Find the range of values of x that satisfy the inequality (x - 3)^2

Question

Wavy Line Method Application - Exercise Question #2

Find the range of values of x that satisfy the inequality \((x - 3)^2 (x + 1)^5 (x^2 - 9) < 0\)

A. x > 1
B. x > -3
C. -1 < x < 3
D. x < -3 or -1 < x < 3
E. x < -3

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Wavy Line Method Application has been explained in detail in the following post:: Wavy Line Method Application - Complex Algebraic Inequalities

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Official Answer

D

EgmatQuantExpert · Accepted Answer

Solution   :

Hey Everyone,

Please find below, the solution of the given problem.

Rewriting the inequality to easily identify the zero points

(x-3)^2 (x+1)^5 (x^2-9)<0

Since

(x^2-9)=(x+3)*(x-3)

The given inequality can be written as

(x-3)^3 (x+1)^5 (x+3)<0

Plotting the zero points and drawing the wavy line  :

https://e-gmat.com/blogs/wp-content/uploads/2016/11/S2.jpg

Required Range : x < -3 or -1 < x < 3

Correct Answer : Option D

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-Algebra-e1533292163516.png

abhimahna · Answer

Please post the options also.

Solving the inequality, we will get (x-3)^2(x+1)^5(x+3)(x-3) < 0

or (x-3)^3 (x+1)^5 (x+3) < 0.

Solving on the number line, we will get the points -3,-1 and 3.

Since the power of each is ODD. hence, the range of x will be (-infinity, -3) U (-1,3).

Please correct me if I am missing anything.

hatemnag · Answer

I guess the final answer is : -1< X <3 & x <-3
since the zero points are 3 , -1 , -3 
if we try each number within the range , the inequality will hold true, using the detailed equation :
(x-3)(x-3)(x+1)(x+3)(x-3)<0
But I propose to ignore the even powered expression like the case of (x-3) since the product of such expression must be positive.
Please Payal correct my answer if something is wrong

Shiv2016 · Answer

@e-gmat team

Is it by any chance possible that all the powers are even and hence the wavy line bounces back for all the points?

naveens222 · Answer

If all the powers are zero then the equation never becomes less than zero.
e.x: (x-3)^4(x+6)^6(x-1)^2<0
here the minimum value of the inequality is 0, for no value of x the equation becomes less than zero.
no solution is the answer in such case.

kunalbean · Answer

Hello! Following this method, eqns will be (x-3)^3 (x+1) ^5 (x+3) < 0 => x < 3 or x<-3 or x<-1 (x+1) = 0 So how did you decipher that x >-1 ?

divyanshag95 · Answer

how to decide the direction of the graph? whether it would start from the negative or positive?

aggarwalneha91 · Answer

How to decide the direction of graph? 
Can't we have the ranges as -3<x<-1 or x>3 ?

agar123 · Answer

Hi Payal,
Can you please help me? According to egmat tutorial whenever power is even the wavy line bounce back and it will not cross the zero point, but here the power of (x-3) is even, so if we start from the top right corner how it can cross the zero point and came into the -ve region?

Regards,

Thelegend2631 · Answer

Wavy line approach is pretty cool.

My question is..

After i arrange the intersections points aka Roots, 
How do i start the wave.

Is it always starting from below X axis - upward curve?

I tried few sources but everyone has a different approach and this concept is not yet clear.
 Bunuel   VeritasKarishma

Posted from my mobile device

Thelegend2631 · Answer

That's because ( X²- 9 ) =( x-3)(x+3) hence the power of

(X-3)³

Posted from my mobile device

KarishmaB · Answer

Hemanthdasu13 Follow one style else you will mix them up. Here is the step by step process I follow: https://youtu.be/PWsUOe77__E

KarishmaB · Answer

Responding to a pm: First check out how the inequalities are represented on the number line here: https://www.youtube.com/watch?v=VnEVS8kmWa8 Next, we can use the wavy line method to solve this. (x - 3)^2 (x + 1)^5 (x^2 - 9) < 0 (x - 3)^2 (x + 1)^5 (x + 3)(x - 3) < 0 Factors with even exponents are taken as constants. Factors with positive odd exponents are taken as if they have an exponent of 1. Hence transition points are -3, -1 and 3 only. Since we are looking for negative values of the expression, we see that x < -3 or -1 < x < 3 Answer (D)

luisdicampo · Answer

Deconstructing the Question

Factor first:
 x^2-9=(x-3)(x+3)

So the inequality becomes:
 (x-3)^3(x+1)^5(x+3)<0

Critical points:
 -3,\,-1,\,3

All exponents are odd, so the sign changes at each critical point.

Step-by-step

For  x<-3 , the product is negative.

For  -3<x<-1 , the product is positive.

For  -1<x<3 , the product is negative.

For  x>3 , the product is positive.

Since the inequality is strict,  x=-3,-1,3  are excluded.

Solution:
 x<-3	ext{ or }-1<x<3

Answer: D

Find the range of values of x that satisfy the inequality (x - 3)^2

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GMAT Problem Solving (PS) Questions

Find the range of values of x that satisfy the inequality (x - 3)^2

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards