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What is the sum of all possible solutions of the equation |x + 4|^2 - [#permalink]
29 Oct 2009, 02:11

9

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Expert's post

4

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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]
29 Oct 2009, 05:37

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]
29 Oct 2009, 12:32

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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]
31 Oct 2009, 20:59

2

This post received KUDOS

Expert's post

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]
07 Oct 2010, 11:19

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]
08 Oct 2010, 00:56

-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]
08 Dec 2010, 08:38

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]
18 Apr 2012, 03:12

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]
08 Sep 2012, 00:10

1

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]
08 Sep 2012, 02:20

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]
25 Sep 2012, 09:42

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]
03 Jul 2013, 06:43

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

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...