Question: If \(x^2-5x+6=2-|x-1|\), what is the product of all possible values of x?

1. 1
2. 3
3. 4
4. 5
5. 15

Answer:

Solution with Explanation:
Approach Solution (1):

One critical point at x = 1. So two ranges will be there

(1) x < 1

\(x^2-5x+6=2-|x-1|=x^2-5x+6=2--(x-1)\)
\(x^2-5x+6=2+x-1\)
\(\Rightarrow x^2-6x+5=0\)
OR
\(x^2-5x-x+5=0\)
OR
(x-5)(x-1)=0

But since we are looking at the range x < 1, both 1 and 5 are not valid.

(2) x >= 1

\(x^2-5x+6=2-|x-1|=x^2-5x+6=2-(x-1)\)
\(x^2-5x+6=2-x+1\)
\(\Rightarrow x^2-4x+3=0\)

OR

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

OR

(x-3)(x-1)=0
x = 3 or x = 1

As we are looking at the range x >= , both 3 and 1 are valid solution
So product = 1 * 3 = 3

Correct Option: B

Approach Solution (2):

We have found the different values of x correctly as 5, 1, 3, 1…

But which are the valid values- you will have to substitute and check it.

Since you are dealing with modulus.

\(x^2-5x+6=2-|x-1|\)

(1) 5 –

\(5^2-5*5+6=2-|5-1|\)
25 – 25 + 6 = 2 – 4 or 6 = -4

No, so 5 is not valid.

(2) 1 –

\(1^2-5*1+6=2-|1-1|\)
2 = 2...

Valid

(3) 3 –

\(3^2-5*3+6=2-|3-1|\)
9 – 15 + 6 = 2 – 2

OR

0=0...

Valid

So there are two solutions 1 and 3.

Product = 1 * 3 = 3

Correct Option: B

Approach Solution (3):

We can solve it by isolating the modulus from the rest of the equation and solving two different quadratic equations:

\(x-1 = x^2-5x+4 \rightarrow x^2-6x+5=0 \rightarrow x_1 = 5, x_2 = 1\)
\(x-1 = -x^2-5x+4 \rightarrow -x^2-4x-3=0 \rightarrow x_1 = -1, x_2 = -3\)

Product = 3 * 5 = 15

Correct Option: E

“If, what is the product of all possible values of x?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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