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Re: sum of all roots of the equation [#permalink]
29 Oct 2009, 12:32

15

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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: sum of all roots of the equation [#permalink]
29 Oct 2009, 02:11

7

This post received KUDOS

Expert's post

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:

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.

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

Re: sum of all roots of the equation [#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: sum of all roots of the equation [#permalink]
31 Oct 2009, 20:59

2

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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 roots of the equation [#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: sum of all roots of the equation [#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: sum of all roots of the equation [#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: sum of all roots of the equation [#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 roots of the equation [#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 roots of the equation [#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 roots of the equation [#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...

Re: What is the sum of all roots of the equation [#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