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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 30 Sep 2021, 06:30
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58% (02:02) correct 42% (01:50) wrong based on 628 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


M37-03

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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 18 Nov 2021, 03:00
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Bunuel
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


M37-03
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Official Solution:

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. \(4\)


The right hand side of the equation (\(2x-1\)), is equal to the absolute value of some number (\(|x^2-x+1|\)), so \(2x-1\) must be non-negative: \(2x-1 \geq 0 \). This gives \(x \geq \frac{1}{2}\).

Now, work on the expression in modulus. Re-write \(x^2-x+1\) by completing the square: \(x^2-x+1=(x-1)^2+x\). Since from above \(x\) is positive, then \(x^2-x+1=(x-1)^2+x=nonnegative + positive=positive\). So, \(|x^2-x+1|=x^2-x+1\).

So, we'd have \(x^2-x+1=2x-1\);

\(x^2-3x+2=0\);

\((x - 2)(x - 1) = 0\);

\(x=2\) or \(x=1\);.

The sum of all possible values of \(x\) is therefore, \(2+1=3\).


Answer: D
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 30 Sep 2021, 06:44
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Bunuel
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
Show more

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
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 30 Sep 2021, 07:11
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Bunuel
If \(|x^2-x+1|=2x-1\), then what is the sum of all possible values of x ?
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You might notice if 2x - 1 equals the absolute value of something, then 2x - 1 must be at least zero, and x must be at least 1/2, so certainly x must be positive. But on the left side, we're taking the absolute value of x^2 - x + 1, which is equal to x^2 - 2x + 1 + x = (x - 1)^2 + x. So we're taking the absolute value of a square, which must be zero or greater, plus some positive value x, and the expression inside the absolute value must be positive (there are several other ways to see that, some simpler if you use slightly more advanced math). So the absolute value isn't doing anything here, because we're taking the absolute value of something that must already be positive, and we can just ignore the absolute value and solve the equation

x^2 - x + 1 = 2x - 1
x^2 - 3x + 2 = 0
(x - 2)(x - 1) = 0

and x = 2 or x = 1, and the sum of the solutions is 3.
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 30 Sep 2021, 09:20
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Bunuel
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
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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
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 01 Oct 2021, 06:14
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Bunuel
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
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When the eqn is positive through simplification we get
=>x^2 -3x +2 =0
=> x=2 ,1

When the eqn is negative
=>x^2 +x =0
=>x=0 ,x=-1

HOwever when resubstitute and check for the answer we get
x=2,1,0

Therefore IMO D
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 04 Oct 2021, 01:12
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Since 2x-1 >= 0, x > 1/2 ....................inferred from the question stem itself.

1. When |x^2 - x + 1| > 0
|x^2 - x + 1| = 2x -1
Solving this we get x = 2 or x = 1 ...........Possible values, Sum = 3

2. When |x^2 - x + 1| < 0
- |x^2 - x + 1| = 2x -1
Solving this we get x = 0 or x = -1. But from our inference earlier, x > 1/2.

Hence sum of all possible values of x = 2+1 =3
IMHO Answer D
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 03 Nov 2021, 07:54
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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
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This question is a part of Are You Up For the Challenge: 700 Level Questions collection.
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 25 Oct 2022, 02:48
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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


M37-03
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There are 2 possible cases:

Case 1, \(x^2-x+1\) is positive,

Equation becomes,
\(x^2-x+1=2x-1\)
=> \(x^2-3x+2=0\)
=> x = 2,1

Case 2, \(x^2-x+1\) is negative

Equation becomes
\(x^2-x+1=-(2x-1)\)
=> x=0,-1 Both doesn't satisfy the original equation

Since, only x=2 and x=1 satisfy the equation \(|x^2-x+1|=2x-1\)
Therefore, the sum of all possible values of x is 2+1 = 3

D is Correct
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 18 May 2025, 02:43
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I followed this approach but did not get 1 as the answer. Can anyone tell where I am going wrong?
|x^2 - x + 1| = 2x - 1 can be rewritten as
|x^2 - x + 1 - 3| = 2x - 1 -3
|x^2 - x - 2| = 2(x-2)
|(x-2)(x+1)| = 2(x-2)
When x>2, (x-2)(x+1)=2(x-2) => x=1(but this does not fall within x>2, hence is not considered in this case)
When -1<x<2, (2-x)(x+1)=2(x-2) => x=-2(again this cannot be considered)
When x<-1, (2-x)(-x-1)=2(x-2) => x=1(cannot be considered as x<-1)
I know i am making a mistake somewhere but can anyone tell me where?

Thanks in advance.
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If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values 18 May 2025, 03:07
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Bunuel
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

Official Solution:

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. \(4\)


The right hand side of the equation (\(2x-1\)), is equal to the absolute value of some number (\(|x^2-x+1|\)), so \(2x-1\) must be non-negative: \(2x-1 \geq 0 \). This gives \(x \geq \frac{1}{2}\).

Now, work on the expression in modulus. Re-write \(x^2-x+1\) by completing the square: \(x^2-x+1=(x-1)^2+x\). Since from above \(x\) is positive, then \(x^2-x+1=(x-1)^2+x=nonnegative + positive=positive\). So, \(|x^2-x+1|=x^2-x+1\).

So, we'd have \(x^2-x+1=2x-1\);

\(x^2-3x+2=0\);

\((x - 2)(x - 1) = 0\);

\(x=2\) or \(x=1\);.

The sum of all possible values of \(x\) is therefore, \(2+1=3\).


Answer: D
I followed this approach but did not get 1 as the answer. Can anyone tell where I am going wrong?
|x^2 - x + 1| = 2x - 1 can be rewritten as
|x^2 - x + 1 - 3| = 2x - 1 -3
|x^2 - x - 2| = 2(x-2)
|(x-2)(x+1)| = 2(x-2)
When x>2, (x-2)(x+1)=2(x-2) => x=1(but this does not fall within x>2, hence is not considered in this case)
When -1<x<2, (2-x)(x+1)=2(x-2) => x=-2(again this cannot be considered)
When x<-1, (2-x)(-x-1)=2(x-2) => x=1(cannot be considered as x<-1)
I know i am making a mistake somewhere but can anyone tell me where?

Thanks in advance.
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Unfortunately, nothing in that approach is correct.

You started with |x^2 - x + 1| = 2x - 1 and rewrote it as |x^2 - x + 1 - 3| = 2x - 1 - 3, which is wrong. If you subtract something from one side of the equation, you must subtract it from the other outside the absolute value. That would give |x^2 - x + 1| - 3 = 2x - 1 - 3, which doesn’t simplify anything useful. So the main error is that you incorrectly inserted -3 into the modulus.

Next, you are also solving |(x - 2)(x + 1)| = 2(x - 2) incorrectly. You should include each critical point in any of the ranges:

  • When x ≥ 2, we get (x - 2)(x + 1) = 2(x - 2), which gives x = 1 or x = 2. Since we consider x ≥ 2 range, only x = 2 is valid.
  • When -1 < x < 2, we get (2 - x)(x + 1) = 2(x - 2), which gives x = -3 or x = 2. Since we consider -1 < x < 2, none of the values is valid.
  • When x ≤ -1, we get (x - 2)(x + 1) = 2(x - 2), which gives x = 1 or x = 2. Since we consider x ≤ -1 range, none of the values is valid.

Thus, |(x - 2)(x + 1)| = 2(x - 2) has only one root: x = 2.
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