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.

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:

Join the webinar and learn time-management tactics that will guarantee you answer all questions, in all sections, on time. Save your spot today! Nov. 14th at 7 PM PST

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 15684 times ]

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:

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.
_________________

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

Show Tags

10 May 2015, 01:47

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 Answer D

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:

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.

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:

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.

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 ?

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

Show Tags

05 Jun 2017, 23:33

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

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

Show Tags

06 Jun 2017, 02:31

6

2

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
_________________

Karishma Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

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:

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.

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

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

Show Tags

07 Jun 2017, 10:01

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

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

(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. _________________

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

Show Tags

10 Sep 2018, 07:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________