Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

29 Oct 2009, 03:11

9

This post received KUDOS

Expert's post

7

This post was BOOKMARKED

tejal777 wrote:

What is the sum of all roots of the equation \(|x + 4|^2 - 10|x + 4| = 24?\)

Please help me find my mistake: Let \(x+4=y\) Now we get two cases, Case1: \(y^2-10y-24=0\) Solving we get -2,12

Case2: \(-y^2+10y-24=0\) where we get 6,4

This is a good question.

Let me show you how I've solved, maybe it'll help:

APPROACH #1: We have |x + 4|^2 - 10|x + 4| = 24

|x + 4| flip sign at x=-4, so we should check two ranges:

1. x<=-4 (x+4)^2 + 10x+40=24 ((x+4)^2 as it's square will be the same in both ranges)

x^2+8x+16+10x+16=0 --> x^2+18x+32=0. Solving for x: x=-16 or x=-2. x=-2 won't work as x<=-4 (see the defined range), hence we have only one solution for this range x=-16.

2. x>-4 (x+4)^2 - 10x-40=24 --> x^2-2x-48=0. Solving for x: x=-6 or x=8. x=-6 wont work as x>-4, hence we have only one root for this range x=8.

-16+8=-8.

APPROACH #2:

|x + 4|^2 - 10|x + 4| = 24

Solve for \(|x+4 |\) --> \(|x+4 |=12\) OR \(|x+4 |=-2\), BUT as absolute value never negative thus -2 is out. Solving \(|x+4 |=12\) --> \(x_1=8\) or \(x_2=-16\) --> \(x_1+x_2=8-16=-8\).

Answer: the sum of all roots of the equation is -8. _________________

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

29 Oct 2009, 06:37

3

This post received KUDOS

Correct me:

I solved from where the author of the problem left it. that is: y = -2 or 12 Hence, considereding + values of |x+4|, i.e. x+4 = -2 or 12, which gives us x = -6 or 8

Considering - values of |x+4|, i.e. -x-4 = -6 or 4, which gives us x = -2 or 8.

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

29 Oct 2009, 13:32

20

This post received KUDOS

2

This post was BOOKMARKED

I solve it by replacing |x+4| as k

so k^2-10k-24=0 (k-12)(k+2) = 0 k = 12 or -2 k can not be -2 because it is an absolute value so k = 12 = |x+4| then x+4 = 12, x = 8 x+4 = -12, x = -16 sum is -8

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

31 Oct 2009, 21:59

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Economist wrote:

IMO -16.

Take y = |x+4 | and solve for y, then solve for |x+4| , we get x=-16,8,-2,-6, sum = -16. OA?

Economist the problem is that -2 and -6 doesn't satisfy the equation. Thus only two values of x are left -16 and 8: -16+8=-8.

Consider this: |x + 4|^2 - 10|x + 4| = 24 Solve for \(|x+4 |\) --> \(|x+4 |=12\) OR \(|x+4 |=-2\), BUT as absolute value never negative thus -2 is out. Solving \(|x+4 |=12\) --> \(x_1=8\) or \(x_2=-16\) --> \(x_1+x_2=8-16=-8\).

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

07 Oct 2010, 12:19

mxgms wrote:

jzd wrote:

I solve it by replacing |x+4| as k

so k^2-10k-24=0 (k-12)(k+2) = 0 k = 12 or -2 k can not be -2 because it is an absolute value so k = 12 = |x+4| then x+4 = 12, x = 8 x+4 = -12, x = -16 sum is -8

I liked that approach, is this always true?

thanks.

Not sure what part you are questioning about (1) You can always do a variable switch in an equation (|x+4|=y) (2) |Any expression| is always greater than or equal to 0 _________________

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

08 Oct 2010, 01:56

-8 for me. Once you solve the QE you get |x+4| = 6 or -4. -4 is not possible so take the case |x+4| = 6 which means x = -10 or 2. So the sum is -8. _________________

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

08 Dec 2010, 09:38

shrouded1 wrote:

mxgms wrote:

jzd wrote:

I solve it by replacing |x+4| as k

so k^2-10k-24=0 (k-12)(k+2) = 0 k = 12 or -2 k can not be -2 because it is an absolute value so k = 12 = |x+4| then x+4 = 12, x = 8 x+4 = -12, x = -16 sum is -8

I liked that approach, is this always true?

thanks.

