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What is the sum of all solutions to the equation |x^2 4x + 4| = x^2
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07 Nov 2016, 05:57
07 Nov 2016, 05:57
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What is the sum of all solutions to the equation |x^2 – 4x + 4| = x^2 + 10x – 24?
A) -5
B) -3
C) -2
D) 2
E) 5
A) -5
B) -3
C) -2
D) 2
E) 5
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Joined: 12 Sep 2015
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Location: Canada
Re: What is the sum of all solutions to the equation |x^2 4x + 4| = x^2
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07 Nov 2016, 16:25
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
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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Re: What is the sum of all solutions to the equation |x^2 4x + 4| = x^2
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09 Nov 2016, 08:09
09 Nov 2016, 08:09
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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
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
