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# The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|

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Math Expert
Joined: 02 Sep 2009
Posts: 46284
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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08 May 2015, 06:01
9
25
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Difficulty:

45% (medium)

Question Stats:

70% (01:54) correct 30% (02:42) wrong based on 940 sessions

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The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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08 May 2015, 09:18
2
3
Two cases: modulus function is positive or modulus function is negative

Positive case:

x^2-8x+21 = (x-4)+5
=> x^2-9x+20 = 0
=> x = 4 or 5

Negative case:

x^2-8x+21 = -(x-4)+5
=> x^2-7x+12 = 0
=> x = 3 or 4

=>Unique solutions: 3, 4, 5
=>Sum = 12?
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Joined: 10 Jul 2014
Posts: 14
Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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08 May 2015, 09:31
2
1
|x-4|. |x-4| - |x-4|=0
|x-4|=0 then x=4
|x-4|=1
x=3,5
sum= 4+3+5=12
ans c
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Joined: 06 Apr 2015
Posts: 5
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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Updated on: 10 May 2015, 02:29
1
x^2 - 8x + 16 = |x-4|

RHS can be -ve or +ve

x^2 - 9x + 20 = 0
x^2 - 7x + 12 = 0

x= 5,4,3

We test all 3 values in original equation, all ok.

Thus, Sum = 5 + 4 +3 = 12

Ans (D)

Originally posted by rajeev90 on 10 May 2015, 02:16.
Last edited by rajeev90 on 10 May 2015, 02:29, edited 1 time in total.
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The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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10 May 2015, 02:25
2
x^2 - 8x + 21 = |x-4| +5

I: Positive Case
x^2 - 8x + 21 = x - 4 + 5
x^2 - 9x + 20 = 0
(x - 4) (x - 5)

x1 = 4
x2 = 5

II: Negative Case

x^2 - 8x + 21 = - |x-4| +5
x^2 - 8x + 21 = -x + 4 + 5
x^2 - 7x + 12 = 0
(x - 3) (x - 4)
x3 = 3
x4 = 4 (cancel out as 3 is already listed as x1)

3 + 4 + 5 = 12

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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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10 May 2015, 02:47
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

Now, we can have 2 situations here
Situation 1:x^2-8x+21=x-4+5
x^2-8x+21-x-5+4=x^2-9x+20
x^2-5x-4x+20
x=5 or x=4
Testing both values in the equation, both satisfy

Situation 2:x^2-8x+21=-x+4+5
x^2-7x+12=0
x^-3x-4x+12=0
x=3 or x=4
Testing both values in the equation, both satisfy
Therefor sum of all solutions=3+4+5=12
Math Expert
Joined: 02 Sep 2009
Posts: 46284
Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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11 May 2015, 07:05
11
3
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

a. Unknowns: x
Given: x2 – 8x + 21 = |x – 4| + 5.
Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation.
Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.
Attachment:

2015-05-11_1802.png [ 55.23 KiB | Viewed 12722 times ]

Sum of the different solutions: 5 + 4 + 3 = 12.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

x^2 – 8x + 21 = |x – 4| + 5
x^2 – 8x + 16 = |x – 4|
(x – 4)(x – 4) = |x – 4|
(x – 4)2 = |x – 4|

“Something squared equals its absolute value. 1 squared equals 1, 0 squared equals 0…hmm, –1 squared equals 1…”

x – 4 = –1, 0, or 1
x = 3, 4, or 5

Sum of the different solutions: 5 + 4 + 3 = 12.

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need
not always be the case.
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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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10 Oct 2016, 08:19
why do we take x-4>=0 and x-4<=0 .....and not x-4>0 and x-4<0....thanks in advance.

Bunuel wrote:
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

a. Unknowns: x
Given: x2 – 8x + 21 = |x – 4| + 5.
Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation.
Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.
Attachment:
2015-05-11_1802.png

Sum of the different solutions: 5 + 4 + 3 = 12.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

x^2 – 8x + 21 = |x – 4| + 5
x^2 – 8x + 16 = |x – 4|
(x – 4)(x – 4) = |x – 4|
(x – 4)2 = |x – 4|

“Something squared equals its absolute value. 1 squared equals 1, 0 squared equals 0…hmm, –1 squared equals 1…”

x – 4 = –1, 0, or 1
x = 3, 4, or 5

Sum of the different solutions: 5 + 4 + 3 = 12.

