2 Jan 1999

### ome Important Properties

1. Let a < b.
1. If n > 0, then
2.  na < nb, a/n < b/n, an < bn (if a, b > 0).
3. If n < 0, then
4.  na > nb, a/n > b/n, an > bn (if a, b > 0).

1. ab > 0  Û  (a > 0 and b > 0)  or  (a < 0 and b < 0)
2. ab < 0  Û  (a > 0 and b < 0)  or  (a < 0 and b > 0)

### olving An Inequalitiy

Solving an inequality in one real variable means to find (the set of) all real values which satisfy the given inequality.

Some points to take note when solving an inequality.

1. Never cross-multiply (unless you know that all are non-negative).
2. Make one side of the inequality to zero.
3. Simplify and factorise as far as possible into linear factors.
4. For quadratic factors which are cannot be or difficult to factorise, use completing square.
5. Non-negative factors should be ignored.  Eg: x4, 2x2 + 3, (x + 2)2, (x - 1)2 + 5, or even positive constants.
6. Find all the critical values where the factors are zero.
7. Use number line or graph to determine the solution.
8. The solution must not include values which make the denominators of the original inequality zero.
9. Answer must be given in set notation or interval notation when the solution set is asked for.

### nterval Notation & Set-builder Notation

 Inequality Interval Notation Set-builder Notation a < x < b (a, b) {x Î R : a < x < b} a < x £ b (a, b] {x Î R : a < x £ b} a £ x < b [a, b) {x Î R : a £ x < b} a £ x £ b [a, b] {x Î R : a £ x £ b} x > a (a, ¥) {x Î R : x > a} x ³ a [a, ¥) {x Î R : x ³ a} x < b (-¥, b) {x Î R : x < b} x £ b (-¥, b] {x Î R : x £ b}

#### 1  Factorisable

Example 1:  Find the solution set of  x2 - 6x + 8 > 0.

Solution:

Factorising, (x - 2)(x - 4) > 0.

Graphically:

 \  x < 2 or x > 4. The solution set is {x Î R : x < 2 or x > 4}

Number line:

 \  x < 2 or x > 4. The solution set is {x Î R : x < 2 or x > 4}

Example 2:  Find the solution set of  (2x - 1)(x + 2) < x(4 + x).

Solution:

 2x2 + 3x - 2 < 4x + x2 x2 - x - 2 < 0 (x + 1)(x - 2) < 0

 \ -1 < x < 2. The solution set is {x Î R : -1 < x < 2}

#### 2  Not Factorisable

Example 3:  Solve x2 - 4x + 1 > 0.

Solution:

Method 1:  Completing Square

 x2 - 4x + 1 = x2 - 4x + 4 - 3 = (x - 2)2 - 3 (x - 2)2 - 3 > 0 (x - 2)2 > 3 |x - 2| > Ö3 x - 2 > Ö3 or x - 2 < -Ö3 x > 2 + Ö3 or x < 2 - Ö3

Method 2:  Use formula to find the roots of the associated equation

The roots of x2 - 4x + 1 = 0 are
x  =
 ______ 4 ± Ö 16 - 4
¾¾¾¾¾¾
2
=  2 ± Ö3

 \ for x2 - 4x + 1 > 0, we have x > 2 + Ö3 or x < 2 - Ö3.

Example 4:  Solve x2 + 2x + 2 < 0.

Solution:

By completing square, (x + 1)2 + 1 < 0.
But (x + 1)2 + 1 > 1 for all real values of x.
Hence there are no real solution.
The solution set is Æ (the empty set).

Example 5:  Solve x2 + 2x + 2 > 0.

Solution:

By completing square, (x + 1)2 + 1 > 0.
But (x + 1)2 + 1 > 1 for all real values of x.
Hence the inequality is always true for any real values of x.
The solution set is R (the set of all real numbers).

### ther Orders

Example 6:  Find the solution set of  (x + 3)(x - 1)(x - 2) ³ 0.

Solution:

 \  -3 £ x £ 1  or x ³ 2. The solution set is {x Î R : -3 £ x £ 1  or x ³ 2}

Example 7:  Solve x2(x2 - 1) ³ 0.

Solution:

 Note that         x2 ³ 0 \        x2(x2 - 1) ³ 0 Þ             x2 - 1 ³ 0 (x + 1)(x - 1) ³ 0 \  x £ 1  or x ³ 1

Example 8:  Solve
 (x + 2)(2x - 5) ¾¾¾¾¾¾ x - 3
³  0.

Solution:

 Note that x ¹ 3. \ -2 £ x £ 5/2  or  x > 3.

Example 9:  Solve
 x + 1 ¾¾¾ x - 1
£
 6 ¾ x
.

Solution:

 x + 1 6 ¾¾ - ¾ x - 1 x
£  0

 x(x + 1) - 6(x - 1) ¾¾¾¾¾¾¾¾ x(x - 1)
£  0

 (x - 2)(x - 3) ¾¾¾¾¾¾ x(x - 1)
£  0

Note that x ¹ 0, 1.

\  0 < x < 1 or 2 £ x £ 3.

### iscellaneous

Example 10:  Find the values of x for which the following expression is a real number.

 æ è x ¾¾ x + 1 ö1/2 ø

Solution:

The expression is a real number when

 x ¾¾ x + 1 ³ 0.

Thus the solution set is {x Î R : x < -1  or  x ³ 0}.

Example 11:  Find the solution set of

 2 < x ¾¾ x + 1 < 3.

Solution:

We solve two inequalities and take the intersection.

 x ¾¾ x + 1
> 2             and
 x ¾¾ x + 1
< 3

 x - 2x - 2 ¾¾¾¾¾ x + 1
> 0
 x - 3x - 3 ¾¾¾¾¾ x + 1
< 0

 x + 2 ¾¾¾ x + 1
< 0
 2x + 3 ¾¾¾ x + 1
> 0

\  -2 < x < -1
\  x < -1  or  x > -3/2

Taking the intersection, the solution set is {x Î R : -2 < x < -1}.

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