inequalities with absolute values

[sachin_yadav]

Hello everyone,

Please help in solving the following questions, and do these kind of questions appear in the GMAT exam.

(1). |x² - 5x| < 6
My approach:- -6 < x² - 5x < 6

There are two inequalities:-
(a). x² - 5x < 6 (b) x² - 5x > -6
x² - 3x - 2x - 6 < 0 x² - 3x - 2x + 6 >0
I am stuck over here x(x - 3) - 2(x - 3)
(x - 3) and (x - 2) > 0
x = 3 and x = 2
x > 3 and x < 2 (not sure if it is correct)


(2). |x² - 2x| < x
(same kind of explanation as given above)

Please advice

Regards
Sachin

[Anju@Gurome]

|x² - 5x| < 6
My approach:- -6 < x² - 5x < 6

There are two inequalities:-
(a) x² - 5x < 6
--> x² - 5x - 6 < 0
--> I am stuck over here

(b) x² - 5x > -6
--> x² - 3x - 2x + 6 > 0
--> x(x - 3) - 2(x - 3) > 0
--> (x - 3) and (x - 2) > 0
--> x = 3 and x = 2
--> x > 3 and x < 2 (not sure if it is correct)

You're correct about the two inequality part.
Let me show how to solve them,
(a) x² - 5x - 6 < 0
--> x² - 6x + x - 6 < 0
--> x(x - 6) + (x - 6) < 0
--> (x + 1)(x - 6) < 0
--> Either {(x + 1) > 0 and (x - 6) < 0} or {(x + 1) < 0 and (x - 6) > 0}
--> Either {x > -1 and x < 6} or {x < -1 and x > 6}
--> As the second set is not possible for any x, our final solution is {x > -1 and x < 6}, i.e. -1 < x < 6

(b) x² - 5x > -6
--> x² - 3x - 2x + 6 > 0
--> x(x - 3) - 2(x - 3) > 0
--> (x - 2)(x - 3) > 0
--> Either {(x - 2) > 0 and (x - 3) > 0} or {(x - 2) < 0 and (x - 3) < 0}
--> Either {x > 2 and x > 3} or {x < 2 and x < 3}
--> Either x > 3 or x < 2

Hence, together our final solution : -1 < x < 6 AND {x < 2 or x > 3} ---> -1 < x < 2 OR 3 < x < 6

Note : It is very much unlikely that you'll see absolute value problems like this in GMAT. Most absolute value problems in GMAT can be solved by the concept of distance on the number line or minimal algebra.

[Anju@Gurome]

This problem can solved easily with wavy curve method too (Read here : https://www.beatthegmat.com/gmat-prep-ea ... tml#602154)

Draw the wavy curves of (a) and (b) as follows,

(a) --> Blue
(b) --> Green

Now their solution regions are as follows,

Hence, the final solution for x is where the green and blue regions overlap, i.e. -1 < x < 2 or 3 < x < 6

[Anju@Gurome]

(2). |x² - 2x| < x

Note that, |x² - 2x| is a positive quantity as absolute value cannot be negative and if x = 0, the inequality will not hold. Hence, x > |x² - 2x| > 0 --> x > 0 (Remember this as I won't consider negative values of x anymore.)

Now, |x² - 2x| = |x(x - 2)|
As x > 0, the value of |x² - 2x| depends upon whether (x - 2) > 0 or not

Case #1 : 0 < x < 2

• (x - 2) < 0 ---> x(x - 2) < 0 ---> |x² - 2x| = -x(x - 2) = -(x² - 2x)
  So, -(x² - 2x) < x
  --> x² - 2x + x > 0
  --> x² - x > 0
  --> x(x - 1) > 0
  --> x > 0 and x > 1
  --> x > 1
  So, final solution, 1 < x < 2

Case #2 : x > 2

• (x - 2) > 0 ---> x(x - 2) > 0 ---> |x² - 2x| = x(x - 2) = (x² - 2x)
  So, (x² - 2x) < x
  --> x² - 2x - x < 0
  --> x² - 3x < 0
  --> x(x - 3) < 0
  --> x > 0 and x < 3
  --> x < 3
  So, final solution, 2 < x < 3

Our final solution : 1 < x < 2 or 2 < x < 3 ---> 1 < x < 3 excluding 2

[GMATGuruNY]

Most problems of this type can be solved efficiently by comparing the answer choices.
Check my posts below:

https://www.beatthegmat.com/inequality-c ... 89803.html
https://www.beatthegmat.com/og-130-t27196.html
https://www.beatthegmat.com/knewton-q-t89317.html
https://www.beatthegmat.com/modulus-ps-t112185.html
https://www.beatthegmat.com/which-of-the ... 13190.html