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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??

(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.

Correct Answer: Option C

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.

(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.

Correct Answer: Option C

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.

Re: Wavy Line Method Application - Exercise Question #7 [#permalink]

Show Tags

04 Dec 2016, 01:45

[/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.

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.

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

Re: Wavy Line Method Application - Exercise Question #7 [#permalink]

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08 Oct 2017, 09:04

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).

(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.

Correct Answer: Option C

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?

(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.

Correct Answer: Option C

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?

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