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Solving Inequalities [#permalink]
19 Jun 2012, 02:34

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Solving Inequalities

I was going through the posts on inequalities and found that many good concepts are explained here, but still people do have trouble solving the question using these concept. In these posts, there were quadratic equations, curves, graphs and other mathematical stuff. With this post, I am trying to provide a simple method to solve such questions quickly. I won't be writing the concepts behind it. Remember this is the same OLD concept, it's just presented differently.

Case 1: Multiplication

for example: (x-1)(x-2)(x-3)(x-7) \leq 0

To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression

Thus, inequality would hold true in the intervals: 1 \leq x \leq 2 3 \leq x \leq 7,

Note that intervals are inclusive of 3 & 7

Case 2: Division

In case of division: \frac{(x-1)(x-2)}{(x-3)(x-7)} \leq 0 Using the same approach as above; 1 \leq x \leq 2 3 < x < 7,(x\neq3,7)

Since (x-3)(x-7) is in denominator, its value can't be 0.

Following things to be kept in mind while using above method: 1. Cofficient of x should be positive: for ex - (x-a)(b-x)>0, can be written as (x-a)(x-b)<0 2. Even powers: for ex - (x-9)^2(x+3) \geq 0, (x-9)^2 is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: (x-a)^3(x-b)^5>0, will be same as (x-a)(x-b)>0 4. Cancelling the common terms:

for ex - \frac {(x^2+x-6)(x-11)}{(x+3)} >0, it can be simplified as (x-2)(x-11)>0 or, ---(+)-----2---(-)-------------11----(+)------ thus x <2 or x>11, but since at x = -3 (in the original expression), we get undefined form, so, x

\neq -3

A question for you: For what values of x, does the following inequality holds true? (x-a)(x-b)...(x-n)...(x-z) \geq 0, where {a, b, c,...} are integers.

Re: Solving Inequalities [#permalink]
20 Jun 2012, 00:24

So that problem that you posted, let me give it a try:

A question for you: For what values of x, does the following inequality holds true? (x-a)(x-b)...(x-n)...(x-z) \geq 0, where {a, b, c,...} are integers.

Roots are: a,b,c,d...z The condition will hold true for these intervals: x\geq{z} x =x {x}\leq{x}\leq{y} {v}\leq{x\leq{w} {s}\leq{x\leq{t} {p}\leq{x\leq{t} {m}\leq{x\leq{n} {j}\leq{x\leq{k} {g}\leq{x\leq{h} {d}\leq{x\leq{e} {a}\leq{x\leq{b}

Re: Solving Inequalities [#permalink]
20 Jun 2012, 00:33

macjas wrote:

So that problem that you posted, let me give it a try:

A question for you: For what values of x, does the following inequality holds true? (x-a)(x-b)...(x-n)...(x-z) \geq 0, where {a, b, c,...} are integers.

Roots are: a,b,c,d...z The condition will hold true for these intervals: x\geq{z} x =x {x}\leq{x}\leq{y} {v}\leq{x\leq{w} {s}\leq{x\leq{t} {p}\leq{x\leq{t} {m}\leq{x\leq{n} {j}\leq{x\leq{k} {g}\leq{x\leq{h} {d}\leq{x\leq{e} {a}\leq{x\leq{b}

Is this correct??

Hi,

You have identified everything, but give a close tought again. The answer is pretty much straight forward.

Re: Solving Inequalities [#permalink]
20 Jun 2012, 00:48

cyberjadugar wrote:

macjas wrote:

So that problem that you posted, let me give it a try:

A question for you: For what values of x, does the following inequality holds true? (x-a)(x-b)...(x-n)...(x-z) \geq 0, where {a, b, c,...} are integers.

Roots are: a,b,c,d...z The condition will hold true for these intervals: x\geq{z} x =x {x}\leq{x}\leq{y} {v}\leq{x\leq{w} {s}\leq{x\leq{t} {p}\leq{x\leq{t} {m}\leq{x\leq{n} {j}\leq{x\leq{k} {g}\leq{x\leq{h} {d}\leq{x\leq{e} {a}\leq{x\leq{b}

Is this correct??

Hi,

You have identified everything, but give a close tought again. The answer is pretty much straight forward.

Re: Solving Inequalities [#permalink]
25 Jun 2012, 04:05

nice one - and a great link .. inequalities

kudos to cyberjadugar as well as gurpreet singh, veritas karishma...

is the answer to macjas quest is - 4 ( -2, -3, 3,4) -? kindly correct with explaination.

and to the solution to cyberjadugar - is it as follows? x<=a = +ve a>=x>=b = -ve b>=x>=n = +ve n>=x>=z = -ve x>=z = +ve

i may have done above wrong. pls do correct. and if the above is right , how the whole thing can be put together? like is it possible to write - z<=x<=a ot like z<=x<=a is +ve.

Re: Solving Inequalities [#permalink]
05 Nov 2012, 05:23

gpk wrote:

Edit: The 'x' before the inequality caused the confusion. I checked and that 'x' doe not exist.

What about '0' and '-4', '-5' '-6'...? 0<5, -4<5 ...

macjas wrote:

This one is a really useful trick cyberjadugar. Kudos to you and gurpreetsingh.

