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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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A. 0
B. 1
C. 2
D. 3
E. 4
M37-03
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Bunuel
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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General Discussion
Posts: 6,727
Bunuel
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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Bunuel
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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Expert reply
Bunuel
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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Bunuel
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
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
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
Bunuel
This question is a part of Are You Up For the Challenge: 700 Level Questions collection.
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Bunuel
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
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.
|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.
Vasavan
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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I didn't get into the absolute value ruckus only
(x-1/2)^2 + 3/4 is the = x2−x+1 - the absolute term - this will always be positive come what may
so get rid of the absolute issue
now
x2−x+1=2x−1
x2 - 3x +2 = 0
Sum of all roots is -b/a = +3 Ans is option D Bunuel - Let me know if this is in-correct.
(x-1/2)^2 + 3/4 is the = x2−x+1 - the absolute term - this will always be positive come what may
so get rid of the absolute issue
now
x2−x+1=2x−1
x2 - 3x +2 = 0
Sum of all roots is -b/a = +3 Ans is option D Bunuel - Let me know if this is in-correct.
Is this method also foolproof?
Splitting it into positive & negative? So, we get:
[ltr]x2−x+1=2x−1[/ltr]
and
[ltr]x2−x+1=−(2x−1)[/ltr]
Let's solve each equation separately
Splitting it into positive & negative? So, we get:
[ltr]x2−x+1=2x−1[/ltr]
and
[ltr]x2−x+1=−(2x−1)[/ltr]
Let's solve each equation separately
Bunuel
eaat
Yes, however, with this method, you should verify the validity of each solution you get by plugging it back into the original equation to check whether it satisfies it.
From x^2 - x + 1 = 2x - 1, we get x = 1 or x = 2, and both work.
From x^2 - x + 1 = -(2x - 1), we get x = 0 or x = -1, but neither of them satisfies |x^2 - x + 1| = 2x - 1, so they are not valid solutions.
So the valid values are 1 and 2, and the sum is 3.
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