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1.5 QUADRATIC INEQUALITIES

In Section 1.3 we solved inequalities of the type


In this section we shall learn about procedures for solving


First, let us consider the simple quadratic inequality x2 b. Since x2 ≥ 0, if b < 0, there is no solution for this inequality. Here, we must point out that the statement p < q has no meaning if p and q are complex numbers.

So, let us assume that b ≥ 0. By taking square roots on both the sides, we obtain Have we solved the inequality? No. Take b = 25. Let us consider


The above procedure tells us that


Now, x = -6 satisfies equation (1.27), but x = -6 does not satisfy equation (1.26) since (-6)2 > 25. In other words, the logical implication


is false.

The actual solution to equation (1.26) is given by -5 ≤ x ≤ 5. This is illustrated in Figure 1.3.

Generalizing from the above example, we obtain the first rule for a quadratic inequality as follows. Assuming b ≥ 0,


Now, let us consider the set of all solutions of


Figure 1.3. Solution of x2 ≤ 25


If a < 0, the condition a < x2 always holds and equation (1.28) simply reduces to that of x2 < b for which we already know the solutions. If b is negative, then clearly there is no solution to this inequality. So, let us now consider equation (1.28) with the conditions that a ≥ 0 and b ≥ 0 and naturally with a < b. A little bit of thought tells us that the solutions of equation (1.28) are given by


and


Figure 1.4 solves this problem graphically.

Figure 1.4. Solutions of a < x2 < b


From the graph, we find that the solutions of equation (1.28) are given by


and


We can easily see how one has to modify the solutions for inequalities like


The solutions are given by


and


The inequalities involving the expression ax2 + bx + c can be simplified by the method of completion of square.

Example 1.8.

Solve x2 + 6x - 9 < 17.

SOLUTION


In this example, we have rewritten x2 + 6x - 8 as (x + 3)2 - 17. This is the procedure known as completion of square.

PRACTICE 1.5

Let a ≥ 0. Show graphically that the solutions of


are given by

and


Remark 1.1

If a < 0, any real number x is a solution to the inequality x2 > a.


Example 1.9.

Solve x2 - 9x + 15 < 0.

SOLUTION


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