# solve for x values: x^2 - 5x + 6 > 0

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### 3 Answers

x^2 - 5x + 6 > 0

Let us factor:

==> (x-2)(x-3) > 0

Now we have 2 cases:

case(1) :

x-2 > 0 and (x-3) > 0

==> x>2 and x > 3

==> x = (2, inf) intersects (3, inf) = ( 3, inf)

Case(2):

x-2 < 0 and x-3 < 0

x < 2 and x < 3

==> x = (-inf, 2) and (-inf, 3) = (-inf, 2)

**==> x = ( -inf, 2) U ( 3, inf)**

**OR **

** x = R - [2, 3] **

x^2-5x+6 >0

Factorise the left:

x^2-3x-2x+6 > 0

x(x-3)- 2(x-3) > 0

(x-3)(x-2) > 0.

Both factors are positive or negative for the product to be poaitive:

So x >3 Or x <2.

Or x should not belong to the interval (2, 3).

The given inequality is x^2 - 5x + 6 > 0

x^2 - 5x + 6 > 0

=> x^2 - 3x - 2x +6 >0

=> x(x-3) - 2 (x-3) >0

=> (x-2) (x-3) >0

Now this is true only if both (x-2) and (x-3) are positive or both (x-2) and (x-3) are negative.

If both (x-2) and (x-3) are positive

=> x -2 >0 and x -3>0

=> x> 2 and x > 3

This gives x >3

If both (x-2) and (x-3) are negative

=> x-2 <0 and x-3 <0

=> x < -2 and x < -3

=> x is less than -2

**Therefore x is either less than -2 or greater than 3**