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
