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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values
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10 Nov 2021, 01:33
10 Nov 2021, 01:33
Expert Reply
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Question Stats:
55% (02:19) correct
44% (01:55) wrong based on 18 sessions
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If |x2−x+1|=2x−1, then what is the sum of all possible values of x ?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values
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10 Nov 2021, 03:28
10 Nov 2021, 03:28
1
Expert Reply
The right side tells us 2x−1>0 and x>12.
Also note the left side must be positive no matter what real x you plug in. We can see that from the determinant b2−4ac=1−4=−3<0 or x2−x+1=(x−1)2+x>0 since x > 0.
Therefore x2−x+1 is always positive and we yield x2−x+1=2x−1, which is x2−3x+2=0
x=1 or x=2, we may plug in to double-check but they both satisfy x > 0.5 so we accept both.
Ans: D
Also note the left side must be positive no matter what real x you plug in. We can see that from the determinant b2−4ac=1−4=−3<0 or x2−x+1=(x−1)2+x>0 since x > 0.
Therefore x2−x+1 is always positive and we yield x2−x+1=2x−1, which is x2−3x+2=0
x=1 or x=2, we may plug in to double-check but they both satisfy x > 0.5 so we accept both.
Ans: D
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Joined: 10 Apr 2015
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Re: If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values
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10 Nov 2021, 05:56
10 Nov 2021, 05:56
2
Carcass wrote:
If |x2−x+1|=2x−1, then what is the sum of all possible values of x ?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots
So, we get: x2−x+1=2x−1 and x2−x+1=−(2x−1)
Let's solve each equation separately
First equation: x2−x+1=2x−1
Set equal to zero: x2−3x+2=0
Factor: (x−2)(x−1)=0
This gives us two possible solutions: x=2 and x=1
Determine whether x=2 is an extraneous root by plugging it into the original equation to get: |22−2+1|=2(2)−1. WORKS!
Determine whether x=1 is an extraneous root by plugging it into the original equation to get: |12−1+1|=2(1)−1. WORKS!
Second equation: x2−x+1=−(2x−1)
Simplify: x2−x+1=−2x+1
Set equal to zero: x2+x=0
Factor: x(x+1)=0
This gives us two possible solutions: x=0 and x=−1
Determine whether x=0 is an extraneous root by plugging it into the original equation to get: |02−0+1|=2(0)−1. DOESN'T WORK
Determine whether x=−1 is an extraneous root by plugging it into the original equation to get: |(−1)2−(−1)+1|=2(−1)−1. DOESN'T WORK
So, the only two solutions that work are x=2 and x=1
The SUM =2+1=3
Answer: D
