I don't understand this part alone. Can you please explain?
_________________

My gmat story
MGMAT1 - 630 Q44V32
MGMAT2 - 650 Q41V38
MGMAT3 - 680 Q44V37
GMATPrep1 - 660 Q49V31
Knewton1 - 550 Q40V27

Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.



This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values.
What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change?

Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0
Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2.

Now let me add another factor: (x+8)(x+2)(x-1)(x-7)
Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes.

So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region.

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

e.g. (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.
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.

Similarly for (x-a)^2(x-b) > 0, x > b

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.
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression.
e.g. \((x-4)^2(x - 9)(x+11) < 0\)
We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this:

positive... -11 ... negative ... 9 ... positive

Since we need the region where x is negative, we get -11 < x < 9.
Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression.

I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them.
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

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 - inequalities-and-roots-118619.html#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
_________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button
http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

Correct me if I'm wrong.
If the lower part of the equation\(\frac {(x+2)(x-1)}{(x-4)(x-7)}\) were \(4\leq x \leq 7\), than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.
_________________

Kudos if you like the post!

Failing to plan is planning to fail.

Yes, you are right in both the cases.

Also, if you want to verify that the range you have got is correct, just plug in some values to see. Put x = 0, the expression is -ve. Put x = 2, the expression is positive.
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

IMHO, it should be x<b and also x is not equal to a .
so, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0
correct me, if I am wrong

my question is -
do we always have a sequence of + and - from rightmost to the left side. I mean is it possible to have + and then + again ?
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth