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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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Ashiquzzaman wrote:
Assuming |x² – 4x + 4| negative:
-(x² – 4x + 4) = x² + 10x – 24
=>-x²+4x-4=x²+10x-24
=>2x²+6x=20
=>x²+3x=10
=>x²+5x-10=0
=>(x+5)(x-2)=0
we have either x=-5 or x=2
Assuming |x² – 4x + 4| positive:
x² – 4x + 4=x² + 10x – 24
=>10x+4x=24+4
=>14x=28
=>x=2
Since both solutions have x=2 in common, the answer will be D.
Although I got it wrong first time.
Awaiting my first Kudos in the forum.


Can u please explain how the answer is 2.
I couldn't understand the highlighted part in your solution.
IMO we have to add all the solutions.
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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Ashiquzzaman wrote:
Assuming |x² – 4x + 4| negative:
-(x² – 4x + 4) = x² + 10x – 24
=>-x²+4x-4=x²+10x-24
=>2x²+6x=20
=>x²+3x=10
=>x²+5x-10=0
=>(x+5)(x-2)=0
we have either x=-5 or x=2
Assuming |x² – 4x + 4| positive:
x² – 4x + 4=x² + 10x – 24
=>10x+4x=24+4
=>14x=28
=>x=2
Since both solutions have x=2 in common, the answer will be D. WRONG
Although I got it wrong first time.
Awaiting my first Kudos in the forum.




your calculations are good; but your answer is right for the wrong reason.

after you find the solutions of the equation assuming that the absolute value is negative ,and then positive, youhave to check whether those solutions are make the value inside the absolute value of the sign you assumed.

Assuming |x² – 4x + 4| positive:
x² – 4x + 4=x² + 10x – 24
=>10x+4x=24+4
=>14x=28
=>x=2

now sobstituting x=2 in |x² – 4x + 4| we find that the expression is positive ==> the solution is valid

Assuming |x² – 4x + 4| negative:
-(x² – 4x + 4) = x² + 10x – 24
=>-x²+4x-4=x²+10x-24
=>2x²+6x=20
=>x²+3x=10
=>x²+5x-10=0
=>(x+5)(x-2)=0
we have either x=-5 or x=2

now sobstituting x=-5 and x=2 in |x² – 4x + 4| we find that the expression is positive in both the cases ==> the solution are NOT valid as we assumed that the espression was negative to find these solutions.

Thus, the only valid solution was the first one x=2

answer D
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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GMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5


When solving equations involving ABSOLUTE VALUE, there are 3 steps:
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

So, we have two equations to solve: x² – 4x + 4 = x² + 10x – 24 and x² – 4x + 4 = -(x² + 10x – 24)

x² – 4x + 4 = x² + 10x – 24
Subtract x² from both sides: –4x + 4 = 10x – 24
Rearrange: 28 = 14x
Solve: x = 2

x² – 4x + 4 = -(x² + 10x – 24)
Simplify right side: x² – 4x + 4 = -x² - 10x + 24
Add x² to both sides: 2x² – 4x + 4 = -10x + 24
Add 10x to both sides: 2x² + 6x + 4 = 24
Subtract 24 from both sides: 2x² + 6x - 20 = 0
Factor: 2(x² + 3x - 10) = 0
Factor again: 2(x - 2)(x + 5) = 0
Solve: x = 2 and x = -5

So, we have two solutions to consider: x = 2 and x = -5
Plug solutions into original equation to check for extraneous roots

x = 2
|2² – 4(2) + 4| = 2² + 10(2) – 24
Evaluate: |0| = 0
This works, so keep this solution

x = -5
|(-5)² – 4(-5) + 4| = (-5)² + 10(-5) – 24
Evaluate: |49| = -49
Doesn't work. So, x = -5 is NOT a solution

Since there's only one valid solution (x = 2), the sum of all solutions is 2.
Answer:

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What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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GMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5

*Kudos for all correct solutions


|x² – 4x + 4| = x² + 10x – 24

Open modulus

x² – 4x + 4 = x² + 10x – 24..........14X-28=0 ......x-2=0 then x=2

x² – 4x + 4 = -(x² + 10x – 24)......x² – 4x + 4 = -x² - 10x + 24.........2x²+6x-20=0.....x²+3x-10=0

(x+5)(x-2)=0

x=2 or x =-5

When checking back in the original equation, we find:

X=2 is valid. it gives zero in both sides.

X=-5 is NOT valid as LHS MUST be positive while put x=-5 gives Negative value. NO need to check the RHS.

Answer: D

Brent,

When I tried to apply critical point, it does not work well. I know the critical point is when x=2 but could not move from there.

