Find the range of values of x such that ((x-3)(x-4))/((x+1)(x+2))0

Question

Solving inequalities- Number Line Method - Practice Question #3

Find the range of values of x such that  \frac{(x-3)(x-4)}{(x+1)(x+2)}>0

A.  x > 4
 B.  (-1 ,4)
 C.  (-2, -1) U (3,4)
 D.  (-1,3) U (4, +inf)
 E.  (-inf, -2) U (-1, 3) U (4, +inf)

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EgmatQuantExpert · Answer

Solution Given: An inequality, \frac{(x-3)(x-4)}{(x+1)(x+2)}>0 To find: The range of x, that satisfies the above inequality Approach and Working: The first step is to multiply the inequality by (x+1)^2 and (x+2)^2 on both sides Thus, we get, \frac{(x+1)^2 (x+2)^2 (x-3)(x-4))}{((x+1)(x+2)}>0 Which implies, (x + 1) * (x + 2) * (x - 3) * (x - 4) > 0, and x ≠ {-1, -2}, since the denominator cannot be zero. The zero points of this inequality are, x = -2, x = -1, x = 3, x = 4 Plot these points on the number line. Since, 4 is the greatest among all, the inequality will be positive, for all the points to the right of 4, on the number line And, it is negative, in the region, between 3 and 4 It is again positive in the region, between -1 and 3 And, it is again negative in the region, between -2 and -1 And, it is positive for all the values less than -2 https://e-gmat.com/blogs/wp-content/uploads/2018/08/PQ3-e1535358703538.png Therefore, the range of x is x > 4 and -1 < x < 3 and x < -2 Hence, the correct answer is option E. Answer: E https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg

PKN · Answer

\frac{((x-3)(x-4))}{((x+1)(x+2))>0} using wavy curve method:- x<-2 or -14. So, ranges of x in interval form: (-inf, -2) U (-1, -3) U (4, +inf) Ans. (E)

soleimanian · Answer

E. (-inf, -2) U (-1,  -3 ) U (4, +inf)

-3 must be changed to 3

PKN · Answer

Hi  soleimanian ,
You have pointed out correctly.

Hi  EgmatQuantExpert ,
Hope you will do the needful.

EgmatQuantExpert · Answer

Hey  soleimanian ,
Apology for the typo.

We have updated the option.
Regards,
Ashutosh
e-GMAT

stne · Answer

Dear Moderator, 
Can we format the question properly, seems as though only the denominator, is > 0, Thank you.

Bunuel · Answer

______________
Done. Thank you.

voisinduzero · Answer

Can someone please explain how the correct answer can include -2 and -1 given that they would make the denominator = 0? Shouldn't these values always be discarded when working with fractions?

bv8562 · Answer

Hi  PKN ,

Could you please tell me how the above highlighted range for 'x' is valid here? Because if I take x=-2 which is within the given range then the denominator becomes zero and thus the whole expression becomes undefined.

PKN · Answer

Please note that

-1<x<3

x=-2 doesn't satisfy the above inequality. Hence x=-2 can't be checked.

Integer values within -1 and 3 are 0,1, and 2.

Hope it helps.

bv8562 · Answer

PKN , Oh! Yes, you are right "-2" doesn't lie within the given range. I didn't read it carefully. May be I interchanged the 3 and 1 in my mind keeping the signs intact. Silly me!!

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Find the range of values of x such that ((x-3)(x-4))/((x+1)(x+2))>0

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