2

$$ x>-2 \sqrt{2x-4}$$

I want to solve this inequality use pure algebra but I struggle when dealing inequality which is of the form 'positive number> negative number'.

To illustrate:

Consider, $$ 2>-1$$

If I square this,

$$ 2^2 > (-1)^2$$

but if I have,

$$ 2> -3$$

Then,

$$ 2^2 < (-3)^2$$

So, I have two distinct cases to deal with.

An attempt to solve:

$$x>-2\sqrt{2x-4}$$ Case 1:

$$|x|> | - 2 \sqrt{2x-4} |$$

then,

$$ x^2> 4(2x-4)$$

Case-2 :

$$|x|<| - 2 \sqrt{2x-4} |$$

$$ x^2 < 4 (2x-4)$$

Both these cases seem horrible to solve..

Another attempt::

The square root can't take negative values, so, $ 2x-4>0$ and hence $ x>2$.. however I want to derive a simplified expression of this by squaring. I can't figure out how to separate out the two cases.

NadAlaba
  • 321
tryst with freedom
  • 11,538

3 Answers3

6

I assume you are asking how to solve inequalities by squaring both sides in general regardless of this particular example.

You can consider this as a general rule:

You can never square both sides of an inequality, unless both sides are positive.
You already noticed why this rule must be followed, when you tried to square both sides of this inequality: $$2 > -3$$

So when you're asking: "When to flip the inequality sign?". The answer is:
You never flip the inequality sign, when squaring both sides in accordance with the aforementioned rule.

Then you may ask: "How do I solve an inequality which contains a square root (needs squaring in order to solve it), and which does not have 2 positive sides?"

There are 3 elementary cases for an inequality with a square root:

  1. Both sides are positive:
    Square both sides of the inequality, and continue accordingly, ex: \begin{align*} 5 &> \sqrt{-x} &&\text{:Defined only when } x\leq0\\ 25 &> -x &&\text{:Square both sides}\\ x &> -25 &&\text{:The Pre-Solution}\\ -25 &< x \leq 0 &&\text{:Final solution is the intersection of the definition and the pre-solution}\\ \end{align*}
  2. Both sides are negative:
    Multiply both sides by $-1$ (and flip the inequality sign), and it's reduced to case 1. Ex: \begin{align*} -5 &< -4\sqrt{1+x^2} &&\text{:Defined when } x \in \mathbb{R}\\ 5 &> 4\sqrt{1+x^2} &&\text{:Multiply both sides by }-1\\ 25 &> 16(1+x^2) &&\text{:Square both sides}\\ 9 &> 16x^2\\ x^2 &< \frac{9}{16}\\ -\frac{3}{4} < x &< \frac{3}{4} &&\text{:Pre-solution}\\ x \in &\left(-\frac{3}{4},\frac{3}{4}\right) &&\text{:Final solution} \end{align*}
  3. One side is positive and one is negative:
    Here the inequality is either impossible or always true, because positive is always greater than negative, ex: \begin{align*} -3&<\sqrt{x}&&\text{:Defined when }x\geq0 \text{ and always true when }x\geq0\\ 5&<-\sqrt{1-x^2}&&\text{:Impossible} \end{align*}

P.S, When you have a side that is not entirely positive or negative, you divide the domain of definition into parts where it is either entirely positive or entirely negative, to reduce the problem to one or more of the cases above. Ex: $$ x > \sqrt{x+6} \quad \text{:Defined when }x \geq -6\\ $$ Here, LHS is negative when $x \in [-6,0)$, and positive when $x \in [0,+\infty)$.
So, when $x \in [-6,0)$ we have case 3, and the inequality is impossible.

When $x \in [0,+\infty)$ both sides are positive and we have case 1, so we can square both sides: \begin{align*} x^2 &> x+6 && \text{:Square both sides}\\ x^2-x-6 &> 0\\ \end{align*} \begin{align*} &x \in (-\infty,-2) \cup (3,+\infty) &&\text{:Pre-solution}\\ &x \in (3,+\infty) &&\text{:Final solution is the intersection of the Pre-solution with }[0,+\infty) \end{align*}

Now, to address your example: $$x>-2\sqrt{2x-4} \qquad \text{:Defined when }x \geq 2$$ Here, when $x \geq 2$ LHS is positive, and RHS is negative. So this is case 3 (always true when defined).

So this inequality is always true when $x \geq 2$. (No need to square both sides) $$x \in [2,+\infty) \quad \text{:Final Solution}$$

NadAlaba
  • 321
3

This is just one of those cases which you should not overthink...

Start with checking the domain in which the inequality makes sense in the first place: $2x-4\ge 0$, i.e. $x\ge 2$. However, for each $x\ge 2\gt 0$ we have that the left side is positive and the right side is negative or zero, so the inequality is satisfied. Conclusion: the set of solutions is $[2,+\infty)$.

This won't always be the case, and in more complex cases you would need to distinguish the cases, however, this example almost looks crafted to demonstrate the basic point that the domain where the inequality is defined matters too.

In general, if we are planning to square both sides of the inequality (say, $A<B$), it is enough to distinguish the cases "$A$ and $B$ both negative and $A^2>B^2$", "$A$ and $B$ both positive and $A^2<B^2$" and "$A$ negative, $B$ positive". (To be pedantic: the case where one of $A$ or $B$ is zero can usually be treated together with either "positive" or "negative" case - rarely those cases need to be spelled out separately.)

As an example, let us look into a slightly modified inequality, e.g. $x<-2\sqrt{4-2x}$, which as a domain has $4-2x\ge 0$, i.e. $x\le 2$. As we know that the right side is negative or zero, the only two cases remaining here will be:

  • $x\le 0$ and $x^2\gt (-2\sqrt{4-2x})^2$
  • $x>0$.