Can you help please?
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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Mo2men wrote:
GMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5

*Kudos for all correct solutions


|x² – 4x + 4| = x² + 10x – 24

Open modulus

x² – 4x + 4 = x² + 10x – 24..........14X-28=0 ......x-2=0 then x=2

x² – 4x + 4 = -(x² + 10x – 24)......x² – 4x + 4 = -x² - 10x + 24.........2x²+6x-20=0.....x²+3x-10=0

(x+5)(x-2)=0

x=2 or x =-5

When checking back in the original equation, we find:

X=2 is valid. it gives zero in both sides.

X=-5 is NOT valid as LHS MUST be positive while put x=-5 gives Negative value. NO need to check the RHS.

Answer: D

Brent,

When I tried to apply critical point, it does not work well. I know the critical point is when x=2 but could not move from there.

Can you help please?



In your solution you found that x = 2 is the only solution, so you're done.
Can you elaborate what you mean by "I know the critical point is when x=2 but could not move from there. "?
I'm not sure where you're having troubles.

Cheers,
Brent
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
GMATPrepNow wrote:


In your solution you found that x = 2 is the only solution, so you're done.
Can you elaborate what you mean by "I know the critical point is when x=2 but could not move from there. "?
I'm not sure where you're having troubles.

Cheers,
Brent



Hi Brent,

I will try to explain my steps.

|x² – 4x + 4| = x² + 10x – 24

The critical point is x=2 that makes the modulus equal to zeros.

X>=2 .......... x² – 4x + 4 = x² + 10x – 24.........then x=2

X<2 .............x² – 4x + 4 = x² + 10x – 24......... then x=2 or x=-5. However, when I put x=0 , it does not turn the term to negative, actually if I put x= negative number, it always positive because it is raise to power of 2 as it is (x-2)^2.

so I do not know. should I discard situation X<2???

I hope you can help
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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Mo2men wrote:
GMATPrepNow wrote:


In your solution you found that x = 2 is the only solution, so you're done.
Can you elaborate what you mean by "I know the critical point is when x=2 but could not move from there. "?
I'm not sure where you're having troubles.

Cheers,
Brent



Hi Brent,

I will try to explain my steps.

|x² – 4x + 4| = x² + 10x – 24

The critical point is x=2 that makes the modulus equal to zeros.

X>=2 .......... x² – 4x + 4 = x² + 10x – 24.........then x=2

X<2 .............x² – 4x + 4 = x² + 10x – 24......... then x=2 or x=-5. However, when I put x=0 , it does not turn the term to negative, actually if I put x= negative number, it always positive because it is raise to power of 2 as it is (x-2)^2.

so I do not know. should I discard situation X<2???

I hope you can help


You are solving a different kind of question when you focus solely on |x² – 4x + 4| (and ignore the right side of that equation)
It just happens to turn out that x = 2 is an x-value that makes x² + 10x – 24 = 0, but that's purely coincidental (which is unfortunate, since it may have led you to believe that you were on the right track).
Please review the solutions posted above. They model the steps that need to be taken for this kind of question.

Cheers,
Brent
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
GMATPrepNow wrote:
Mo2men wrote:
GMATPrepNow wrote:


In your solution you found that x = 2 is the only solution, so you're done.
Can you elaborate what you mean by "I know the critical point is when x=2 but could not move from there. "?
I'm not sure where you're having troubles.

Cheers,
Brent



Hi Brent,

I will try to explain my steps.

|x² – 4x + 4| = x² + 10x – 24

The critical point is x=2 that makes the modulus equal to zeros.

X>=2 .......... x² – 4x + 4 = x² + 10x – 24.........then x=2

X<2 .............x² – 4x + 4 = x² + 10x – 24......... then x=2 or x=-5. However, when I put x=0 , it does not turn the term to negative, actually if I put x= negative number, it always positive because it is raise to power of 2 as it is (x-2)^2.

so I do not know. should I discard situation X<2???

I hope you can help


You are solving a different kind of question when you focus solely on |x² – 4x + 4| (and ignore the right side of that equation)
It just happens to turn out that x = 2 is an x-value that makes x² + 10x – 24 = 0, but that's purely coincidental (which is unfortunate, since it may have led you to believe that you were on the right track).
Please review the solutions posted above. They model the steps that need to be taken for this kind of question.

Cheers,
Brent


I apologize but I do not understand what step should I review. How i ignored the right side?
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GMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5

*Kudos for all correct solutions

There is absolutely no need to solve for 2 cases.

