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Here is what i did in this question==>

Critical points => Equate the numerator and denominator terms to zero and obtain the critical points.

In here => x^2-4=0 => x=2 and x=-2
Also x-5=0 => x=5
And finally x^2-9=0=> x=3 and x=-3

Critical points => -3,-2,2,3,5

Now mark these on the number line and pick up a number in each boundary.
If the value makes the original inequality true => Pick that boundary else discard

=> x=> (-∞,-3)U(-2,2)U(3,5)


Personal Opinion-> This question might be good for practise but is most certainly not a GMAT like Question.


The wavy figure that is attached above makes it look scary.In reality we don't need that crazy figure.
Just plug in numbers in a simple plain straight line by marking the critical points.
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Does the range include 2 and -2 because the points are not circled on the number line like other points?
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Does the range include 2 and -2 because the points are not circled on the number line like other points?

The range would have included any of the critical points -3, -2, 2, 3 and 5 if the equation read \(\frac{(x^2-4)}{(x-5)(x^2-9)} <= 0\) (notice the equal sign at the right)

So, to answer your question, it doesn't include 2 and -2.
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Solution:

Hey Everyone,

Please find below the solution of the given problem.

Rewriting the inequality to easily identify the zero points

We know that \((x^2 – 4) = (x+2)(x-2)\)

And similarly, \((x^2-9) = (x+3)(x-3)\)

So, the given inequality can be written as:

\((x+2)(x-2)/(x – 5)(x+3)(x-3)<0\)

Plotting the zero points and drawing the wavy line:



Required Range:

x < -3 or -2 < x < 2 or 3 < x < 5

Correct Answer: Option D


Hi, since we have 5 zero points and no number with an even power, can we deduce that the must be 3 or 4 intervals for x? Since there is no answer with 4 intervals and only one with 3 intervals, this must be the right answer, true?
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EgmatQuantExpert
Solution:

Hey Everyone,

Please find below the solution of the given problem.

Rewriting the inequality to easily identify the zero points

We know that \((x^2 – 4) = (x+2)(x-2)\)

And similarly, \((x^2-9) = (x+3)(x-3)\)

So, the given inequality can be written as:

\((x+2)(x-2)/(x – 5)(x+3)(x-3)<0\)

Plotting the zero points and drawing the wavy line:



Required Range:

x < -3 or -2 < x < 2 or 3 < x < 5

Correct Answer: Option D


Can you please explain, how we have come to the required range?
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arorni
Can you please explain, how we have come to the required range?

This is exactly what she did :-D I think you have to specify your question a bit. Which part did you not understand?
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arorni
Can you please explain, how we have come to the required range?

This is exactly what she did :-D I think you have to specify your question a bit. Which part did you not understand?

I saw Payal's other post on wavy line https://gmatclub.com/forum/wavy-line-me ... 24319.html. It indeed cleared my doubt about this question, but I am a little confused about drawing wavy lines. Do we always have to start drawing the wavy line from the -ve region (ie Is it the upper right corner)? How the wavy line of (x-1)^2 will look, I got the idea that for even powers the line will remain in the same region but why we have started drawing from the -ve region?
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arorni
Can you please explain, how we have come to the required range?

This is exactly what she did :-D I think you have to specify your question a bit. Which part did you not understand?

I saw Payal's other post on wavy line https://gmatclub.com/forum/wavy-line-me ... 24319.html. It indeed cleared my doubt about this question, but I am a little confused about drawing wavy lines. Do we always have to start drawing the wavy line from the -ve region (ie Is it the upper right corner)? How the wavy line of (x-1)^2 will look, I got the idea that for even powers the line will remain in the same region but why we have started drawing from the -ve region?

Ok i got you. As you can see in her approach, the largest zero point for x is 5 (from all the others:-3,-2,2,3,5). So its easy to check if the equation gives a positive or negative result, if you plug in numbers that are less than the smallest zero point (here: -3) or greater the largest zero point (5).

So let's try all number greater the largest zero point, which is 5. For example, plug in \(x=10\). Then all terms in the brackets will be positive, so for all \(x>5\) the equation is \(>0\).

Remember:
\(+*+=+\)
\(-*+=-\)
\(-*-=+\)

Further on, for even powers, the line will not cross the axis, but bounces back, for odd powers it does cross the axis.
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