M37-03

Question

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

Bunuel · Answer

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 key to solving this question is deducing that  x^2-x+1  is positive for any  x  value. To see this, we can rewrite  x^2-x+1  by completing the square: 
 
 x^2-x+1 =  
 
 =x^2 - x + \frac{1}{4} +\frac{3}{4}= 
 
 =(x - \frac{1}{2})^2 + \frac{3}{4}= 
 
 =nonnegative + positive=positive .
 
Therefore,  |x^2-x+1|=x^2-x+1 .
 
So, we'd get:  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

r_putra_rp · Answer

Hi  Bunuel

Why don't we consider the negative part of the absolute?
 -(x^2-x+1) = 2x-1

Thanks!

Bunuel · Answer

|a| = -a, when a <= 0.

However, in this question x^2 - x + 1 is never negative, thus |x^2 - x + 1| = x^2 - x + 1 only.

phmahi1997 · Answer

I like the solution - it’s helpful.

Aakash7 · Answer

Given: |x^2 -x+1| = 2x-1
The LHS will always be positive. Notice that for x<0and x=0, the RHS becomes negative which is not possible.
Hence, 2x-1 > 0
 => x>1/2 . In other words, x is positive

x^2 - x + 1 = 2x - 1
Solving quadratic equation, we get x=1 or x=2 (for x>1/2)
Therefore, sum of possible values of x is 1+2 = 3 Ans.

Bunuel  - Please let me know if there are any flaws in this solution/ approach. Thanks!

Bunuel · Answer

Most of your solution is correct, except for one thing.

When you get x ≥ 1/2, it does not automatically mean that the expression inside the modulus is positive and will come out of the modulus with a + sign. We should have checked whether x^2 - x + 1 is positive for x ≥ 1/2 (and in fact, it turns out that x^2 - x + 1 is positive for any value of x, but I’m talking about the principle).

Finally, if you had shown that x^2 - x + 1 is positive for all x, then you would not even need to consider the x ≥ 1/2 part and could directly solve x^2 - x + 1 = 2x - 1.

MegB07 · Answer

Hey  Bunuel , is it not negative because if we take a negative value for x in RHS it will not hold true or be equal to LHS since its in Mod?

Bunuel · Answer

It is not negative because the expression on the LHS, \(x^2 - x + 1\), is always positive for any value of \(x\). Since the inside of the modulus never becomes negative, \(|x^2 - x + 1|\) is simply equal to \(x^2 - x + 1\) (because |a| = a when a is positive).

MegB07 · Answer

But when we solve similar questions we do consider +- values of Mod. Why arent we applying that here? Thank you for the prompt response, as always.

Bunuel · Answer

We only split into  +a  and  -a  cases when the inside of the modulus can actually be either positive or negative. Here, the key difference is that  x^2 - x + 1  is always positive for every real  x .

Because the inside never goes below zero, there’s no “-a” case to consider.

shaliny · Answer

hii Bunuel,
 how would i know that any equation inside modulus is always positive as in this ques. because generally in modulus i solve by taking both + and - sign of equation.
So is there any quick method like plugin (may take time) or some specific points to notice in equations or strategy to know that these equations would be positive for any values?

Bunuel · Answer

It totally depends on the question, how else? Sometimes it’s straightforward, for example, if we had something like  x^2 + 1  inside the modulus, that’s always positive, so we can directly write  |x^2 + 1| = x^2 + 1  without needing cases. And sometimes it’s a little bit trickier, like in this question, where you need to complete the square to see that  x^2 - x + 1  is always positive. So the approach varies with the expression.

lehung97 · Answer

I like the solution - it’s helpful.

janhvi167 · Answer

Hi  Bunuel

x^2-x+1 =  
 
 =x^2 - x + \frac{1}{4} +\frac{3}{4}= 
 
 =(x - \frac{1}{2})^2 + \frac{3}{4}= 
 
 =nonnegative + positive=positive .

This way is not very intuitive to find out whether the equation would always be positive or not.
Very difficult to adjust the original equation to this : =(x - \frac{1}{2})^2 + \frac{3}{4}=  
Is there any other way to find out?

Bunuel · Answer

Yes, you can check it without completing the square. For x^2 - x + 1 = 0, the discriminant is b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3. Since the discriminant is negative and the coefficient of x^2 is positive, the expression is always positive for all x. Check alternative solutions here: https://gmatclub.com/forum/if-x-2-x-1-2 ... 70453.html

NupurNeha123 · Answer

Hi , how do we know x^2-x +1 is never negative ; while solving absolute we usually make two cases, please clarify

Bunuel · Answer

It is explained  HERE  and  HERE . Hope it helps.

M37-03

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