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Wavy Line Method Application - Complex Algebraic Inequalities

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Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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Added the PDF of the article at the end of the post! :)

Wavy Line Method Application - Complex Algebraic Inequalities


Many of you are familiar with the Wavy Line Method used to solve inequalities containing algebraic expressions in single variable. This method helps identify the range of values satisfying an inequality.

For instance, consider the following question:

Question: Find the range of values of x satisfying the inequality \((x – 3)(x – 6) < 0\).

To solve this inequality, you draw the wavy line as follows:

Image



And your inferences regarding the expression \((x – 3)(x – 6)\) would be:

    1. The expression will be positive for the range of values for which the curve is above the number line.
      a. In the above example, the pertinent ranges are \(x < 3\) and \(x > 6\). So, any value of \(x\) either less than \(3\) or greater than \(6\) will make the above expression positive in sign.
    2. The expression will be equal to zero for the values at which the curve intersects the number line.
      a. In the above example, the pertinent points are \(x = 3\) and \(x = 6\). So, both these values make the expression zero.
    3. The expression will be negative for the range of values for which the curve is below the number line.
      a. In the above example, there is only one portion where the curve is below the number line and this portion corresponds to the range \(3 < x < 6\). So, any value of \(x\) strictly between \(3\) and \(6\) will make the above expression negative in sign.

Since the above question asks for the range of values of x for which the expression is negative, the answer would be \(3 < x < 6\).


Well, that was a simple question, so application of the wavy line method was fairly straight-forward.


Now, for many test takers, things can seem complicated as exponents and algebraic fractions are added into the mix. However, the application of this method is pretty straightforward in such scenarios too.

In this article we’ll be focusing on such complex application of the wavy line method and list out the two fundamental rules that are used to draw the wavy line to solve any inequalities testing such algebraic expressions on the GMAT.

(This article benefits those who are not familiar with the wavy line method. The rules provided in this article are generic and can be utilized for all cases. :) )


When there are multiple instances of the same root:

Try to solve the following inequality using the Wavy Line Method:

\((x−1)^2 (x−2) (x−3) (x−4)^3 < 0\)

Did you notice how this inequality differs from the example above?

Notice that two of the four terms had an integral power greater than 1.


How to draw the wavy line for such expressions?

Let me directly show you how the wavy line would look and then later on the rule behind drawing it.

To know how you did, compare your wavy line with the correct one below.

Image


Notice that the curve bounced back at the point \(x = 1\). (At every other root, including \(x = 4\) whose power was \(3\), it was simply passing through them.)

Can you figure out why the wavy line looks like this for this particular inequality?

(Hint: The wavy line for the inequality \((x−1)^{38} (x−2)^{57} (x−3)^{15} (x−4)^{27} < 0\) is also the same as above)

Come on! Give it a try.

If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line.

Remember: We’ll refer the region above the number line as positive region and the region below the number line as negative region.

Image


What do you need before drawing the wavy line?

    1. You need to draw a horizontal line – this will be our Number Line to identify the ranges.
    2. Mark the “zero points”: Note the values at which at least one of the factor terms in the expression become zero.
      a. Eg: For instance, in the above expression, the term \(x – 1 = 0\) at the point \(x = 1\), the term \(x – 2 = 0\) at the point \(x = 2\), and so on. So, we need to mark the points \(1\), \(2\), \(3\), and \(4\) on the number line.


How to draw the wavy line?

    1. How to start: Once you draw the number line and mark the zero points on the number line, it’s time to draw the wavy line. Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression.

    2. How to alternate:
    if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region).

For instance, in the above example, observe the behavior of the wave at the points \(2\), \(3\), and \(4\).

Image



However, if the power of a term is even, then the wave bounces back into the same region. Observe the behavior of the wave at the point \(x = 1\) (the term \((x-1)\) has an even power in the above expression).

Image


Now look back at the above expression and analyze your wavy line. (Refer to the complete wavy line provided above if required)

Solution

Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to identify the range where the above expression would be less than zero, look for the area(s) in your wavy line diagram where the curve is below the number line.

So the correct solution set would simply be {\(3 < x < 4\)} \(U\) {{\(x < 2\)} – {\(1\)}}

In words, it is the Union of two regions – region1 between \(x = 3\) and \(x = 4\) and region2 which is \(x < 2\), excluding the point \(x = 1\).

Image


Food for Thought

Now, try to answer the following questions:

    1. Why did we exclude the point \(x = 1\) from the solution set of the last example? (Easy Question)
    2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains?


What if the algebraic expression contains a fraction?

Consider the following inequality.

\(\frac{(x – 3)}{(x – 6)} ≤ 0\)

How would you apply the wavy line method to solve the above inequality?

The good news is it’s still the same process!
    • Draw the number line
    • Mark the zero points
    • Draw the wavy line per the above rules

However, we need to keep in mind, the following things.

    1. The inequality asks for “\(≤ 0\)” (less than or equal to) and not just “\(< 0\)” (less than)
      a. This is an indicator that the required range is a combination of two ranges: range of values of \(x\) for which the expression satisfies “\(<0\)’ condition and the values of \(x\) for which the expression satisfies “\(= 0\)” condition.
    2. The term \((x – 6)\) is in the denominator and we know that we should never consider cases where the denominator becomes zero in any fraction.
      a. So, we need to exclude the point \(x = 6\) from of our solution set.


Image


So our solution set becomes \(3 ≤ x < 6\)

(Once again, observe that the point \(x = 3\) is included in the solution while \(x = 6\) is excluded.)


Image


Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases. :)

- Krishna
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Originally posted by EgmatQuantExpert on 26 Aug 2016, 02:04.
Last edited by EgmatQuantExpert on 07 Aug 2018, 02:39, edited 8 times in total.
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New post Updated on: 07 Aug 2018, 02:50
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Originally posted by EgmatQuantExpert on 26 Aug 2016, 02:35.
Last edited by EgmatQuantExpert on 07 Aug 2018, 02:50, edited 4 times in total.
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 06 Sep 2016, 19:15
Excellent post. This is one level more comprehensive than the course video itself, I feel.

One question though : How is 6 a zero point of the inequality \(\frac{(x-3)}{(x-6)}<=0\)?

I get only 3, if we were to solve the equation to find zero points. I do understand that if I substituted x with a number very close to 6, I see that the equation will be less than 0. But is there a process to do this?
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 16 Nov 2016, 21:02
EgmatQuantExpert wrote:
Exercise Questions


Detailed solutions will be posted soon.


Happy Learning! :)



Perhaps it is time to post the solutions?
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Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post Updated on: 07 Aug 2018, 02:50
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bb wrote:
EgmatQuantExpert wrote:
Exercise Questions


Detailed solutions will be posted soon.


Happy Learning! :)



Perhaps it is time to post the solutions?


Hey bb,


The solutions of all the 7 problems have been posted today.


Any feedback on them would be appreciated :)


Thanks,
e-GMAT Team

Image

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Originally posted by EgmatQuantExpert on 18 Nov 2016, 10:12.
Last edited by EgmatQuantExpert on 07 Aug 2018, 02:50, edited 1 time in total.
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 26 Mar 2017, 07:01
Hi Payal..
Nice explation.... really helpful, but m really confused between the application of critical key points and wavy line method.
could you please explain.
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 08 May 2017, 08:09
2
Hello,

I found this process of yours extremely helpful and intuitive. Thanks!

A point has been bothering me a lot and I have not been able to answer it myself:

When drawing the wavy line, why do we always start from the top right corner rather than the bottom right corner? Maybe this is a dumb question and the answer is obvious, but I can't figure it out.

Can a curve never end sloping downwards?

Thanks!
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 08 May 2017, 08:24
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kaaaaku wrote:
Hello,

I found this process of yours extremely helpful and intuitive. Thanks!

A point has been bothering me a lot and I have not been able to answer it myself:

When drawing the wavy line, why do we always start from the top right corner rather than the bottom right corner? Maybe this is a dumb question and the answer is obvious, but I can't figure it out.

Can a curve never end sloping downwards?

Thanks!


Though I have not read the initial post because i heavily rely on this method only learnt in school days :D , It does not matter you start from top right or top bottom. You place +ve sign on the extreme right i.e. the last DOT on the right (or say intersection of wave with line) and then a -ve sign on left of it and then keep alternating with signs moving towards left.

Hope it is clear :)
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 31 Jul 2017, 00:21
@e-gmat team

We did not take point where x=1 because then the expression would be equal to zero and not less than zero. Am I right?

Also we have to just exclude x=1, we can take all other values less than 1 ?

For example, Let x= 0

(0-1)^2 * (0-2) * (0-3) * (0-4)^3 <0

1*-2*-3*-32 <0

-192 which is less than 0
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 31 Jul 2017, 00:25
@e-gmat team

When we say that region 1 union region 2, that just simply means the values of both the regions work or common values between the two regions?
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Re: Wavy Line Method Application - Complex Algebraic Inequalities  [#permalink]

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New post 27 Oct 2018, 21:41
EgmatQuantExpert wrote:
Added the PDF of the article at the end of the post! :)

Wavy Line Method Application - Complex Algebraic Inequalities


Many of you are familiar with the Wavy Line Method used to solve inequalities containing algebraic expressions in single variable. This method helps identify the range of values satisfying an inequality.

For instance, consider the following question:

Question: Find the range of values of x satisfying the inequality \((x – 3)(x – 6) < 0\).

To solve this inequality, you draw the wavy line as follows:

Image



And your inferences regarding the expression \((x – 3)(x – 6)\) would be:

    1. The expression will be positive for the range of values for which the curve is above the number line.
      a. In the above example, the pertinent ranges are \(x < 3\) and \(x > 6\). So, any value of \(x\) either less than \(3\) or greater than \(6\) will make the above expression positive in sign.
    2. The expression will be equal to zero for the values at which the curve intersects the number line.
      a. In the above example, the pertinent points are \(x = 3\) and \(x = 6\). So, both these values make the expression zero.
    3. The expression will be negative for the range of values for which the curve is below the number line.
      a. In the above example, there is only one portion where the curve is below the number line and this portion corresponds to the range \(3 < x < 6\). So, any value of \(x\) strictly between \(3\) and \(6\) will make the above expression negative in sign.

