Find the range of values of x that satisfy the inequality (x+1)(x-2)

Question

Wavy Line Method Application - Exercise Question #4

Find the range of values of x that satisfy the inequality \(\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0\)

A. x < -3
B. -1 < x < 2
C. x < -3 or -1 < x < 2 or x > 5
D. -1 < x < 2 or x > 5
E. x < -3 or -1 < x < 2

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Official Answer

C

abhimahna · Accepted Answer

Equation is  \frac{(x+1)(x-2)}{(x-5)(x+3)} > 0

We have the values of x as -3,-1,2 and 5.

Since x cannot be equal to 5 or -3 we need to exclude those values(they are at the denominator and we cannot have 0 at the denominator)

So, substituting these values of the number line, we will the range of x as

(-infinity,-3) U (-1,2) U (5,infinity)

Please correct me if I am missing anything.

EgmatQuantExpert · Answer

Solution  :

Hey Everyone,

Please find below the solution of the given problem,

Plotting the zero points and drawing the wavy line:

https://e-gmat.com/blogs/wp-content/uploads/2016/11/S4.jpg

Required Range  :

x < -3 or -1 < x < 2 or x > 5

Correct Answer  : Option C

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

gmat800live · Answer

VeritasKarishma  apologies for bothering you. I read your inequality series and it's been enlightening.

I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".

Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.

Would that work? Thank you.

KarishmaB · Answer

You don't need to make it so complicated though what you suggest works too. The reason why factors in the denominator work is simply this:

When is abcd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc

When is ab/cd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc

Aren't the two cases the same? (Except that c and d cannot be 0 in the second case) Anyway, since the expression needs to be positive, a, b, c and d cannot be 0 in first case either.

Does it matter whether c and d are in numerator or denominator? No. 
The signs of a, b, c and d decide the sign of the expression irrespective of whether they are in numerator or denominator. The case is the same here.

uttam94317 · Answer

We have the values of x as -3,-1,2 and 5.

So, substituting these values of the number line, we will the range of x as

kunalbean · Answer

Hello! Can you please explain how did you get x < -3? The eqn = x -5 > 0 i.e. x >5. So how come for x+3 > 0 it becomes x<-3? Shouldnt it be x>-3?

Stanindaw · Answer

Where can i get the wavy ine method?

KarishmaB · Answer

Check this post: https://gmatclub.com/forum/inequalities ... 91482.html

bumpbot · Answer

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Find the range of values of x that satisfy the inequality (x+1)(x-2)

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Find the range of values of x that satisfy the inequality (x+1)(x-2)

Prep Toolkit

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