# 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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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**

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**