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

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

What is the sum of all possible solutions of the equation |x + 4|^2 - 10|x + 4| = 24?

A. -16
B. -14
C. -12
D. -8
E. -6

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.
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|x + 4|² - 10|x + 4| = 24
Let's simplify matters by using some u-substitution

Let u = |x + 4| and then replace |x + 4| with u to get: u² - 10u = 24
Subtract 24 from both sides to get: u² - 10u - 24 = 0
Factor to get: (u - 12)(u + 2) = 0
So, u = 12 or u = -2

Now let's replace u with |x + 4|.
This means that |x + 4| = 12 or |x + 4| = -2

If |x + 4| = 12, then x = 8 or -16
If |x + 4| = -2, then there are NO SOLUTIONS, since |x + 4| will always be greater than or equal to zero.

So, there are only 2 solutions: x = 8 and x = -16
We're asked to find the SUM of all possible solutions
x = 8 + (-16) = -8

Answer: D

Cheers,
Brent

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.

Sum of all, -6+8-2+8 = 8.

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 \(x_2=-16\) --> \(x_1+x_2=8-16=-8\).

Hope it's clear.
_________________

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
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-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.

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
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|x+4| = y
gives y^2 -10y -24 = 0

y = -2 and 12

|x+4| = 12 gives x = 8 and -16.

sum is -8.

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.

Solution in this post might be easier to understand: what-is-the-sum-of-all-roots-of-the-equation-85988.html#p645659

For basic understanding check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

DS question on absolute values: search.php?search_id=tag&tag_id=37
PS question on absolute values: search.php?search_id=tag&tag_id=58

Hope it helps.
_________________

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

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 \(x\)...
_________________

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.

It also took me time to understand how to get solutions for absolute values. But thanks to GMATClub...

Here is a detailed explanation on how you could solve for the roots https://burnoutorbreathe.blogspot.com/2012/12/how-to-get-solution-for-absolute-values.html

My answer: -8

Have a question. There are two scenarios -
1. x<=-4

2. x>-4

How do we determine whether to include equal sign (=) in first equation or second equation or does it not matter and can be included in any?

It does not matter in which range you include 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


logic and basic = magic in gmat