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

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

[(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??

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
Quant Expert
e-GMAT

[/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.

Oreo

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


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


You could find more about this tool in this link:
fator-table-with-sign-the-useful-tool-to-solve-polynomial-inequalities-229988.html#p1771368

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

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 \(-2<x<2\) & \(x>5\)

Hi,

Just a heads up .

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 .

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?

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

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

\(\frac{(x-2)^3(x+2)^3}{(x-5)^5(x-3)^4(x+3)^4} >0\)
if equivalent to:
\((x-2)(x+2)/(x-5) >0 & x \neq -3 & x \neq 3\)

x = {-1,0,1,6,.....}

Product of all integer values of x = 0*(-1)*1*6*... = 0

IMO C

I was wondering, if the first value here was 3 with an even exponent, would the line start at exactly point 3 instead of starting from top right?