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Solving Quadratic Inequalities Using Wavy Curve Methods

Posted on - 24-02-2017

JEE Math QE

IIT JEE

The Method of Intervals (Wavy Curve Method)

In order to solve inequalities of the form

≥ 0, £ 0,

where P(x) and Q(x) are polynomials, we use the following result:

If x1 and x2 (x1 < x2) are two consecutive distinct roots of a polynomial equation, then within this interval the polynomial itself takes on values having the same sign. Now find all the roots of the polynomial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write.

= f(x) = ,

where a1 , a2 , ....., an , b1, b2 , ...., bm are distinct real numbers.

Then f(x) = 0 for x = a1 , a2 , ...., an, .

and f(x) is not defined for x = b1 , b2 , ......, bm.

Apart from these (m + n) real numbers f(x) is either positive or negative.

Now arrange a1 , a2 , ...., an , b1 , b2 , ......, bm in an increasing order.

say c1 , c2 , ......, cm+n . Plot them on the real line.

Draw a curve starting from right along the real line which alternately changes its position at these points. This curve is known as the wavy curve.

The intervals in which the curve is above the real line will be the intervals for which f(x) is positive and intervals in which the curve is below the real line will be the intervals in which f(x) is negative. .

Example 1

Find the set of all x for which

Solution:

We have

There are five intervals x < -2, -2<x<-1,-1<x<-2/3, -2/3<x<-1/2, x>-1/2

The inequality (i) will hold for -2<x<-1 and for -2/3<x<-1/2

Hence -2 < x <-1 and -2/3 < x < -1/2.

Example 2

Let f(x) = . Find the intervals where f(x) is positive or negative.

Solution:

For x > 7 , f(x) > 0 . It is clear from the figure that.

f(x) > 0 " x ∈ (-5, -2) È (-1, 3) È(7, ¥)

and f(x) < 0 " x ∈ (-¥ ,-5)È(-2, -1)È(3, 7)

 
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