The second case is impossible as the left side will be positive and the right side will be negative.

Thus, the only case we need to consider is the first case, which is equivalent to $x^2+8x-16\gt0$. This is a quadratic inequality, and there is a standard procedure to solve it: I will assume that you are familiar with this procedure. Shortly, you first solve the quadratic equation $x^2+8x-16=0$ (which has solutions $x_{1,2}=-4\pm4\sqrt{2}$), and observing the sign of the factor multiplying $x^2$ in the inequality, you reach the solutions: $(-\infty, -4-4\sqrt{2})\cup(-4+4\sqrt{2}, +\infty)$. However, we restricted ourselves to $x\le 0$ only, so the actual set of solutions is only $(-\infty, -4-4\sqrt{2})$.

  • Could you please illustrate a complex case? I have tried really hard to split up the cases but I just can't figure out how to do it – tryst with freedom Sep 07 '20 at 19:48
  • @Buraian Please bear with me, it's a bit late for me now, but I will try to edit my answer tomorrow with a slightly different problem. In fact, I think I could just change it to $x\gt-2\sqrt{4-2x}$, the domain will be $x\le 2$, and the cases will be $0\le x\le2$ and $x\lt0$. –  Sep 07 '20 at 21:05
  • looking forward to it – tryst with freedom Sep 08 '20 at 09:33
  • Still you haven't done it using pure algebra, you've given gone back to arguing it. So does this mean you can't solve it without usage of like logical arguements ( not direct algebra methods that is) – tryst with freedom Sep 10 '20 at 17:27
  • @Buraian I may not have understood your concerns properly. Of course, the very moment you start distinguishing cases, you stop using pure algebra. You basically switch to some sort of logic (such as $p\iff p\land(q\lor\lnot q)\iff (p\land q)\lor(p\land\lnot q)$) - in your case $p$ is your equation and $q$ is "case $x>0$" and $\lnot q$ is "the opposite case". I could've written it using first order logic (instead of English) - but this won't make it easier to read, thus I preferred English. In both cases it won't be a pure algebraic proof (e.g. not one like $2x=4 \iff x=4/2\iff x=2$). –  Sep 10 '20 at 18:09
  • so, does this mean this is impossible to do using algebra? – tryst with freedom Sep 10 '20 at 18:11
  • @Buraian Can this problem be solved using pure algebra? Probably not, but this question also misses the point. Part of studying equalities and inequalities is studying how to use logic (distinguishing cases, following implications/equivalences). There may be a trick that would work in some special case (and would fail in a case that looks almost identical to the first one). Again, the point of practising inequalities is not to find those tricks but to be familiar with the (more powerful) technique of using logic. –  Sep 10 '20 at 18:15
  • (Cont'd) After all, how do you solve a simpler case $x^2-1\ge 0$ using "pure algebra"? (Answer: by distinguishing cases $x\le -1, -1\lt x \lt 1, x\ge 1$, i.e. not even that is "pure algebra") There is a trick there: $x^2-1\ge0\iff x^2\ge 1\iff \sqrt{x^2}\ge 1\iff |x|\ge 1$ - but note that distinguishing cases is built in here at the lower level (why is $\sqrt{x^2}=|x|$? Checking some cases is hidden in this equality.) Thus I would not run away from distinguishing cases and discussion - I would try to embrace this method, and I suggest you do as well. –  Sep 10 '20 at 18:29
  • I think if you could write an abstract of this discussion in comments and put it in your answer, then it would stand as treatsie for inequalities :D – tryst with freedom Sep 10 '20 at 18:49
  • Maybe, but I keep writing similar stuff on MSE (and others too!), so I am not sure how much of that would be a repeat of what's been said elsewhere. Have a look at https://math.stackexchange.com/a/3575316/700480, https://math.stackexchange.com/a/3780331/700480. In the first of those links you can also see an "inventive" (more algebraic) solution that was the accepted answer - I merely wanted to add there (as I am trying to say now) - sometimes you will find it, but if you cannot, the pedestrian method still works! –  Sep 10 '20 at 21:52
2

The square root on the right-hand side of the inequality allows you to restrict the domain to a subset of the real numbers. $\sqrt{2x-4}$ is only defined when $2x-4 \ge 0$. Therefore, the domain is from $[2,+\infty)$.

There is no need to square the inequality and consider separate cases. As suggested in the comments, you can rewrite the inequality as

$$-\frac{x}{2}<\sqrt{2x-4}$$

in which if $x\ge 2$ then the left-hand side of the inequality is strictly negative while the right-hand side is greater than or equal to zero. Therefore, the inequality is satisfied on $[2,+\infty)$.

Axion004
  • 10,056
  • Ok, my question is when you squared the expression [ $ x> - 2 \sqrt{2x-4} $ ] how did you know not to flip the inequality sign? – tryst with freedom Sep 07 '20 at 19:50
  • Look at the example I gave in the post originally, sometimes it turns out the equality flip when you square an inequality of form 'positive> negative' – tryst with freedom Sep 07 '20 at 19:54
  • $$a^2 > b^2 \iff a^2-b^2 > 0 \iff (a-b)(a+b) > 0$$ In your case, $a=x>0$ and $b=-2\sqrt{2x-4}<0$. So $(a-b)>0$ by definition but we require that $(a+b)>0$. This means that $a > |b|$. Although, I don't see why you would take this approach when you can analyze the inequality directly. – Axion004 Sep 07 '20 at 20:26