If |x + 5|

Question

If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ? A. x >= 10 B. -5 <= x<= 0 or x >= 10 C. x <= -5 or x >= 10 D. x <= 0 or x >= 10 E. none.

Bunuel · Accepted Answer

APPROACH #1 |x + 5| \leq |2x - 5| ; Square both sides: x^2 + 10x + 25 \leq 4x^2 - 20x + 25 Re-arrange and simplify: x^2 - 10x \geq 0 Factor x: x(x - 10) \geq 0 ; x \leq 0 or x \geq 10 Answer: D. APPROACH #2 Transition points of |x + 5| \leq |2x - 5| are -5 and 5/2. Consider three ranges: When x \leq -5 , then x + 5 \leq 0 and 2x - 5 \leq 0 , so |x + 5| \leq |2x - 5| becomes -(x + 5) \leq -(2x -5) . This gives x \leq 10 . Since we consider the case when x \leq -5 then the actual solution for this range is x \leq -5 (the overlap of x \leq -5 and x \leq 10 is x \leq -5 ); When -5 < x < \frac{5}{2} , then x + 5 > 0 and 2x - 5 < 0 , so |x + 5| \leq |2x - 5| becomes x + 5 \leq -(2x -5) . This gives x \leq 0 . Since we consider the case when -5 < x < \frac{5}{2} then the actual solution for this range is -5 < x \leq 0 (the overlap of x \leq 0 and -5 < x < \frac{5}{2} is -5 < x \leq 0 ); When x \geq \frac{5}{2} , then x + 5 > 0 and 2x - 5 \geq 0 , so |x + 5| \leq |2x - 5| becomes x + 5 \leq 2x -5 . This gives x \leq 10 . Since we consider the case when x \leq -5 then the actual solution for this range is x \geq 10 (the overlap of x \geq 10 and x \geq \frac{5}{2} is x \geq 10 ). The three ranges x \leq -5 , -5 < x \leq 0 and x \geq 10 give the final range of x as x \leq 0 or x \geq 10 . Answer: D. Hope it's clear.

DreamerHBS · Answer

Use options. It would hardly take a minute.

Nikhil30 · Answer

hey , i did solved the question , i am getting only x>=10 , not getting x<=0 , thats why i have posted. please come forward with your answer , rather just asking me to solve with the help pf option , if i could have solved the question at first , i wouldn't have posted here .

DreamerHBS · Answer

|x+5|<=|2x-5| case 1) both sides + x+5 <=2x-5 x-2x+5+5<=0 -x+10<=0---> x>=10 case 2) both sides negative -x-5<= -2x+5 -x+2x-5-5<=0 x-10<=0 x<=10 case 3) LHS positive and RHS negative x+5<= -2x+5 x+2x+5-5<=0 3x<=0---> x<=0

Nikhil30 · Answer

hey , can we solve this question by squaring both side ,
as |a| <= |B| , can also be write as a^2<=b^2

and than proceed , by that i am getting only x>=10. can you please tell me whats wrong with that method.

Kinshook · Answer

Asked: If |x + 5| <= |2x - 5|, which of the following specifies all the value of x ?

|x + 5| <= |2x - 5|

Case 1: x<=-5
|x + 5| <= |2x - 5|
-x-5 < = 5-2x
x <= 10
x<=-5<=10
The inequality is valid for all x such that x<=-5

Case 2: -5<x<=2.5
|x + 5| <= |2x - 5|
x+5 <= 5-2x
3x <=0
The inequality is valid for all x such that -5<x<=0

Case 3: x>2.5
|x + 5| <= |2x - 5|
x+5 <= 2x-5
x >=10
The inequality is valid for all x such that x >=10

Combing the 3 results
x<=0 or x>=10

IMO D

Nikhil30 · Answer

hey i got your point thanks for your reply,
i am again getting confused here , 
x(x-10) >=0,
it means , x>=0 and x>=10, 
would you please explain how you have written x<=0. (in approach 1)

Bunuel · Answer

The product x(x - 10) will be non-negative ( \geq 0 ) in the following two cases: both x and x - 10 are non-negative ( \geq 0 ). Here notice that x - 10 is always less than x, so if x - 10 is positive, x will also be positive. Thus, the product will be non-negative if x -10 >= 0, or if x >= 10. both x and x - 10 are non-positive ( \leq 0 ). Here notice that x is always more than x - 10, so if x is negative, x - 10 will also be negative. Thus, the product will be non-positive if x <= 0. Else you can use the method explained here: https://gmatclub.com/forum/solving-quad ... 70528.html Also, you might find the following links useful: 9. Inequalities Theory Inequalities Made Easy! Inequalities: Tips and hints Arithmetic with Inequalities Wavy Line Method Application - Complex Algebraic Inequalities Solving Quadratic Inequalities: Graphic Approach Inequalities - Quadratic Inequalities Graphic approach to problems with inequalities Inequalities trick INEQUATIONS(Inequalities) Part 1 INEQUATIONS(Inequalities) Part 2 Questions Inequality and absolute value questions from my collection DS Questions PS Questions 10. Absolute Value Theory Math: Absolute value (Modulus) Absolute Value: Tips and hints Properties of Absolute Values on the GMAT Absolute modulus : A better understanding GMAT Question Patterns: Abolute Value The Reason Behind Absolute Value Questions on the GMAT Questions Inequality and absolute value questions from my collection The E-GMAT Question Series on ABSOLUTE VALUE DS Questions PS Questions For more check Ultimate GMAT Quantitative Megathread Hope it helps.

somyamehta777 · Answer

APPROACH #1

|x + 5| \leq |2x - 5| ;

Square both sides:  x^2 + 10x + 25 \leq 4x^2 - 20x + 25

Re-arrange and simplify:   x^2 - 10x \geq 0

Factor x:   x(x - 10) \geq 0 ;

x \leq 0  or  x \geq 10

Answer: D.

Hey  Bunuel , in approach 1 when you rearrange wouldn't it be   3x^2 - 10x \geq 0  and wouldn't answer choices change a bit because of it ?

Bunuel · Answer

Here is how it goes:

x^2 + 10x + 25 \leq 4x^2 - 20x + 25 ;

0 \leq 3x^2 - 30x

0 \leq x^2 - 10x

So, you should get  0 \leq x^2 - 10x  (  x^2 - 10x \geq 0 ) not   3x^2 - 10x \geq 0

Ujjawalm8shra · Answer

My query is how is the first result of x=<10 included here?

Posted from my mobile device

bumpbot · Answer

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If |x + 5| <= |2x - 5|, which of the following specifies all the value

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