math question
Started by spoog74
x^2 - 6x + 8 < 0
(x-4) (x-2) < 0
In order for this to hold true one of the values must be (+) while the other is (-)
(+) (-) < 0
(-) (+) < 0
A. x - 4 > 0 and x - 2 < 0
or
B. x -4 < 0 and x -2 > 0
In A: x > 4 and x < 2 [This is impossible so this combination is not an option]
In B: x < 4 and x > 2 thus 2 < x < 4
(x-4) (x-2) < 0
In order for this to hold true one of the values must be (+) while the other is (-)
(+) (-) < 0
(-) (+) < 0
A. x - 4 > 0 and x - 2 < 0
or
B. x -4 < 0 and x -2 > 0
In A: x > 4 and x < 2 [This is impossible so this combination is not an option]
In B: x < 4 and x > 2 thus 2 < x < 4
This is the exact question that was on the Quantitative Reasoning of the Kaplan Diagnostic DAT so here is their explanation word for word: The question for the similar problem was: " If x^2-9<0, which of the following is true?"
If x^2-9<0, then x^2<9. If x^2<9 and x is positive, then x<3. If x^2<9 and x is negative, then x>-3. So it must be true that -3<x<3.
It is good to remember that x^2<T is equivalent to -sqrtT<x<sqrtT if T is a positive number. This would be applicable in this question, where x^2<9 is equivalent to -sqrt9<x<sqrt9, or -3<x<3.
It is also good to remember that x^2>T is equivalent to x>sqrtT or x<-sqrtT if T is a positive number
tooth, while i appreciate your effort in trying to explain the problem but i have no clue what you just wrote lol. So basically in the inequality such as these, one is the opposite of the other one, but it doesnt matter which one?
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