Example for practice from OG13 PS229:

How many of the integers that satisfy the inequality \frac{{(x+2)(x+3)}}{{x-2}}x\geq{0} are less than 5?

A 1 B 2 C 3 D 4 E 5

OldFritz wrote:

-3,-2,3,4

Therefore, there are 4 integers less than 5 that satisfy the inequality.

Cheers, Der alte Fritz.

Yes, the extra 'x' is a typo. Of course, none of the options match if that x was still present in the inequality. _________________

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You can have anything you want if you want it badly enough. You can be anything you want to be and do anything you set out to accomplish, if you hold to that desire with the singleness of purpose. ~William Adams

Many of life's failures are people who did not realize how close to success they were when they gave up. ~Thomas A. Edison

Wir müssen wissen, Wir werden wissen. (We must know, we will know.) ~Hilbert

Re: Solving Inequalities [#permalink]
05 Nov 2012, 05:33

cyberjadugar wrote:

2. Even powers: for ex - (x-9)^2(x+3) \geq 0, (x-9)^2 is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: (x-a)^3(x-b)^5>0, will be same as (x-a)(x-b)>0

Hello cyberjadugar,

Could you illustrate (2) and (3) with examples. Though (3) is kind of clear, I do not understand what exactly you mean in (2). _________________

Kudos is the currency of appreciation.

You can have anything you want if you want it badly enough. You can be anything you want to be and do anything you set out to accomplish, if you hold to that desire with the singleness of purpose. ~William Adams

Many of life's failures are people who did not realize how close to success they were when they gave up. ~Thomas A. Edison

Wir müssen wissen, Wir werden wissen. (We must know, we will know.) ~Hilbert

2. Even powers: for ex - (x-9)^2(x+3) \geq 0, (x-9)^2 is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: (x-a)^3(x-b)^5>0, will be same as (x-a)(x-b)>0

Hello cyberjadugar,

Could you illustrate (2) and (3) with examples. Though (3) is kind of clear, I do not understand what exactly you mean in (2).

Re: Solving Inequalities [#permalink]
15 May 2013, 22:55

shelrod007 wrote:

closed271 wrote:

cyberjadugar wrote:

2. Even powers: for ex - (x-9)^2(x+3) \geq 0, (x-9)^2 is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: (x-a)^3(x-b)^5>0, will be same as (x-a)(x-b)>0

Hello cyberjadugar,

Could you illustrate (2) and (3) with examples. Though (3) is kind of clear, I do not understand what exactly you mean in (2).

Could some one please explain point 2 ??

Hi,

Sorry for the late reply.

For even powers, such as (x-9)^2, if you check for various values of x, for example, x = 1, (x-9)^2 = 64(>0) x=-1, (x-9)^2 = 100(>0) x =10, (x-9)^2=1(>0) but for x = 9, (x-9)^2=0 so, for every value of x, the even powers will always be greater or equal to 0, i.e. the sign of the expression doesn't change from positive to negative.

I was going through the posts on inequalities and found that many good concepts are explained here, but still people do have trouble solving the question using these concept. In these posts, there were quadratic equations, curves, graphs and other mathematical stuff. With this post, I am trying to provide a simple method to solve such questions quickly. I won't be writing the concepts behind it. Remember this is the same OLD concept, it's just presented differently.

Case 1: Multiplication

for example: (x-1)(x-2)(x-3)(x-7) \leq 0

To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression

Thus, inequality would hold true in the intervals: 1 \leq x \leq 2 3 \leq x \leq 7,

Note that intervals are inclusive of 3 & 7

Case 2: Division

In case of division: \frac{(x-1)(x-2)}{(x-3)(x-7)} \leq 0 Using the same approach as above; 1 \leq x \leq 2 3 < x < 7,(x\neq3,7)

Since (x-3)(x-7) is in denominator, its value can't be 0.

Following things to be kept in mind while using above method: 1. Cofficient of x should be positive: for ex - (x-a)(b-x)>0, can be written as (x-a)(x-b)<0 2. Even powers: for ex - (x-9)^2(x+3) \geq 0, (x-9)^2 is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: (x-a)^3(x-b)^5>0, will be same as (x-a)(x-b)>0 4. Cancelling the common terms:

for ex - \frac {(x^2+x-6)(x-11)}{(x+3)} >0, it can be simplified as (x-2)(x-11)>0 or, ---(+)-----2---(-)-------------11----(+)------ thus x <2 or x>11, but since at x = -3 (in the original expression), we get undefined form, so, x

\neq -3

A question for you: For what values of x, does the following inequality holds true? (x-a)(x-b)...(x-n)...(x-z) \geq 0, where {a, b, c,...} are integers.

PS: I hope you find this post useful, please provide feedback to improve the quality of the post.

Thanks,

Hi,

I have question with the below one,

[list]for ex - \frac {(x^2+x-6)(x-11)}{(x+3)} >0, it can be simplified as (x-2)(x-11)>0 or, ---(+)-----2---(-)-------------11----(+)------ thus x <2 or x>11, but since at x = -3 (in the original expression), we get undefined form, so, [m]x

using the plot of + ,- , how to identify x<2 or x>2 . I have a impression like if > is the used then it should be always x>2 & x>11 but the answer u have mentioned again confused me. Please help

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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