The higher order inequalities can be solved by waving curve method. We will understand this method with the help of an example. Let us we have an inequality (x – 3) (x – 2) (x + 1) ≤ 0
We will understand it in following steps:
Step 1: First find the zero’s of all factors (x – 3) (x – 2) (x + 1). Zeroes are found by putting all the factors equal to zero. So zeroes in this case are 3, 2 & - 1
Now mark the above said zero’s on a number line as shown below
Now draw a curve starting from the greatest point (Extreme right hand side) in this case it is 3 and the curve should start from the upper side of number line and should cross the each of the marked point as shown below.
The regions covered by the curve above the number line are marked as +ve and the region covered by the curve below the number line are marked as –ve as shown below.
Now define the region as per the question given. In our case we have to find the region (less than or equal to zero), so the region is (- , - 1] [2, 3]. As infinity is not defined so we have put a round bracket.
Now, one should be cautious that waving method is applicable only when following conditions are met:
- This is applicable when the inequality can be factorized in the form (x – a) (x – b) (x – c) where a, b, c are integers.
- On the right hand side of the inequality, there should be zero.
- Negative sign should not be there with x. If it is there, then remove it by changing the sign of inequality.
e.g. If the inequality is of the form (x + 3) (- x + 2) ≤ 0, then first you should change it to (x + 3) (x – 2) ≥ 0 and then solve it.
Now we will solve the inequalities where even power of factors is given and some factors are given in denominator. E.g. if we are to solve (x-3)(x+2)(x+1)2
/(x-2) ≤ we will solve it by waving curve To solve this type of inequality following steps are involved.
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Step 1: Firstly check that both the basic conditions i.e. right hand side of the inequality must be zero and no negative coefficient of x, are satisfied.
Step 2: Mark the zero’s on the number line as already discussed and shown below
Put a circle at a point that is derived from the factor which is in the denominator. Mark X on a point that is derived from a factor having even power. In this case it is -1 as shown below. The relevance of this exercise will be explained in the following step.
Now draw the curve keeping in mind that the curve will retrace its path where X is marked (It will remain on upper side of the number line if it already on the upper side and it will remain on lower side of number line if it is already on the lower side). Rest process remains the same as shown below.
Step 5: Now we will define the regions as per the requirement of the question. In this case it is 0. One thing is to be remembered here that we will not include the points where circles are marked because they are derived from the denominator and by putting those values will make the denominator as zero and the inequality will become not defined.
So the required region is (-∞ - 1] ∪ (2, 3].
(Here - ∞ is not included as ∞ in not defined and 2 is not included as already explained.)
Now one important fact is also to be kept in mind that all even powers and all odd powers are handled in the same manner as we have done in the 2nd example. It means that if there is a factor of the form (x – 3)5 it will be considered of the form (x – 3) and if there is factor of the form (x – 2)8 it will be considered of the form (x – 2)2 and so we will solve it accordingly.
Illustration : Solve (x + 5) (2x – 3) (x – 1) < 0
We have (x + 5) (2x – 3) (x – 1) < 0
The critical points are x = -5, 1, 3/2
Now plot these points on the number line and draw curve.
Since our inequality is negative, so the solution is x ε (-∞, -5) ∪ (1,3/2)