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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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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

Answer:

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

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

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


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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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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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/
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Re: The sum of all solutions for x in the equation x^2 8x + 21 = |x 4| [#permalink]
1
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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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Re: The sum of all solutions for x in the equation x^2 8x + 21 = |x 4| [#permalink]
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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]
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]
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 =
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Re: The sum of all solutions for x in the equation x^2 8x + 21 = |x 4| [#permalink]
Damn!
I made a silly mistake!
I added 4 twice! :D
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Re: The sum of all solutions for x in the equation x^2 8x + 21 = |x 4| [#permalink]
Bunuel egmat VeritasKarishma GMATNinja


Quote:
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


I have a conceptual question here

If it says
\((x-4)^2 = |x-4|\)

Doesnt this mean that \(x-4 >= 0\) all the time? (as \((x-4)^2\) is always positive)
Why are we considering\( x-4\) can be negative?

Can anybody please explain if my understanding is wrong.
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Re: The sum of all solutions for x in the equation x^2 8x + 21 = |x 4| [#permalink]
Expert Reply
rsrighosh wrote:
Bunuel egmat VeritasKarishma GMATNinja


Quote:
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


I have a conceptual question here

If it says
\((x-4)^2 = |x-4|\)

Doesnt this mean that \(x-4 >= 0\) all the time? (as \((x-4)^2\) is always positive)
Why are we considering\( x-4\) can be negative?

Can anybody please explain if my understanding is wrong.


Because y^2 is positive does not mean y needs to be positive. Because |y| is positive, doesn't mean y needs to be positive too.
If y is -5, y^2 = 25 (positive) and |y| = 5 (positive) but y is still negative.

In above, (x - 4)^2 and |x - 4| are positive but x - 4 may not be positive. It could be positive or negative.
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