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We need to find the range of $x$ for the following problem:

$$\frac{(x+6)(x-4)^2}{(x-6)(x+4)^2} + \frac{(x-6)(x+9)^2}{(x+6)(x-9)^2} < \frac{2x^2 + 72}{x^2 - 36}$$

They invlove the use of squares, which we have to avoid while solving inequalities.....And sadly, I cannot think of ways for trying this question. Can someone please give me hints/suggestions for solving? Formatting edits/suggestions are gladly welcomed! Question edits/suggestions are welcomed too!



EDITs: So as discussed in the comments, I had got this for problem 1: $$\frac{(x+6)(x-4)^2}{(x-6)(x+4)^2} + \frac{(x-6)(x+9)^2}{(x+6)(x-9)^2} < \frac{2x^2+72}{x^2-36}$$ which exapands to $$\frac{(x+6)(x-4)^2(x+6)(x-9)^2}{(x-6)(x-4)^2(x+6)(x-9)^2} + \frac{(x-6)(x+9)^2(x-6)(x+4)^2}{(x+6)(x-9)^2(x-6)(x+4)^2} = \frac{2x^2+72}{x^2-36}$$ which simplifies to $$\frac{(x+6)(x-4)^2(x+6)(x-9)^2 + (x-6)(x+9)^2(x-6)(x+4)^2}{(x-6)(x+4)^2(x+6)(x-9)^2} = \frac{2x^2+72}{x^2-36}$$ and right now, I am stuck. I don't dare to open up that numerator (I have no guts in front of those brackets). There has to be a much elegant and simple looking solution than this brute-force.....But the problem is.....I can't find it! Anyone who would like to show some sympathy on me after this :(

Cuckoo Beats
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  • "They invlove the use of square roots and squares, which we have to avoid while solving inequalities" : What does this mean? – user2661923 Apr 23 '23 at 08:58
  • Please see this article on MathSE protocol. As onerous as the article may appear to you, it provides a defense mechanism against the MathSE forum being used as a do my homework forum. In particular, please see the Edit-Tools section of the article, and the portion of the article that discusses showing work. It is irrelevant whether the problem is homework. What counts is whether the protocol is observed. – user2661923 Apr 23 '23 at 08:59
  • In problem (1), what happens when you multiply everything out? Please show your work. – user2661923 Apr 23 '23 at 09:00
  • Going forward with your future MathSE questions, in a similar situation, please post the problems as separate questions. In general, there should be one Math problem presented for each posted question. – user2661923 Apr 23 '23 at 09:01
  • @user I meant that in general we tend to eliminate square roots and squares first in an inequality, to make it easier to comprehend and all...... – Cuckoo Beats Apr 23 '23 at 15:03
  • @user thank you for your suggestions, i clearly understand i should post out anything that i have tried so that time is not wasted. basically i just used brute force and tried to simplify all those terms but in the end it was too overwhelming so i just stopped going further. there has to be some better approach than just brute force...... – Cuckoo Beats Apr 23 '23 at 15:06
  • Are you saying that in problem (1), you have already attempted to manipulate the three fractions so that they each have the common denominator of $$(x-6) \times (x+6) \times (x+4)^2 \times (x-9)^2 ~?$$ If so, what was the result? – user2661923 Apr 23 '23 at 16:44
  • @user , yes, i did attempt (i wont deny that) but there are a few reasons why i did not mention it at the first place - 1) I am not happy with what I got, a long chain of terms it feels as if i have made a new polymer! (2) the place where i did all this "rough" requires extensive amounts of glue as every single page comes off that notebook (3) this is again, not a very efficient thing to do and chances of getting it incorrect is hugeeee, also i donot feel like it is even possible to solve since i have already encountered fourth degree terms on the way – Cuckoo Beats Apr 23 '23 at 17:04
  • I don't understand what you mean by "...requires extensive amounts of glue". You simply proofread/organize your work, and then type it directly into the posting via MathJax. Also, often, even when there is a more elegant approach, the brute force method will reveal the corresponding simplifications. That is, you may wish to double check whether arithmetic errors have prevented you from finding a simplification. Personally, while there may be a more elegant approach, I am unsure why you would feel that a more elegant approach must exist. – user2661923 Apr 23 '23 at 17:28
  • @user I am sorry for any incovience caused....i have re-edited the question now.....expanding the brackets does help simplifying a lot of things and sometimes it also reveals what the question wants! sadly, i have almost lost hope in this :( – Cuckoo Beats Apr 25 '23 at 09:16
  • It's not clear to me what you mean by 'the range of $x$ for the following problem'. Are you interested in the domain, i.e. a set of $x$-es for which the inequality makes sense (aka it is defined)? Or rather in the set of $x$-es which satisfy the inequality (values, for which it is true)...? – CiaPan Mar 01 '24 at 07:35