The trick here is that the value in the absolute is always positive, since x^2 -4x + 4 = (x-2)^2, which is always a positive value, thus we can simply remove the absolute and solve for x, we get x=2.
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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I found the correct answer using a different method and I would like to know whether it is or not.

from the equation: |x² – 4x + 4| = x² + 10x – 24 => \(|(x-2)^2| = x² + 10x – 24\) => the argument of the modulus is always \(>= 0\) => I can eliminate the absolute value => x² – 4x + 4 = x² + 10x – 24 => solve for x => x= +2 => there are no other roots to this equation thus the sum of all solutions must be equal to 2
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
Sir but here they have asked the sum of all the equations. How can we conclude the ans is 3. The ans should be -5+2=-3.Kindly help me.
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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Raj94* wrote:
Sir but here they have asked the sum of all the equations. How can we conclude the ans is 3. The ans should be -5+2=-3.Kindly help me.


We have two POSSIBLE solutions to consider: x = 2 and x = -5
However, when it comes to absolute value equations, we must plug solutions into original equation to check for extraneous roots
When we do this, we see that x = -5 is NOT a solution

Here's why:
Plug in x = -5
we get: |(-5)² – 4(-5) + 4| = (-5)² + 10(-5) – 24
Simplify: |25 – (-20) + 4| = 25 + (-50) – 24
Evaluate: |49| = -49
As we can see, |49| does NOT equal -49
So, x = -5 is NOT a valid solution.

If we check the other solution (x = 2) we see that this is, indeed, a solution.

Since x = 2 is the ONLY valid solution, the sum is 2 (answer choice D)
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
GMATPrepNow wrote:
GMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5


When solving equations involving ABSOLUTE VALUE, there are 3 steps:
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

So, we have two equations to solve: x² – 4x + 4 = x² + 10x – 24 and x² – 4x + 4 = -(x² + 10x – 24)

x² – 4x + 4 = x² + 10x – 24
Subtract x² from both sides: –4x + 4 = 10x – 24
Rearrange: 28 = 14x
Solve: x = 2

x² – 4x + 4 = -(x² + 10x – 24)
Simplify right side: x² – 4x + 4 = -x² - 10x + 24
Add x² to both sides: 2x² – 4x + 4 = -10x + 24
Add 10x to both sides: 2x² + 6x + 4 = 24
Subtract 24 from both sides: 2x² + 6x - 20 = 0
Factor: 2(x² + 3x - 10) = 0
Factor again: 2(x - 2)(x + 5) = 0
Solve: x = 2 and x = -5

So, we have two solutions to consider: x = 2 and x = -5
Plug solutions into original equation to check for extraneous roots

x = 2
|2² – 4(2) + 4| = 2² + 10(2) – 24
Evaluate: |0| = 0
This works, so keep this solution

x = -5
|(-5)² – 4(-5) + 4| = (-5)² + 10(-5) – 24
Evaluate: |49| = -49
Doesn't work. So, x = -5 is NOT a solution

Since there's only one valid solution (x = 2), the sum of all solutions is 2.
Answer:

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Do we need to test both +ve and -ve scenario in this case? As the expression in LHS is a square, the modulus is meaningless because the LHS expression will always be positive?
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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|(x-2)^2|= (x+12)(x-2)
x=2 will make both sides zero
Therefore option D is the answer.

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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
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LHS is a perfect square. So can't be negative. So we can directly eliminate modulous and equate both equations and only possible value is 2. Kudos if you like/did in the same approach ?

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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
BrentGMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5

*Kudos for all correct solutions

The key to solving the equation is getting all the possibilities and plugging in and checking whether these value satisfy

Therefore case1)
x2-4x+4=x2+10x -24
=>24x=28 x=2

Case2)
2x2+6x-20=0
=>x2+3x-10=0
=>x+5*x+2=0
=>x=-5 and x=2

However when we substitute for x we get 9 in lhs and -49 in rhs which doesn't match
and the only option that matches is x=2
Therefore IMO D
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| = [#permalink]
BrentGMATPrepNow wrote:
What is the sum of all solutions to the equation |x² – 4x + 4| = x² + 10x – 24?

A) -5
B) -3
C) -2
D) 2
E) 5

*Kudos for all correct solutions

The key to solving the equation is getting all the possibilities and plugging in and checking whether these value satisfy

Therefore case1)
x2-4x+4=x2+10x -24
=>24x=28 x=2

Case2)
2x2+6x-20=0
=>x2+3x-10=0
=>x+5*x+2=0
=>x=-5 and x=2

However when we substitute for x we get 9 in lhs and -49 in rhs which doesn't match
and the only option that matches is x=2
Therefore IMO D
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