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# Find the product of the integer values of x that satisfy the inequalit

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Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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Updated on: 23 Dec 2018, 03:58
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65% (02:12) correct 35% (02:24) wrong based on 277 sessions

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Exercise Question #7 of the Wavy Line Method Application

Find the product of the integer values of x that satisfy the inequality $$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} > 0$$

A. -2
B. -1
C. 0
D. 1
E. 2

Apologies for the inconvenience. Question edited appropriately to correct the earlier typo.

Detailed solution will be posted soon.

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Originally posted by EgmatQuantExpert on 26 Aug 2016, 02:51.
Last edited by Bunuel on 23 Dec 2018, 03:58, edited 3 times in total.
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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18 Nov 2016, 03:33
1
5
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, $$(x^2-4)^3/((x – 5)^5 (x^2-9)^4 )>0$$ can be written as

$$((x+2)^3 (x-2)^3)/((x – 5)^5 (x+3)^4 (x-3)^4 )>0$$

(x + 3) and (x-3) are the terms with even power. So, the wavy line should bounce back at the corresponding zero-points x = -3 and x = 3.

Plotting the zero points and drawing the wavy line:

Required Range:

-2< x < 2 or x > 5

Calculations:

We need the product of all integers in this range. However, note that the integer 0 is a part of the range (-2 < x < 2).

Therefore, the product will be 0.

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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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26 Aug 2016, 10:17
1
EgmatQuantExpert wrote:
Wavy Line Method Application - Exercise Question #7

Find the product of the integer values of x that satisfy the inequality $$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} < 0$$

Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html

Detailed solution will be posted soon.

The inequality is $$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} < 0$$

or $$\frac{(x-2)^3 (x+2)^3}{(x-5)^5 (x - 3)^4 (x + 3)^4} < 0$$

Now, we have (x-3) and (x+3) with EVEN powers, so we will not move the curve direction to the upper side.

So, the zero points of x will be -3,-2,2,3 and 5

But we cannot have x = 5,3 or -3 as they are at the denominator.

So, the range of x will be (-infinity, -2) U (2,5).

Please correct me if I am missing anything.
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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04 Dec 2016, 05:27
1
EgmatQuantExpert wrote:
Find the product of the integer values of x that satisfy the inequality $$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} > 0$$

We could solve this question easily with the tool called: factor table with sign

$$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} > 0 \iff \frac{(x-2)^3(x+2)^3}{(x-5)^5(x-3)^4(x+3)^4} > 0$$

First, find the value that all factors are equal to 0

$$\begin{split} x-2=0 &\iff x=2 \\ x+2=0 &\iff x=-2\\ x-5=0 &\iff x=5\\ x-3=0 &\iff x=3\\ x+3=0 &\iff x=-3 \end{split}$$

Now, make a factor table with sign like this
Attachment:

Capture table 1.PNG [ 12.77 KiB | Viewed 6235 times ]

In line (1), we review all possible values of $$x$$ that make each fator equal to 0. The order is ascending (from lowest to highest)

Next, from line (2) to (6), we review the sign of each raw factor without power that based on the range value of $$x$$. Character "|" means that we no need to care about the specific value of them, just the sign positive (+) or negative (-).

Next, from line (7) to (11), we review the sign of each fator with power that exists in the expression.

Finally, in line (12), we could quickly review the sign of the expression.

Hence, we have $$\frac{(x-2)^3(x+2)^3}{(x-5)^5(x-3)^4(x+3)^4} > 0 \iff x \in (-2,2) \cup (5, +\infty)$$

fator-table-with-sign-the-useful-tool-to-solve-polynomial-inequalities-229988.html#p1771368
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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08 Oct 2017, 10:04
1
Given inequality $$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} > 0$$

Since only expressions with odd powers play role in determining the sign of the equation,
1.) we can remove the even power expressions and
2.) make all odd powers equivalent to 1.
We also note that x cannot be equal to 5, 3, -3 as these will make the numerator '0'.

