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
inequalities with absolute values
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- sachin_yadav
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You're correct about the two inequality part.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)
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 Agarwal
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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
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 Agarwal
Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
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- Anju@Gurome
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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.)sachin_yadav wrote:(2). |x² - 2x| < x
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
- (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
Anju Agarwal
Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
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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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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