as it doesn't satisfy (1) and (3)). Clearly, if $x \not \in (-1, 1)$, it is true, so for me that's actually true. $\endgroup$ – servabat Feb 19 '15 at 14:47
  • 1
    $\begingroup$ @servabat: nothing satisfies all 4, since (2) and (3) are mutually exclusive. So taking the conjunction is still wrong: the original has solutions and the "result" doesn't. And anyway (1) is redundant given (2), so there's something wrong there too in the way the result is given. "Simplify" could be ambiguous, but anything with redundant conditions can be further simplified. If you're saying it's not wrong I don't really see your point, unless you refer to some meaning of the word "wrong" that, I'd claim, fails to characterise what the questioner is trying to do :-) $\endgroup$ – Steve Jessop Feb 19 '15 at 15:00
5
$\begingroup$

You can just go from $x^2 > 1$ directly to $|x| > |1|$.

Now obviously $|1| = 1$, so $|x| > 1$, therefore either $x > 1$ or $x < -1$.

$\endgroup$
15
$\begingroup$

As $\sqrt\cdot$ is increasing and $\sqrt{(\cdot)^2}=|\cdot|$, $$a^2>b^2\iff|a|>|b|.$$

$\endgroup$
14
$\begingroup$

We need $x^2-1=(x-1)(x+1)>0$

As the product is positive, multiplicands should either be both positive or both negative.

First we consider the scenario where they are both positive:

If $x-1>0\iff x>1\ \ \ \ (1)$

and $x+1>0\iff x>-1\ \ \ \ (2)$

$(1),(2)\implies x>$max$(1,-1)$

Test the negative case similarly

$\endgroup$
  • $\begingroup$ So we have to take Max for +x and Min -x ? $\endgroup$ – Robby Feb 19 '15 at 8:48
  • $\begingroup$ @Robby, Intuitively so. $\endgroup$ – lab bhattacharjee Feb 19 '15 at 8:49
  • $\begingroup$ @Robby: Yes, but only in this case. If it were $<$ you couldn't necessarily take the maximum of the two. In the general case the solution is the intersection of all the particular solutions. $\endgroup$ – rubik Feb 19 '15 at 9:29
11
$\begingroup$

In solving quadratic inequalities, it must be the case that the right hand side of the equation is zero. So we have:

$$x^2-1>0.$$

Then we find the critical numbers, these are values of $x$ that will make our inequality above zero.

We have $x=1$ and $x=-1$ as critical numbers.

My critical numbers then partitioned my real number line into 3 parts/subintervals. Namely:

$$(-\infty,-1),\quad(-1,1)\quad\mbox{and}\quad(1,\infty).$$

In each part it is advisable to get a test value, a number that lies on the subintervals and substitute it on the LHS above and we must note the sign, this are the sign of the subintervals relative to our inequality.

If $x=-2$ we have positive sign so for all $x$ in the first subinterval we have $x^2-1$ is positive.

If $x=0$ we have a negative sign so for all $x$ in the second subinterval $x^2-1$ is negative.

Lastly if $x=2$ we have a positive sign, so for all $x$ in the third subinterval we have $x^2-1$ is positive.

Originally we have $x^2-1>0$ this means that $x^2-1$ is positive so our answer is $$(-\infty,-1)\cup (1,+\infty).$$

This is the general way to solve quadratic inequality.

$\endgroup$
  • $\begingroup$ great. thanks for the detailed explanation. $\endgroup$ – Robby Feb 19 '15 at 9:14
  • $\begingroup$ Your welcome @Robby $\endgroup$ – Jr Antalan Feb 19 '15 at 9:15
  • $\begingroup$ I would suggest an edit--I thought you meant "positive" when you said "values of x that will make our inequality above zero". Probably best to just say to find the values of x that will make the $x^2 - 1$ part equal to zero. $\endgroup$ – msouth Feb 20 '15 at 6:28
  • $\begingroup$ That is just right @msouth the purpose of that is to find the critical points. $\endgroup$ – Jr Antalan Feb 20 '15 at 6:30
  • $\begingroup$ I think this answer is the best one, really. It is also a good thing to visualize the graph of this as a parabola, because then you can see that it crosses the x axis at the roots, and therefore you just need to do a quick test in each region to see which way it's oriented. Understanding this takes some of the mystery out of why it's done that way. $\endgroup$ – msouth Feb 20 '15 at 6:33
2
$\begingroup$

When you apply square root operation to both sides of inequation like $f^2(x) < c$ or $f^2(x) > c$, then $c < 0$ gives you no solutions for $f^2(x) < c$ and any $x \in \mathscr D(f)$ for $f^2(x) > c$.
If $c \geq 0$, then $f^2(x) < c \rightarrow |f(x)| < \sqrt{c}$ and $f^2(x) > c \rightarrow |f(x)| > \sqrt{c}$ (same for $\leq$ and $\geq$).
In your case $c = 1$ and $f(x) = x$, so $f^2(x) > c \rightarrow |x| > \sqrt{c}, c = 1$, so $|x| > 1$, or $x \in (-\infty,1)\cup(1,+\infty)$.

$\endgroup$

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.