# Linier Inequalities and Quadratic Inequalities

Linier Inequalities and Quadratic Inequalities | Open sentences of the form x + 3 < 10 , , etc are called inequalities. For appropriate values of the variable, one member of an inequality represents a real number that is less than (<), less than or equal to (=<), greater than or equal to (>=), or greater than (>) the real number represented by the other member.

Any element of  the replacement set of the variable for which an inequality is true is called a solution, and the set of all solutions of an inequality is called the solution set of inequality

Fundamental properties of inequalities

1. The addition of the same expression representing a real number to each member of an

inequality produces an equivalent inequality in the same sense

Illustration : Notice that 2 < 3

And that 2 + 5 < 3 +5

And         2-5 < 3-5

2. If each member of an inequality is multiplied by the same expression representing

a positive number, the result is an equivalent in equality in the same sense

Illustration :  If we multiply each member of  2 < 3 by 2, we have 4 < 6

3. If each member of an inequality is multiplied by the same expression representing a

negative number, the result is an equivalent inequality in the opposite sense

Illustration : If we multiply each member of  2 < 3  by -2, we have -4 < -6

Statements 1, 2, and 3 above can be expressed in symbols as follows :

if  P(x), Q(x) and R(x) are expressions then for all values of x for which these expressions

represent real numbers

P(x) < Q(x) is equivalent to :

1.      P(x) + R(x) < Q(x) + R(x)

2.      P(x) . R(x) < Q(x) . R(x), if R(x) > 0

3.      P(x) . R(x) > Q(x) . R(x), if R(x) < 0

Many kinds of inequalities :

### 1. LINEAR INEQUALITIES

The general form of these inequalities is ax + b > 0. The sign of inequality can be >, <,

Example :  Find the solution set of the following inequalities :

a)      2x – 3 > 7

b)      5(x + 1) < 2(x + 7)

c)    -2 < 3x – 5 < 4

Solution : a) 2x – 3 > 7

2x    > 10 ( add 3 to each member)

x      > 5 ( multiply each member by ... )

Solution set is S = ...

(This symbolism is read,” S is the set of all x such that x is greater than 5,

x element of real number”)
A line graph of solution set is

b) 5(x + 1)  2 (x + 7)

5x + 5     2x + 14

3x   9 ( add -5-2x to each member )

x   3 ( multiply each member by )

Solution set is S =

A line graph of solution set is

c)  <

3(x – 3) < 8 ( multiply each member by 12 )

3x – 9    < 8

3x    < 17 (add 9 to each member )

x    <  ( multiply each member by )

Solution set is S =

A line graph of solution set is

d) -2 < 3x – 5  4

3  <  3x  9 ( add 5 to each expression )

1  <   x   3 ( each expression may be multiplied by )

Solution set is S =

A line graph of solution set is                    ○                      ●

1                       3

EXERCISES

1. Find the solution set of the following inequalities

a) 5x + 4 < 2x -8
b) 2(x + 1) < 3(x-2)
c) 2x – 8  7x + 12
d) 4(4-x) – 3(x – 7) – 16x – 14 < 0
e)  – 1 < x+9 < 3
f) (x – 2) < (x + 10)

2. The inequality 2x – a >  has solution is x>5. Find the value of a.

3. Candra bought 21 books, Doni bought 33 books.  Doni’s books have cost less than

Rp 500.00 each. Candra’s books were Rp 100.00 more expensive than one half of

each Doni’s books. The difference of money  paid was less than  Rp 1,500.00. Find the

price border of  each Doni’s.

4. To give contribution to 1,000 calamity casualties, a public kitchen was  built.  Every

day they need about kg and  kg  of  rice for each casualty. Find the border of rice

quantity which must be available every day

The general form of quadratic inequalities is ax2 + bx +c > 0 of course the sign of  inequality can be <, =< , >=   .

There are two method to solving these inequalities :

1.      Graphically

2.      By using number line / line graph

1. Graphically

We have known that the graph of  y = ax2 + bx + c is a parabola.

If we want to solve the solution set of  :

ax2 +bx + c > 0 , it means we have to find the values of x in which the parabola is above

the x-axis.

ax2 + bx +c<0 , it means we have to find the values of x in which the parabola is under

the x-axis

Examples :

1. 7x + 4 – 2x2 < 0

2x2 - 7x - 4  > 0 ( we have to find the values of x in which the parabola y = 2x2-7x-4

above  the x-axis )

X                                 The graph has x-intercepts at x =  and x = 4.

4     Y     The graph is above the x-axis or y > 0 if x <  or

x > 4

The solution set is

2. –x2 – 7x – 10  0

x2 + 7x + 10  0  ( multiply by -1)

(x + 2)(x + 5)  0   ( we have to find values of x in which the graph is under or at the

x-axis)

Y                        The graph has x-intercepts at x = -2 and x = -5

The graph is under the x-axis when -5 < x < -2.

The graph is the x-axis when x = -5 and x = -2

So the solution set is

2. By using number line

The steps of finding the solution of an inequality :

1.      Find values of x which cause the left side of an inequality is zero, if possible.

2.      Place the x values on the number line

3.      Sign every part of the number line with either positive or negative

4.      Find the solution set

Examples :

1.      3x2 – x – 2 > 0

(3x + 2)(x – 1) > 0               Step 1

x = ,   x = 1

We have to choose positive parts (> 0)        Step 4

So the solution set is

2. x2 – 10x + 24  0

(x – 6)(x – 4)   0

x = 6,   x = 4
+ + +               - - -               + + +                      We have to choose negative parts (<0)

3.  4x2 – 4x + 1 > 0

(2x – 1)(2x – 1) > 0

4. x2 – x + 2 > 0, it is definite positive, because a > 0, D < 0

Solution set is

5. -2x2 + x – 3 < 0, it is definite negative, because a < 0, D < 0

Solution set is

6. -4x2 – 4x – 5 > 0, it is definite negative

Solution set is

EXERCISES

1. Find the solution set of the following quadratic inequalities by graphically

a) x(x + 4) – 12  0                                b) 3x2 – 5x + 2  0

b) 2x2 + x < 10                                        c) x2 < -7x -12

2. Find the solution set of the following quadratic inequalities by using number line

a) 2x2 + 7x – 15  0                               d) 15 – 7x  2x2

b) 2x2 – x – 6 > 0                                    e) 3x2 + 2x < 3 – 6x

c) 3x2 – 7x + 2  0                                 f) 8x2  5x + 3

3. The missile is launched. Its height in air is given h(t) = 30t – t2 , where h is in meters,

t is in seconds. How long will it reach the height of  the greater than 221 meters ?

4. The length of a wire is 20 cm, it will be made a skeleton of rectangle. Its area is less

than or equal is 21 cm2. Find the length of a rectangle.