Thus, we can solve for below inequality instead.
$$\frac{(x^2-4)}{(x-5)} > 0$$

Or,
$${(x^2-4)(x-5)} > 0$$

Points are : -2, 2 and 5.
Thus, Ans: (-2,2) U (5,infinity).

KS
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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22 Oct 2017, 00:09
1
soodia wrote:
EgmatQuantExpert wrote:
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, $$(x^2-4)^3/((x – 5)^5 (x^2-9)^4 )>0$$ can be written as

$$((x+2)^3 (x-2)^3)/((x – 5)^5 (x+3)^4 (x-3)^4 )>0$$

(x + 3) and (x-3) are the terms with even power. So, the wavy line should bounce back at the corresponding zero-points x = -3 and x = 3.

Plotting the zero points and drawing the wavy line:

Required Range:

-2< x < 2 or x > 5

Calculations:

We need the product of all integers in this range. However, note that the integer 0 is a part of the range (-2 < x < 2).

Therefore, the product will be 0.

Hi

for satisfying the inequality (x^2-4)^3 and ((x – 5)^5 should have the same sign
so, they could be both negative or positive
the range will be different
am I wrong?

Hi soodia

the range of $$x$$ is $$-2<x<2$$ & $$x>5$$. In this range the expression $$\frac{(x^2-4)^3}{(x-5)^5}$$ will always be greater than $$0$$. you can test this by using some values from the range

let $$x=-1$$ then $$\frac{(x^2-4)^3}{(x-5)^5} = \frac{(1-4)^3}{(-1-5)^5}>0$$

if $$x=6$$ then $$\frac{(x^2-4)^3}{(x-5)^5}=\frac{(6^2-4)^3}{(6-5)^5}>0$$

Hi EgmatQuantExpert

we know that $$(x^2-9)^4>0$$ and $$x≠3$$,$$-3$$ & $$5$$ because in that case denominator will be $$0$$ which is not possible.

So isn't our equation be simplified as $$\frac{(x^2-4)^3}{(x-5)^5}>0$$ (cross-multiplying $$(x^2-9)^4$$)

And the wavy line for this equation will be as below giving us the same range $$-2<x<2$$ & $$x>5$$

Attachment:

inequality 1.jpg [ 38.38 KiB | Viewed 4987 times ]
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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26 Jan 2019, 00:08
1
Hi,

The solution set is $$(-2,2)$$ U $$(5,\infty)$$
So the product of all integers $$= -1 \times 0 \times 1 \times 2 \times .... 10 \times ... \infty$$
So basically we are multiplying $$\rightarrow 0 \times -\infty = undefined$$

Its better to tweak the question so as to get a finite integer set . Then the answer will be zero .
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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26 Oct 2016, 18:09
My answer is {-3<x<-2} U {2<x<3}U{x>5}.

Not sure if I did it correctly. Please correct me if I am wrong
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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29 Oct 2016, 11:32
EgmatQuantExpert wrote:
Wavy Line Method Application - Exercise Question #7

Find the product of the integer values of x that satisfy the inequality $$\frac{(x^2-4)^3}{(x-5)^5 (x^2 - 9)^4} < 0$$

Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html

Detailed solution will be posted soon.

[(x-2)(x+2)]^3 / (x-5)^5 ( x^2-9)^4

critical values are -2,2,5,

the inequality is valid in ranges x<-3 and 2<x<5 , x not equal 3 , thus +ve integer value of x is 4 , and the -ve integer values belong to (-infinity,-4)

how can i get the product then or where am i going wrong??
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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27 Nov 2016, 11:51
EgmatQuantExpert wrote:
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, $$(x^2-4)^3/((x – 5)^5 (x^2-9)^4 )>0$$ can be written as

$$((x+2)^3 (x-2)^3)/((x – 5)^5 (x+3)^4 (x-3)^4 )>0$$

(x + 3) and (x-3) are the terms with even power. So, the wavy line should bounce back at the corresponding zero-points x = -3 and x = 3.