2 Answers2

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Some thoughts:

Let $u = \frac{x + 6}{x - 6}$. We have $x = \frac{6(u + 1)}{u - 1}$. We have $$\frac{x-4}{x+4} = \frac{u + 5}{5u + 1}, \quad \frac{x + 9}{x - 9} = \frac{5u - 1}{5 - u}, \quad \frac{2x^2 + 72}{x^2 - 36} = \frac{(x+6)^2 + (x-6)^2}{(x - 6)(x + 6)} = u + \frac{1}{u}.$$

The inequality is written as $$\frac{u(u+5)^2}{(5u+1)^2} + \frac{(5u-1)^2}{u(5-u)^2} < u + \frac{1}{u}$$ or $$\frac{1}{u}\left(\frac{(5u-1)^2}{(5-u)^2} - 1\right) < u \left(1 - \frac{(u+5)^2}{(5u+1)^2}\right)$$ or $$\frac{1}{u} \cdot \frac{24(u^2 - 1)}{(5-u)^2} < u \cdot \frac{24(u^2 - 1)}{(5u+1)^2}$$ or $$24(u^2 - 1)\left(\frac{u}{(5u+1)^2} - \frac{1}{u(5-u)^2}\right) > 0$$ or $$\frac{24 (u + 1)u(u-1)(u^2 + 1)(u^2 - 10u - 1)}{(5u+1)^2(5-u)^2u^2} > 0$$ or $$\frac{24(u^2 + 1)}{(5u+1)^2(5-u)^2u^2} (u + 1)(u - 5 + \sqrt{26})u(u-1)(u - 5 - \sqrt{26}) > 0.$$

River Li
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  • we are supposed to deduce that $u^2 + 1$ is always positive and remove all the squares in the denominator with the conditions $u \neq \frac{-1}{5} , 5 , 0$ ?? – Cuckoo Beats Apr 25 '23 at 15:55
  • @CuckooBeats Alternatively, you can insert $u = (x+6)/(x-6)$ back to the last inequality. You will get something like $(x-a)(x-b)(x-c)(x-d)(x-e) > 0$. For example, $(x+3)(x+1)(x-1) > 0$, the solution is $(-3, -1)\cup (1, \infty)$. But don't forget the zeros of the denominator, as you pointed out. – River Li Apr 26 '23 at 00:00
  • well this indeed was a good solution, but can we substitute better variables to the inequality? i just want to know.....although this might be the best known solution for this question, can we still try it out? – Cuckoo Beats Apr 26 '23 at 00:43
  • @CuckooBeats Do you mean that first the last inequality has solution $u \in (-1, -1/5)\cup (-1/5, 5 - \sqrt{26})\cup (0, 1) \cup (5+\sqrt{26}, \infty)$, then convert this solution into $x$? – River Li Apr 26 '23 at 00:48
  • no i meant the whole question......like you did $u = \frac{x+6}{x-6}$, can there be a better substitution as the solution seems lengthier, isn't it? also it would be better to first replace the value of $u$ in your last inequality and then simplify and solve for $x$..... – Cuckoo Beats Apr 26 '23 at 09:31
  • @CuckooBeats OK. Yes, it is simpler to replace $u$ with $x$. – River Li Apr 26 '23 at 12:00
  • hello, i have added my solution to the problem, the trick i found useful was $2a^2 +2b^2 = (a+b)^2+(a-b)^2$ – Cuckoo Beats Apr 30 '23 at 07:45
  • @CuckooBeats It is the same as mine? You may give a complete detailed solution with final answer as an accepted answer. – River Li Apr 30 '23 at 09:47
  • oh definetly it is going to come the same, but i feel mine a bit more comfortable than yours, no offence, but it seems we need to work out more in your approach, although i had never thought the same earlier, thanks! – Cuckoo Beats Apr 30 '23 at 13:51
  • @CuckooBeats Your $\frac{(x+6)}{(x-6)}\cdot\left[\frac{(x-4)^2}{(x+4)^2}-1\right]+\frac{(x-6)}{(x+6)}\cdot\left[\frac{(x+9)^2}{(x-9)^2}-1\right] < 0$ is the same as my $\frac{1}{u}\left(\frac{(5u-1)^2}{(5-u)^2} - 1\right) < u \left(1 - \frac{(u+5)^2}{(5u+1)^2}\right)$. So basically the same thing. – River Li Apr 30 '23 at 14:03
  • oh yess, i did not notice that step, basically we could replace $u$ back and get the same equation.......actually i would want to do the same thing as yours when nothing strikes at my head..... – Cuckoo Beats Apr 30 '23 at 14:27
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Thank you to everyone who took much time and helped me to solve the questions. Here's the solution to the first problem after a lot of research and hints from teachers. Note that this solution is analogus to @RiverLi 's solution....
We can simplify $2x^2 + 72$ as $$2\cdot x^2 + 2\cdot6^2 + 2\cdot x\cdot6 - 2\cdot x\cdot6$$ or $$(x+6)^2 + (x-6)^2$$ now, we can easily simplify the RHS $$\frac{2x^2+72}{x^2-36} = \frac{(x+6)^2+(x-6)^2}{(x+6)(x-6)} = \frac{(x+6)}{(x-6)}+\frac{(x-6)}{(x+6)}$$ sweet! plugging this outcome in the original question: $$\frac{(x+6)(x-4)^2}{(x-6)(x+4)^2} + \frac{(x-6)(x+9)^2}{(x+6)(x-9)^2} < \frac{(x+6)}{(x-6)}+\frac{(x-6)}{(x+6)}$$ or $$\frac{(x+6)}{(x-6)}\cdot\left[\frac{(x-4)^2}{(x+4)^2}-1\right]+\frac{(x-6)}{(x+6)}\cdot\left[\frac{(x+9)^2}{(x-9)^2}-1\right] < 0$$ after that, you can easily simplify and go on to find the range of $x$. Make sure to not encounter fourth degree polynomials and instead eliminate terms like squares, $x^2 + 36$ as they are never negative. Also make sure to intersect the domain of $x$ while cancelling terms in the denominator. here's the working after this:

