Inequalities trick

thank you VeritasKarishma all is clear now just last question

in your last step you have \(2(x – 0)(x – 5/2)(x – 6) < 0\) you have \(2(x-0)\) may i know how you got it, i thought maybe you factored out 2 like this \(2x = 2(x-0)\) but if it is so why are you subtracting 0 and not adding it though yes adding or subtracting zero doesnt change solution ....but still kinda curious
and what role does \(2(x-0)\) play here and why
many thanks

WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.

1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

example -
(x+2)(x-1)(7 - x)<0

Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.

2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)

We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0

3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

BUT if it is a fraction the denominator in the solution will not have = SIGN
example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given \(\sqrt{x}\), x cannot be negative.

\(\sqrt{x}\) < 10
implies 0 < \(\sqrt{x}\) < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.

Refer - https://gmatclub.com/forum/inequalities- ... ml#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate
<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.

I will save this future references....
Please add anything that you feel will help.

Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post

Didn't understand 2 and 3 VeritasKarishma Bunuel could you please explain?

These three posts explain these points comprehensively. Please check:

Hi KarishmaB - I've been going through this thread/your articles (all great resources/posts, so thank you!). One bit of confusion as I'm still wrapping my head around when Division of the roots doesn't change anything, as well as the situations where the inequality includes an "=" sign. When I read the 3rd post you link, I cannot tell what the inequalities are (all I see are "?" where the signs should be), and can't grasp this topic.

As well, I don't see much detail on why the division aspect has no impact on the values. Would you be able to elaborate just a little, or link another resource if easier?

Thanks so much, everything else makes perfect sense so far!

CM12

Hope you understand that the solution of say (x-a)(x-b)(x-c) < 0
is all about the sign of the factors (x-a), (x-b) and (x-c).

(x-a)(x-b)(x-c) < 0 only means that (x-a)(x-b)(x-c) must be negative. Which means that either exactly one of these factors or all three must be negative and our entire method is based on that one concept.

Now notice that if instead the factors are written as (x-a)(x-b) / (x-c) does it change anything regarding the sign of the expression based on the signs of the factors?

Consider two numbers a and b.
The sign of both ab and a/b will depend on the sign of a and sign on b.
If a and b are both positive, ab and a/b both are positive too.
If a is positive and b is negative, ab and a/b both are negative.
If a is negative and b is positive, ab and a/b both are negative.
If a and b are both negative, ab and a/b both are positive.

It doesn't matter whether a and b are multiplied or divided, the sign of ab and a/b will always be the same.

The only difference between ab and a/b is that in a/b, b cannot be 0 because division by 0 is not allowed.

So solution of \(ab \leq 0\) includes a = 0 or b = 0 (because the expression can be 0 too now)

but solution of \(\frac{a}{b} \leq 0\) includes only a = 0 (b cannot be 0)

So solution of (x-a)*(x-b) < 0 is the same as the solution of (x-a) / (x-b) < 0.

But solution of \((x-a)*(x-b) \leq 0\) includes x = a, x = b but solution of \(\frac{(x-a)}{(x-b)} \leq 0\) includes only x = a.

This is the concept in brief.

I have discussed it in more detail in my inequalities and absolute values module here:
https://anglesandarguments.com/study-module