Please explain why x> 3 or x<2 instant of x>3 or x> 2? Consider the question: Solve for x in lR, X^2 -5x + 6 >0
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calendarEducator since 2010
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For the inequation x^2 - 5x + 6 > 0
we first write the quadratic expression as factors
x^2 - 5x + 6 > 0
=> x^2 - 3x - 2x + 6 > 0
=> x( x - 3) - 2(x - 3) > 0
=> (x - 3)(x - 2) > 0
This is possible only if either both x - 3 and x - 2 are positive or both of them are negative.
If they are positive
x - 3 > 0 and x - 2 > 0
=> x > 3 and x > 2
x > 3 satisfies both the conditions
x - 3 < 0 and x - 2 < 2
=> x < 3 and x < 2
Here x < 2 satisfies both the conditions.
The two relations derived above give the solution as x > 3 or x < 2
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
For the inequality:
x^2 - 5x +6 > 0
First we will factor:
==> (x-2)(x-3) > 0
Now we have a product of two function. In order for the product of two numbers to be positive ( > 0) then both numbers should be positive or both numbers should be negative.
Then we have two options:
(x-2) > 0 AND x-3 > 0
We will solve.
==> x > 2 AND x > 3
==> x = (2, inf) n ( 3, inf) = (3,in)
==> x = (3, inf ) ............(1)
For the second option:
(x-2) < 0 AND (x-3) < 0
==> x < 2 AND x < 3
==> x = (-inf , 2) n (-inf, 3) = (-inf, 2)
==> x = (-inf, 2) .............(2)
Then, we have two possible solutions: (1) and (2).
==> x = (-inf , 2) U ( 3, inf)
OR: x <2 OR x > 3