$\frac{(x+6)}{(x-6)}\cdot\left[\frac{((x-4)+(x+4))((x-4)-(x+4))}{(x+4)^2}\right]+\frac{(x-6)}{(x+6)}\cdot\left[\frac{((x+9)+(x-9))((x+9)-(x-9))}{(x-9)^2}-1\right] < 0$

or

$\frac{(x+6)}{(x-6)}\cdot\left[\frac{(2x)(-8)}{(x+4)^2}\right]+\frac{(x-6)}{(x+6)}\cdot\left[\frac{(2x)(18)}{(x-9)^2}\right] < 0$

cancel $4$ both sides, we get

$\frac{9x(x-6)}{(x+6)(x-9)^2} - \frac{4x(x+6)}{(x-6)(x+4)^2} < 0$

this simplied result gives me eternal satisfaction I needed for this question XD

simplifying further, we get

$x\cdot\left[\frac{9(x-6)^2(x+4)^2 - 4(x+6)^2(x-9)^2}{(x+6)(x-6)(x+4)^2(x-9)^2}\right] < 0$

or

$x\cdot\left[\frac{(3x^2-6x-72)^2 - (2x^2-6x-108)^2}{(x+6)(x-6)(x+4)^2(x-9)^2}\right] < 0$

sweet!

$\frac{(5x^2-12x-180)(x^2+36)}{(x+6)(x-6)(x+4)^2(x-9)^2} < 0$

now,

$5x^2-12x-180$ has no integer roots, so we shall apply the quadractic formula to find the roots of x. Don't lose hopes though as we can now freely eliminate x^2 + 36 , (x+4)^2 . (x-9)^2 (they are never negative :D)

this simplifies the equation to

$\frac{(5x^2-12x-180)}{(x+6)(x-6)} < 0$

too many spoilers right? i am providing the solution now :D

domain: $$x \in (-\infty , -4) \cup (-4 , 9) \cup (9 , \infty)$$

range: $$x \in (-\infty , -6) \cup \left(\frac{12 - \sqrt{3744}}{10} , 0\right)\cup \left(6 , \frac{12 + \sqrt{3744}}{10}\right)$$

intersecting the domain and the range, the finalllll answerrrrr: $$x \in (-\infty , -6) \cup \left(\frac{12 - \sqrt{3744}}{10} , -4\right) \cup (-4 , 0) \cup \left(6 , \frac{12 + \sqrt{3744}}{10}\right)$$

eternal satisfaction reached

Cuckoo Beats
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