If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?

I solved it by isolating the modulus from the rest of the equation and solving two different quadratic equations:
x-1 = xˆ2-5x+4 -> xˆ2-6x+5=0 -> x1 = 5, x2 = 1
x-1 = -xˆ2+5x-4 -> -xˆ2+4x-3=0 -> x1 = -1, x2 = -3

Solution: 3x5 = 15, letter E

Try substituting the values from the options

\(x^2−5x+6=2−|x−1|\)

Option (A)

\(1^2−5*1+6=2−|1−1|\)

\(1−5+6=2−|0|\)

\(10 not =2\)

Option (B)

\(3^2−5*3+6=2−|3−1|\)

\(9−5*3+6=2−2\)

\(15−15=2−2\)

Hence this is our answer , you don't need to go beyond that, to mark the answer as (B)

I don't understand !

Here, we find the solutions : (1 or 5) for one case | x - 3| = x - 3
and (1 or 3) for the other case | x - 3| = - (x - 3) = - x + 3

This question teaches us that when we solve an equation with modulus, all the answers have to be double-checked ?

Yes, I would solve by substituting from the options too but note that I would not try option (A) at all and after option (B), I will need to try option (D) too.

Here are the reasons:
Option (A) is 1. If it satisfies, I am not able to eliminate any options because the product of the values of x would be the same whether you multiply by 1 or not. So no point trying out 1.

Option (B) needs to be tried out to find out whether 3 can be a value of x.

x^2−5x+6=2−|x−1|
3^2−5*3+6=2−|3−1|
0 = 0

So 3 is a value of x. Can we say answer must be (B) yet? No. You have 15 as an option too. So you must try 5 too. If that can also be the value of x, then the product will be 15 and answer will be (E).

5^2−5*5+6=2−|5−1|
6 is not equal to -2.
So 5 is not a value of x.

Now we can be sure that answer must be (B).

Since modulus is given, it is prudent to solve this equation in two parts.

(1) x-1 > 0 or equal to 0. so x is greater than or equal to 1
Then |x−1| can be written as x-1.

The equation can be written as:

\(x^2\)−5x+6=2−(x−1)
\(x^2\)−5x+6=2−x+1
\(x^2\)−5x+6=3−x
\(x^2\)−4x+3=0
(x-1)(x-3) = 0 ; x = 1 or 3. This is in line with our assumption (that x is greater than or equal to 1)

(2) x-1 < 0. so x is less than 1
Then |x−1| can be written as -(x-1).

The equation can be written as:

\(x^2\)−5x+6=2−[-(x−1)]
\(x^2\)−5x+6=2+x-1
\(x^2\)−5x+6=1+x
\(x^2\)−6x+5=0
(x-1)(x-5) = 0; x = 1 or 5. This is NOT in line with our assumption (that x is less than 1)

Acceptable values: 1 and 3.

Product of these values: 1*3 = 3

B is the answer.

case 1 x>=1
x^2-5x+6 = 2-x+1
x = 1,3(acceptable)
case 2 x<1
x^2-5x+6 = 2+x-1
x = 5,1 but x<1 so rejected

x²-5x+6=2-|x-1|
Note that |p|²=p² for any real number p.
We rewrite the above equation as
(x-1)²-3x+5= 2-|x-1|
Implies
|x-1|²-3x+3+|x-1|=0
|x-1|²+|x-1|-3(x-1)= 0 ...(★)
We consider two scenarios
when x≥1 and x<1
For x≥1
(x-1)=|x-1|
So the equation (★) becomes

|x-1|²+|x-1|-3|x-1|= 0
Which implies
|x-1|²– 2|x-1|= 0
Which implies
|x-1|=0 or |x-1|=2
Implies x=1 or x=3 or -1
But -1 is discarded because the condition is for x≥1 as stated earlier. Hence x=1 or 3.
For condition the condition x<1
|x-1|=-(x-1)
Hence equation (★) becomes
|x-1|²+|x-1|+3|x-1|= 0
Implies
|x-1|²+ 4|x-1|= 0
|x-1|(|x-1|+4)=0
Implies
|x-1|=0 or
|x-1|=-4(which is not possible as absolute value function can not be negative)
Hence x=1 is the only valid solution for this condition.
Hence x=1 and x=3 are the only possible solution thus their product yields 3 (option B)

Posted from my mobile device

Actually the problem can be solved quickly if we make use of graphs.
We draw the LHS : quadratic equation with 2 roots at 2 and 3 ( parabola)
Next draw the RHS equation.
You wil get something like this:

The only point of intersection is at x = 3 and hence the only value that satisfies the equations. Hence product of values is 3

Hope this helps!

Please correct me if my approach is incorrect !

x<1
x2−5x+6=2+(x−1)
x2−5x+6=2+x−1.
=>x2−6x+5=0=>x2−6x+5=0 (x−5)(x−1)=0
x=5 or x=1
We are eleminating both of them since we are not interested anything above 1 right now


2) x>=1
x2−5x+6=2−(x−1)..
x2−5x+6=2−x+1
=>x2−4x+3
=>(x−3)(x−1)=0
x=3 or x=1
Since both are above 1 we are chosing both of them
Therefore IMO B

If you are new to absolute value/modulus questions (questions which use |x|) I highly recommend that you check out the guide of @walker. (Also give him/her a kudos )

https://gmatclub.com/forum/math-absolute-value-modulus-86462.html

The I first time I encountered absolute value questions on the GMAT was in diffult questions (600 to 700+) and even after looking at the the explanations I did very much. @walkers guide truely helped to understand the fundamentals!

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