Since the above question asks for the range of values of x for which the expression is negative, the answer would be \(3 < x < 6\).


Well, that was a simple question, so application of the wavy line method was fairly straight-forward.


Now, for many test takers, things can seem complicated as exponents and algebraic fractions are added into the mix. However, the application of this method is pretty straightforward in such scenarios too.

In this article we’ll be focusing on such complex application of the wavy line method and list out the two fundamental rules that are used to draw the wavy line to solve any inequalities testing such algebraic expressions on the GMAT.

(This article benefits those who are not familiar with the wavy line method. The rules provided in this article are generic and can be utilized for all cases. :) )


When there are multiple instances of the same root:

Try to solve the following inequality using the Wavy Line Method:

\((x−1)^2 (x−2) (x−3) (x−4)^3 < 0\)

Did you notice how this inequality differs from the example above?

Notice that two of the four terms had an integral power greater than 1.


How to draw the wavy line for such expressions?

Let me directly show you how the wavy line would look and then later on the rule behind drawing it.

To know how you did, compare your wavy line with the correct one below.

Image


Notice that the curve bounced back at the point \(x = 1\). (At every other root, including \(x = 4\) whose power was \(3\), it was simply passing through them.)

Can you figure out why the wavy line looks like this for this particular inequality?

(Hint: The wavy line for the inequality \((x−1)^{38} (x−2)^{57} (x−3)^{15} (x−4)^{27} < 0\) is also the same as above)

Come on! Give it a try.

If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line.

Remember: We’ll refer the region above the number line as positive region and the region below the number line as negative region.

Image


What do you need before drawing the wavy line?

    1. You need to draw a horizontal line – this will be our Number Line to identify the ranges.
    2. Mark the “zero points”: Note the values at which at least one of the factor terms in the expression become zero.
      a. Eg: For instance, in the above expression, the term \(x – 1 = 0\) at the point \(x = 1\), the term \(x – 2 = 0\) at the point \(x = 2\), and so on. So, we need to mark the points \(1\), \(2\), \(3\), and \(4\) on the number line.


How to draw the wavy line?

    1. How to start: Once you draw the number line and mark the zero points on the number line, it’s time to draw the wavy line. Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression.

    2. How to alternate:
    if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region).

For instance, in the above example, observe the behavior of the wave at the points \(2\), \(3\), and \(4\).

Image



However, if the power of a term is even, then the wave bounces back into the same region. Observe the behavior of the wave at the point \(x = 1\) (the term \((x-1)\) has an even power in the above expression).

Image


Now look back at the above expression and analyze your wavy line. (Refer to the complete wavy line provided above if required)

Solution

Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to identify the range where the above expression would be less than zero, look for the area(s) in your wavy line diagram where the curve is below the number line.

So the correct solution set would simply be {\(3 < x < 4\)} \(U\) {{\(x < 2\)} – {\(1\)}}

In words, it is the Union of two regions – region1 between \(x = 3\) and \(x = 4\) and region2 which is \(x < 2\), excluding the point \(x = 1\).

Image


Food for Thought

Now, try to answer the following questions:

    1. Why did we exclude the point \(x = 1\) from the solution set of the last example? (Easy Question)
    2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains?


What if the algebraic expression contains a fraction?

Consider the following inequality.

\(\frac{(x – 3)}{(x – 6)} ≤ 0\)

How would you apply the wavy line method to solve the above inequality?

The good news is it’s still the same process!
    • Draw the number line
    • Mark the zero points
    • Draw the wavy line per the above rules

However, we need to keep in mind, the following things.

    1. The inequality asks for “\(≤ 0\)” (less than or equal to) and not just “\(< 0\)” (less than)
      a. This is an indicator that the required range is a combination of two ranges: range of values of \(x\) for which the expression satisfies “\(<0\)’ condition and the values of \(x\) for which the expression satisfies “\(= 0\)” condition.
    2. The term \((x – 6)\) is in the denominator and we know that we should never consider cases where the denominator becomes zero in any fraction.
      a. So, we need to exclude the point \(x = 6\) from of our solution set.


Image


So our solution set becomes \(3 ≤ x < 6\)

(Once again, observe that the point \(x = 3\) is included in the solution while \(x = 6\) is excluded.)


Image


Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases. :)

- Krishna



Hey egmat team.

Thank you so much for this post. I always had issues understanding this concept (I couldnt understand this in class 9th as well).

Thanks a ton once again. I was able to solve all the questions that you posted.
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New post Updated on: 23 May 2019, 21:18

Originally posted by EgmatQuantExpert on 15 May 2019, 02:13.
Last edited by EgmatQuantExpert on 23 May 2019, 21:18, edited 1 time in total.
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