First for some humour :

*“A chemist, a physicist and a mathematician are stranded on an island when a can of food rolls ashore. The chemist and the physicist comes up with many ingenious ways to open the can. Then suddenly the mathematician gets a bright idea : ‘ Assume we have a can opener …….’ “*

Okay, linear inequalities in one variable is easy. For example if x – 3 ≥ 0, then x ≥ 3. But what about (x − 2)(x + 1) ≥ 0 ? There are two methods to solve this but I prefer the ‘graphical’ method. Because a ‘picture’ is clearer than just words n numbers. Lets sketch the curve of y = (x – 2)(x + 1). When y = 0, (x – 2)(x + 1) = 0 ⇒ x = 2 or x = − 1. Since the coefficient of x² is positive, then the curve has a minimum point.

For (x − 2)(x + 1) ≥ 0, we look at the range of values of x for which the curve is on or above the x-axis. ∴ x ≥ 2 or x ≤ − 1.

*Now try solving this question* :

1) Find the range of values of x for which x² − 5x + 1 lies between − 5 and 25.

Answer :

Let y = x² − 5x + 1

when y = − 5, x² − 5x + 1 = − 5 ⇒ x² − 5x + 6 = 0 ⇒ (x − 2)(x − 3) = 0

∴ x = 2 or 3

when y = 25, x² − 5x + 1 = 25 ⇒ x² − 5x − 24 = 0 ⇒ (x + 3)(x – 8) = 0

∴ x = −3 or 8

Next sketch the curve y = x² − 5x + 1 and the lines y = 25 and y = − 5.

The answer will be for the range of x whereby the curve lies between the two horizontal lines of y = − 5 and y = 25. So** −3 < x < 2 and 3 < x < 8 .**

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