Bunuel wrote:

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

A. 1

B. 3

C. 4

D. 5

E. 15

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.