Plotting the zero points and drawing the wavy line:

Required Range:

-2< x < 2 or x > 5

Calculations:

We need the product of all integers in this range. However, note that the integer 0 is a part of the range (-2 < x < 2).

Therefore, the product will be 0.

just had a question to solidify my understanding of this - if the question was < instead of >, then there would be no ranges that satisfy this equation? Thank you.
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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27 Nov 2016, 13:04
OreoShake wrote:
EgmatQuantExpert wrote:
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, $$(x^2-4)^3/((x – 5)^5 (x^2-9)^4 )>0$$ can be written as

$$((x+2)^3 (x-2)^3)/((x – 5)^5 (x+3)^4 (x-3)^4 )>0$$

(x + 3) and (x-3) are the terms with even power. So, the wavy line should bounce back at the corresponding zero-points x = -3 and x = 3.

Plotting the zero points and drawing the wavy line:

Required Range:

-2< x < 2 or x > 5

Calculations:

We need the product of all integers in this range. However, note that the integer 0 is a part of the range (-2 < x < 2).

Therefore, the product will be 0.

just had a question to solidify my understanding of this - if the question was < instead of >, then there would be no ranges that satisfy this equation? Thank you.

Hey,

If the question was less than (<) instead of greater than(>), then we can still find the range of the range of x.

Look at the wavy curve in the solution given above and think, if we had to find the range of x for which the expression is less than zero, what range will we consider?

Hint: We had considered the positive range in the wavy curve to find the range of x for which the expression was more than zero.

Thanks,
Saquib
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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04 Dec 2016, 02:45
[/quote]

Hey,

If the question was less than (<) instead of greater than(>), then we can still find the range of the range of x.

Look at the wavy curve in the solution given above and think, if we had to find the range of x for which the expression is less than zero, what range will we consider?

Hint: We had considered the positive range in the wavy curve to find the range of x for which the expression was more than zero.

Thanks,
Saquib
Quant Expert
e-GMAT[/quote]

Hi Saquib,

Thank you for your prompt response. If we look at the graph, the for negatives we will get the ranges - x<-3, -2>x>-3, 3>x>2, and 5>x>3. Here the product of the integer values cannot be calcuated. Would these be correct ranges if the question asked for negative? Please confirm. Thanks a ton.

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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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19 Oct 2017, 09:51
EgmatQuantExpert wrote:
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, $$(x^2-4)^3/((x – 5)^5 (x^2-9)^4 )>0$$ can be written as

$$((x+2)^3 (x-2)^3)/((x – 5)^5 (x+3)^4 (x-3)^4 )>0$$

(x + 3) and (x-3) are the terms with even power. So, the wavy line should bounce back at the corresponding zero-points x = -3 and x = 3.

Plotting the zero points and drawing the wavy line:

Required Range:

-2< x < 2 or x > 5

Calculations:

We need the product of all integers in this range. However, note that the integer 0 is a part of the range (-2 < x < 2).

Therefore, the product will be 0.

Hi

for satisfying the inequality (x^2-4)^3 and ((x – 5)^5 should have the same sign
so, they could be both negative or positive
the range will be different
am I wrong?
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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07 Oct 2019, 20:52
Hi
I want to understand how to arrive at 0??
I made the curve line and got the correct ranges: -2<X<2 and X>5
But now how to arrive at the answer 0?
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Re: Find the product of the integer values of x that satisfy the inequalit  [#permalink]

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16 Oct 2019, 09:15
EgmatQuantExpert
Hi can u pls explain why 0 is the answer and not 1 or -1
Because all 3 numbers fall in to the range -2<x<2
Re: Find the product of the integer values of x that satisfy the inequalit   [#permalink] 16 Oct 2019, 09:15
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