# Case 1

*f*(

*x*) =

*x*

^{2}– 5

*x*+ 6 ≥ 0

*f*(

*x*) =

*x*

^{2}– 5

*x*+ 6 = (

*x*–2) (

*x*− 3) ≥ 0

We have got three intervals of values here: −∝ to 2, 2 to 3 and 3 to + ∝

To check the values of x satisfying the above written inequation, any one value of x is taken from any of the intervals. If this value satisfies the inequation, then inequality will be satisfied in the alternate intervals.

Taking
So, inequality is satisfied in this interval. Hence, it will not be satisfied in the interval 2 ≤ x ≤ 3 and again will be satisfied in the interval −∝ < x ≤ 2

*x*= 10,*f*(*x*) = (*x*–2) (*x*−3) = (10−2) (10−3) = +ve# Case 2

*f*(

*x*) =

*x*

^{2}– 5

*x*+ 6 ≤ 0

It can similarly be observed that the values of

*x*are lying in between 2 and 3Hence, 2 ≤

*x*≤ 3In general, if (

*x*−*a*) (*x*−*b*) ≥ 0, then*a*≤*x*≤*b*And if (

*x*−*a*) (*x*−*b*) ≤ 0 and*a*<*b*, then*x*≤*a*or*x*≥*b*Example

What values of

*x*will satisfy the quadratic inequation*f*(*x*) = −*x*^{2}+ 3*x*+ 4 > 0?Solution

*f*(

*x*) = −

*x*

^{2}+ 3

*x*+ 4 > 0, or,

*x*

^{2}− 3

*x*− 4 < 0

*x*+ 1) (

*x*– 4 ) < 0

*x*satisfying

*f*(

*x*) = −1 <

*x*< 4