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If x^2-5x+6=2|x-1|, what is the product of all

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by swerve » Fri Jun 01, 2018 9:51 am

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Answers

A

B

C

D

E

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Difficulty—

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

The OA is B.

I solved it by isolating the modulus from the rest of the equation and solving two different quadratic equations:
x-1 = x^2-5x+4 -> x^2-6x+5=0 -> x1 = 5, x2 = 1
x-1 = -x^2+5x-4 -> -x^2+4x-3=0 -> x1 = -1, x2 = -3

Solution: 3x5 = 15, option E.

Please, can anyone explain how to solve this PS question? Thanks in advance!
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by Brent@GMATPrepNow » Fri Jun 01, 2018 10:13 am
swerve 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

The OA is B.

I solved it by isolating the modulus from the rest of the equation and solving two different quadratic equations:
x-1 = x^2-5x+4 -> x^2-6x+5=0 -> x1 = 5, x2 = 1
x-1 = -x^2+5x-4 -> -x^2+4x-3=0 -> x1 = -1, x2 = -3

Solution: 3x5 = 15, option E.

Please, can anyone explain how to solve this PS question? Thanks in advance!
Your approach is perfect.
HOWEVER, once you determine that the solutions are x = 1, x = 3 and x = 5, you must then plug those values back into the original equation (x² - 5x + 6 = 2 - |x - 1| ) to identify any EXTRANEOUS roots.

Let's do that

x = 1
Plug into the original equation to get: 1² - 5(1) + 6 = 2 - |1 - 1|
Simplify: 1 - 5 + 6 = 2 - |0|
Evaluate: 2 = 2
WORKS!! So, x = 1 is a definite solution

x = 3
Plug into the original equation to get: 3² - 5(3) + 6 = 2 - |3 - 1|
Simplify: 9 - 15 + 6 = 2 - |2|
Evaluate: 0 = 0
WORKS!! So, x = 3 is a definite solution

x = 5
Plug into the original equation to get: 5² - 5(5) + 6 = 2 - |5 - 1|
Simplify: 25 - 25 + 6 = 2 - |4|
Evaluate: 6 = -2
Does NOT work!! So, x = 5 is NOT a solution

So, the solutions are x = 1 and x = 3
Product = (1)(3) = 3

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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