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24 Jul 2018, 06:12
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Bunuel
Given: x³ - x = (x - a)(x - b)(x - c)
Factor x from left side: x(x² - 1) = (x - a)(x - b)(x - c)
x² - 1 is a difference of squares, so we can factor that: x(x + 1)(x - 1) = (x - a)(x - b)(x - c)
Rewrite first 2 terms as: (x - 0)(x - -1)(x - 1) = (x - a)(x - b)(x - c)
Rearrange as: (x - 1)(x - 0)(x - -1) = (x - a)(x - b)(x - c)
Since, we're told that c < b < a, we can see that a = 1, b = 0 and c = -1
Answer: C
Cheers,
Brent
09 Jun 2021, 04:19
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Solution:
x^3 - x = (x - a)(x - b)(x - c)
=>x(x^2-1) = x (x+1)(x-1) using (a+b)(a-b) = a^2 - b^2
=>The critical points are 0,-1 and 1
=>As a > b > c, and we have 1 > 0 >-1 its clear that b is 0. (option c)
Happy studying
!
Devmitra Sen
GMAT SME
x^3 - x = (x - a)(x - b)(x - c)
=>x(x^2-1) = x (x+1)(x-1) using (a+b)(a-b) = a^2 - b^2
=>The critical points are 0,-1 and 1
=>As a > b > c, and we have 1 > 0 >-1 its clear that b is 0. (option c)
Happy studying
!Devmitra Sen
GMAT SME
25 Jun 2023, 22:30
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my question is why or where are people getting -1,0,1 from? I assume they factor …. but then Why do people automatically set the quadratic equation equal to zero after factoring?
nowhere on the question does it say that….(x)(x+1)(x-1) = 0…..
is there a fundamental theorem im missing?
nowhere on the question does it say that….(x)(x+1)(x-1) = 0…..
is there a fundamental theorem im missing?
