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Case 1

f(x) = x2 – 5x + 6 ≥ 0
f(x) = x2 – 5x + 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 x = 10, f(x) = (x –2) (x−3) = (10−2) (10−3) = +ve
 
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
 
So, the values of x satisfying the above written inequality lies outside 2 and 3 i.e., x ≤ 2 or x ≥ 3

Case 2

f(x) = x2 – 5x + 6 ≤ 0
It can similarly be observed that the values of x are lying in between 2 and 3
Hence, 2 ≤ x ≤ 3
 
In general, if (xa) (xb) ≥ 0, then a ≤ x ≤ b
And if (xa) (xb) ≤ 0 and a < b, then x ≤ a or x ≥ b
 
Example
What values of x will satisfy the quadratic inequation f(x) = − x2 + 3x + 4 > 0?
Solution
f(x) = − x2 + 3x + 4 > 0, or, x2 − 3x − 4 < 0
 
Or, (x + 1) (x – 4 ) < 0
 
So the values of x satisfying f(x) = −1 < x < 4
 




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