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inequalities with absolute values

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inequalities with absolute values

by sachin_yadav » Wed Apr 10, 2013 12:39 am
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
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by Anju@Gurome » Wed Apr 10, 2013 12:58 am
sachin_yadav wrote:|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.
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by Anju@Gurome » Wed Apr 10, 2013 1:15 am
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,
Image
(a) --> Blue
(b) --> Green

Now their solution regions are as follows,
Image
Hence, the final solution for x is where the green and blue regions overlap, i.e. -1 < x < 2 or 3 < x < 6
Anju Agarwal
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by Anju@Gurome » Wed Apr 10, 2013 1:30 am
sachin_yadav wrote:(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
Anju Agarwal
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Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by GMATGuruNY » Wed Apr 10, 2013 7:08 am
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
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