Not sure what part you are questioning about (1) You can always do a variable switch in an equation (|x+4|=y) (2) |Any expression| is always greater than or equal to 0

Shrouded: can we do this question by the approach you have mentioned in the walker post. .i.e |x-a|<b => a-b<x<a+b or this approach is for specific questions. Thanks _________________

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

18 Apr 2012, 04:12

Expert's post

shankar245 wrote:

Hi Buneuel,

Please help me with the basic understanding of the mod probs

when we have

|x+4|= |x-5|

We can two solutions

x+4= x-5 and x+4=-x+5

but in this problem why do we check for ranges. I mean what s the step by step approach to attack a modulus question? a few examples would be grateful

|x+4| can expand in two ways: if x<=-4 then |x+4|=-(x+4) and if x>-4 then |x+4|=x+4. So, we expand |x+4| for |x + 4|^2 - 10|x + 4| = 24 according to this and then solve for x.

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

08 Sep 2012, 01:10

2

This post received KUDOS

tejal777 wrote:

What is the sum of all roots of the equation \(|x + 4|^2 - 10|x + 4| = 24?\)

Please help me find my mistake: Let \(x+4=y\) Now we get two cases, Case1: \(y^2-10y-24=0\) Solving we get -2,12

Case2: \(-y^2+10y-24=0\) where we get 6,4

Let's try this with number line. |x+4| = y ==> y^2-10y-24=0 ==> y = 12 or y = -2 Substitute the value of y we have |x+4|=12 or |x+4|= -2 Hmm.. can mod be a negative number? NO ==> Eliminate |x+4|= -2

Now we are left only with |x+4|=12 Lets draw a number line .................................|x+4|................................. <------------------------------------------------------------------------> -16..............................(-4)................................8

Thus, two possible roots are -16 and +8 Sum of roots => -16+8=-8 _________________

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

08 Sep 2012, 03:20

tejal777 wrote:

What is the sum of all roots of the equation \(|x + 4|^2 - 10|x + 4| = 24?\)

Please help me find my mistake: Let \(x+4=y\) Now we get two cases, Case1: \(y^2-10y-24=0\) Solving we get -2,12

Case2: \(-y^2+10y-24=0\) where we get 6,4

Case 1: You mean \(y\geq{0}\), right? Because \(|x+4|=y\) only if \(y\) is non-negative. Only \(y=12\) is acceptable. From \(|x+4|=12\) we obtain \(x=8\) and \(x=-16.\)

Case 2: Now \(y<0,\) so \(|x+4|=-y\). But \(|x+4|^2=(-y)^2\) is still \(y^2\), doesn't matter that \(y\) is negative! Your equation should be \(y^2+10y-24=0,\) solutions \(2, -12\). Now only \(-12\) is acceptable (\(y\) must be negative), and we obtain the same solutions as in Case 1.

It would have been better to denote \(|x+4|=y\geq{0}\) (see other posts above). Then \(|x+4|^2=y^2\), and for the quadratic equation \(y^2+10y-24=0\) you choose only the non-negative root, then find \(x\)... _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

25 Sep 2012, 10:42

tejal777 wrote:

What is the sum of all roots of the equation \(|x + 4|^2 - 10|x + 4| = 24?\)

24 ends with 4 and \(10|x+4|\) ends with 0. So \(|x+4|^2\) should end with 4. Options below 0 are out because of the absolute value. Let's take the squares ending with 4: \(2^2; 8^2; 12^2; 18^2\) etc...

We find \(12^2 - 10*12 = 24\).

From here \(x_1=8\) and \(x_2=-16\), and the sum: -8.

Re: What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]

Show Tags

03 Jul 2013, 07:43

tejal777 wrote:

What is the sum of all roots of the equation \(|x + 4|^2 - 10|x + 4| = 24?\)

Please help me find my mistake: Let \(x+4=y\) Now we get two cases, Case1: \(y^2-10y-24=0\) Solving we get -2,12

Case2: \(-y^2+10y-24=0\) where we get 6,4

dude here the key remember Bodmas childhood rule

now keep lx+3 l as a k and re write equation we get k = 12 or k= -2 and then now substitute the mod value and remember mod can be negative or positive, as we dont know x and we are finding all possible values we get 8 and -16 once and also we get -2 and -6 i guess so now add them all

wish u a very good luck and make a wish for me too

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I may have spoken to over 50+ Said applicants over the course of my year, through various channels. I’ve been assigned as mentor to two incoming students. A...