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need
not always be the case.
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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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05 Jun 2017, 22:04
Bunuel wrote:
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

a. Unknowns: x
Given: x2 – 8x + 21 = |x – 4| + 5.
Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation.
Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.
Attachment:
2015-05-11_1802.png

Sum of the different solutions: 5 + 4 + 3 = 12.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

x^2 – 8x + 21 = |x – 4| + 5
x^2 – 8x + 16 = |x – 4|
(x – 4)(x – 4) = |x – 4|
(x – 4)2 = |x – 4|

“Something squared equals its absolute value. 1 squared equals 1, 0 squared equals 0…hmm, –1 squared equals 1…”

x – 4 = –1, 0, or 1
x = 3, 4, or 5

Sum of the different solutions: 5 + 4 + 3 = 12.

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need
not always be the case.

Hi Bunuel,

I have a conceptual doubt wrt such questions.
Like most people above have, I also solved for the x>0 and x<0 cases and got the roots as 4,5 for x>0 and 3,4 for x<0.
As we were considering case x<0 which doesn't get satisfied with the roots 3 and 4, I discarded those. Is that not right to do ?

Shruti.
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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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06 Jun 2017, 00:33
IMO D
there will be two equations taking into account the two possible value of LHS i.e positive and negative .
The solution we will have 3 different values of the solutions
3,4 and 5
Total =12

Sent from my ONE E1003 using GMAT Club Forum mobile app
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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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06 Jun 2017, 03:31
6
1
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

My first thought here is that I want to separate out the absolute value. I see that I have a constant term on the left hand side too. So the first step is

$$x^2 - 8x + 21 - 5 = |x - 4|$$

$$x^2 - 8x + 16 = |x - 4|$$

$$(x - 4)^2 = |x - 4|$$

which is the same as

$$|x - 4|^2 = |x - 4|$$

|x - 4| * (|x - 4| - 1) = 0

|x - 4| = 0
implies x = 4

or

|x - 4| = 1
implies x = 3 or 5

Total sum = 3+ 4+ 5 = 12
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8102 Location: Pune, India Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink] ### Show Tags 07 Jun 2017, 01:25 2 Shruti0805 wrote: Bunuel wrote: Bunuel wrote: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to: (A) –7 (B) 7 (C) 10 (D) 12 (E) 14 Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION: a. Unknowns: x Given: x2 – 8x + 21 = |x – 4| + 5. Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation. Question: What is the sum of all possible solutions for x? b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer. Attachment: 2015-05-11_1802.png Sum of the different solutions: 5 + 4 + 3 = 12. The answer is D. Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right: x^2 – 8x + 21 = |x – 4| + 5 x^2 – 8x + 16 = |x – 4| (x – 4)(x – 4) = |x – 4| (x – 4)2 = |x – 4| “Something squared equals its absolute value. 1 squared equals 1, 0 squared equals 0…hmm, –1 squared equals 1…” x – 4 = –1, 0, or 1 x = 3, 4, or 5 Sum of the different solutions: 5 + 4 + 3 = 12. d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need not always be the case. Hi Bunuel, I have a conceptual doubt wrt such questions. Like most people above have, I also solved for the x>0 and x<0 cases and got the roots as 4,5 for x>0 and 3,4 for x<0. As we were considering case x<0 which doesn't get satisfied with the roots 3 and 4, I discarded those. Is that not right to do ? Please help understand. Thanks in advance! Shruti. Responding to a pm: The ranges for which you are solving are not x > 0 and x < 0. They are x >= 4 and x < 4 because we say that |a| = a when a >= 0 |a| = -a when a < 0 So |x - 4| = (x - 4) when (x - 4) >= 0 i.e. x >= 4 |x - 4| = -(x - 4) when (x - 4) < 0 i.e. x < 4 For more on this, check: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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07 Jun 2017, 11:01
1
Top Contributor
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

Given: x² – 8x + 21 = |x – 4|+ 5
Subtract 5 from both sides: x² – 8x + 16 = |x – 4|
Apply above rule

