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What is the sum of all solutions to the equation |x² – 4x + 4| =

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What is the sum of all solutions to the equation |x² – 4x + 4| =  [#permalink]

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07 Nov 2016, 05:57
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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

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

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07 Nov 2016, 09:44
8
1
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.
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| =  [#permalink]

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07 Nov 2016, 11:37
1
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.

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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07 Nov 2016, 13:20
5
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.

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

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

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07 Nov 2016, 16:25
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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.

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What is the sum of all solutions to the equation |x² – 4x + 4| =  [#permalink]

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08 Nov 2016, 02:06
1
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.

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.

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

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08 Nov 2016, 06:45
Top Contributor
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.

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.

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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What is the sum of all solutions to the equation |x² – 4x + 4| =  [#permalink]

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Updated on: 08 Nov 2016, 09:18
2
No need to solve for both 2 cases positive and negative.

Just have $$x^2 - 4x + 4 = (x-2)^2 ≥ 0$$

So $$|x² – 4x + 4| = x² – 4x + 4$$

Since $$x² – 4x + 4 = x² + 10x – 24$$ we have $$14x - 28 = 0$$, now $$x = 2$$
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Originally posted by broall on 08 Nov 2016, 09:07.
Last edited by broall on 08 Nov 2016, 09:18, edited 1 time in total.
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Re: What is the sum of all solutions to the equation |x² – 4x + 4| =  [#permalink]

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08 Nov 2016, 09:08
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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Posts: 3009
Re: What is the sum of all solutions to the equation |x² – 4x + 4| =  [#permalink]

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08 Nov 2016, 12:20
Top Contributor
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]

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08 Nov 2016, 12:35
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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09 Nov 2016, 07:19
2
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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09 Nov 2016, 08:09
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]

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29 Oct 2017, 11:26
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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30 Oct 2017, 08:20
Top Contributor
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]

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13 Oct 2018, 17:34
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.

RELATED VIDEO

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?
Re: What is the sum of all solutions to the equation |x² – 4x + 4| = &nbs [#permalink] 13 Oct 2018, 17:34
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