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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 \(|x^2-x+1|=2x-1\), 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: 1053
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 \(x > \frac{1}{2}\).
Also note the left side must be positive no matter what real x you plug in. We can see that from the determinant \(b^2 - 4ac = 1 - 4 = -3 < 0\) or \(x^2 - x + 1 = (x - 1)^2 + x > 0\) since x > 0.
Therefore \(x^2 - x + 1\) is always positive and we yield \(x^2 - x + 1 =2x - 1\), which is \(x^2 - 3x + 2 = 0\)
\(x = 1\) 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 \(b^2 - 4ac = 1 - 4 = -3 < 0\) or \(x^2 - x + 1 = (x - 1)^2 + x > 0\) since x > 0.
Therefore \(x^2 - x + 1\) is always positive and we yield \(x^2 - x + 1 =2x - 1\), which is \(x^2 - 3x + 2 = 0\)
\(x = 1\) or \(x = 2\), we may plug in to double-check but they both satisfy x > 0.5 so we accept both.
Ans: D
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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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 \(|x^2-x+1|=2x-1\), 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: \(x^2-x+1=2x-1\) and \(x^2-x+1=-(2x-1)\)
Let's solve each equation separately
First equation: \(x^2-x+1=2x-1\)
Set equal to zero: \(x^2-3x+2=0\)
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: \(|2^2-2+1|=2(2)-1\). WORKS!
Determine whether \(x = 1\) is an extraneous root by plugging it into the original equation to get: \(|1^2-1+1|=2(1)-1\). WORKS!
Second equation: \(x^2-x+1=-(2x-1)\)
Simplify: \(x^2-x+1=-2x+1\)
Set equal to zero: \(x^2+x=0\)
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: \(|0^2-0+1|=2(0)-1\). 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