We get two cases: x² – 8x + 16 = x – 4 and -(x² – 8x + 16) = x – 4

x² – 8x + 16 = x – 4
Add 4 to both sides: x² – 8x + 20 = x
Subtract x from both sides: x² – 9x + 20 = 0
Factor: (x - 5)(x - 4) = 0
So, x = 5 or x = 4

-(x² – 8x + 16) = x – 4
Simplify: -x² + 8x - 16 = x – 4
Add 4 to both sides: -x² + 8x - 12 = x
Subtract x from both sides: -x² + 7x - 12 = 0
Multiply both sides by -1 to get: x² - 7x + 12 = 0
Factor: (x - 3)(x - 4) = 0
So, x = 3 or x = 4

We have three potential solutions: x = 5, x = 4 and x = 3

Now let's test for EXTRANEOUS ROOTS

x = 5
Plug into original equation to get:
5² – 8(5) + 21 = |5 – 4|+ 5
Evaluate: 6 = 6
WORKS! So, x = 5 IS a valid solution

x = 4
Plug into original equation to get:
4² – 8(4) + 21 = |4 – 4|+ 5
Evaluate: 5 = 5
WORKS! So, x = 4 is a valid solution

x = 3
Plug into original equation to get:
3² – 8(3) + 21 = |3 – 4|+ 5
Evaluate: 6 = 6
WORKS! So, x = 3 is a valid solution

SUM of all solutions = 5 + 4 + 3 = 12

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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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14 Jun 2017, 11:06
let me ask a lame question...

x^2-8x+21=|x-4|+5
x^2-8x+21-5=|x-4|
x^2-8x+16=|x-4|

(x-4)(x-4)=|x-4|

Now Scenario 1 where |x-4| is positive

(x-4)(x-4)=(x-4)
(x-4) = (x-4)/(x-4) ------> (x-4)=1-----> x = 5

Scenario 2 where |x-4| is negative
(x-4)(x-4)=-(x-4)---------> (x-4)=-(x-4)/(x-4)----------> (x-4)=-1--------> x = 3

So i have only two solutions x=5 and x=3...Why am I getting only 2 solutions ? while there are 3...

I know I am doing something very stupid with the basics. Can somebody please point out what is it ?
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Posts: 46284
Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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14 Jun 2017, 11:44
ratinarace wrote:
let me ask a lame question...

x^2-8x+21=|x-4|+5
x^2-8x+21-5=|x-4|
x^2-8x+16=|x-4|

(x-4)(x-4)=|x-4|

Now Scenario 1 where |x-4| is positive

(x-4)(x-4)=(x-4)
(x-4) = (x-4)/(x-4) ------> (x-4)=1-----> x = 5

Scenario 2 where |x-4| is negative
(x-4)(x-4)=-(x-4)---------> (x-4)=-(x-4)/(x-4)----------> (x-4)=-1--------> x = 3

So i have only two solutions x=5 and x=3...Why am I getting only 2 solutions ? while there are 3...

I know I am doing something very stupid with the basics. Can somebody please point out what is it ?

You cannot reduce (x-4)(x-4)=(x-4) by x-4 because x-4 can be 0 and we cannot divide by 0. By doing so you loose a root, namely x = 4.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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15 Jun 2017, 00:02
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

x^2 – 8x + 21 = |x – 4|+ 5

If x>=4,
x^2 – 8x + 21 = x – 4+ 5
-> x^2 – 8x + 21 = x+1
-> x^2 - 9x + 20 = 0
-> (x-4) (x-5) = 0
-> x =4,5

If x<4 ,
x^2 – 8x + 21 = 4-x +5 = 9-x
x^2 - 7x + 12 = 0
(x-4)(x-3) =0
-> x=3

Total sum = 3+4+5= 12
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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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31 Aug 2017, 19:26
Bunuel wrote:
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:

(A) –7
(B) 7
(C) 10
(D) 12
(E) 14

Kudos for a correct solution.

Two basic ways to write out absolute values for the GMAT

l x -4 l

x - 4 =
- l x -4 l =
Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|   [#permalink] 31 Aug 2017, 19:26
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