topic 2 equations and inequalities

8/2/2019 Topic 2 Equations and Inequalities

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22

Equations andInequalities

LEARNING OUTCOMES

By the end of this topic, you should be able to:1. Describe the operations of equations;

2. Form a mathematical equation to daily life mathematics problems;

3. Solve quadratic equations; and

4. Solve inequalities.

INTRODUCTION

An equation is a statement of equality between two algebraic expressions which

holds true for a limited possibility of value for the unknown(s), and the process of

finding such value(s) is called solving an equation. For example 2 x + 1 = 3 x – 1 is

true for x = 2.

If the statement of equality is true for all values of the unknowns, then the

statement is an identity.

EQUATIONS2.1

We have seen that an equation is normally true for several values of unknown

only. Each of the values of unknown that makes an equation valid, if exists, is

called the solution or the root of the equation. Solving an equation means finding

all the possible solutions to an equation.



TOPIC 2 EQUATIONS AND INEQUALITIES26

Example 2.1

The equation x – 3 7 is true when x 10 and false for all other values of x.

Therefore 10 is the solution for x – 3 7. We also said that 10 satisfies the

equation x – 3 7, because when x is replaced by 10, a true statement results.

Example 2.2

We have seen that the equation ( x 1)2 x 7 has two solutions, i.e. x 3 or

x 2.

(a) Equivalent Equations

Two or more equations that have the same solutions are called equivalent

equations. For example, the following equations are equivalent because

each of them has the same solution i.e. x 3:

3 1 8

3 9

3

x

x

x

These three equations demonstrate a way of solving equations, i.e. by

replacing the original equation by equivalent equations continuously until

the solution is obvious (likes x 3).

(b) Steps to Produce Equivalent Equation(s)

How to obtain these equivalent equations? Generally there are five steps.

(i) Interchange both sides of the equation:

Replace 5 x with x 5.

(ii) Simplify both sides of the equations by combining monomials of the

same degree, eliminating the brackets, etc:

Replace ( x 1) 5 2 x ( x 1)

with x 6 3 x 1



TOPIC 2 EQUATIONS AND INEQUALITIES 27

(iii) Add or subtract the same expression to both sides of the equation:

Replace 2 x 3 5

with (2 x 3) 3 5 3.

(iv) Multiply or divide both sides of the equation with the same non-zero

expression:

Replace2 5

, 11 1

x x

x x

with2 5

( 1) ( 1)1 1

x x x

x x

.

(v) When the left hand side or the right hand side equation is equals to

zero, and the other side of the equation can be factorised, we can use

the law of multiplication ** (see notes below) to solve the equation.

We only need to factorise the non-zero side and set each of the factors

to be equalled to 0.

Replace x ( x 3) 0

with x 0 or x 3 0

** Note: The law of multiplication states that for any two real number a

and b:

If ab 0 then a 0 or b 0 or a b 0.

Example 2.3

Solve the equation 3(2 x) 2 x 1.

SolutionObserve that the unknown x exists on both sides of the equation. The first step is

to group all terms to the left hand side of the equation.

Subtract 2 x 1 from both sides of the equation

3(2 – x) – (2 x – 1) 2 x – 1 – (2 x – 1)




we get

or

3(2 ) (2 1) 0

6 3 2 1 0

7 5 0

7

5

x x

x x

x

x

eliminate brackets

add and subtract monomials of the

same degree

Example 2.4

Solve the equation x2 9 x.

Solution

Subtract 9 x from both sides of the equation

2

2

9 9 9

9 0

( 9) 0

x x x

x x

x x

x

factorise

law of multiplication

x 0 or x 9

Example 2.5

Solve the equation3

33 3

x

x x

Solution

Observe that for this case, x 3 because otherwise it means division by zero.

Therefore the possible domain of the unknown x is { x | x 3}. This equation can

be simplified by multiplying both sides with the denominator factor ( x 3). This

is possible because x 3 0. We get,

3 ( 3)3

x x

x

3( 3

3 x

x

)

or 3( 3) x x 3

Our calculation results in the solution of x 3. But, x 3 does not belong to thedomain of unknown x. Therefore we can conclude that this equation has no

solution.




APPLICATION INVOLVING FIRST DEGREEEQUATIONS

2.2

A first degree equation is also called a linear equation. A linear equation is anequation that involves only one variable (unknown) with power not higher than

one. Any equation of the form

ax b 0

where a, b, and c are real numbers and 0a is called a linear equation. A linear

equation is also referred to as simple equation.

There are many applications involving linear equations. Several examples areshown as follows.

Example 2.6

The perimeter of a square is 20 unit. Find the length of the side of the square.

Solution

We know that each side of a square is the same. Let x denote the length, then we

have 4 x = 20. Thus x = 5.

Example 2.7The price for 2 pencils is RM4. Form a linear equation and hence find the price of

the pencil.

SolutionLet x denote the price of the pencils. So we have 2 x = 4. Thus the price of a pencil,

x = 2.

SECOND DEGREE OR QUADRATIC

EQUATIONS

2.3

Any equation of the form

ax2 bx c 0

where a, b and c are real numbers and 0a is called a second-degree or quadratic

equation.




(a) Solving Quadratic Equations by Factoring

When a quadratic equation is written in the standard form ax2

bx c 0,the expression on the left hand side may be factorised into product of two

polynomials of the first degree.

For example, the standard quadratic equation 3 x2 10 x – 8 0 can be solved

by writing

3 x2 10 x – 8 ( x 4)(3 x – 2)

and then applying the law of multiplication of two numbers, that is:

x 4 0, hence 4 x

or 3 x 2 0, hence2

3 x

Therefore the solutions to the equation are x 4 or 2

.3

x

(b) Solving Quadratic Equations that Cannot be Factorised

Consider the equation

3 x2 4 x – 2 0.

Division by 3 gives us

2 4 20

3 3 x x

or

2 4 2 .3 3

x x

Now, let us add

22

3

to both sides of the equation.




This number is specially chosen so that the left hand side of the equation will

become a perfect square of two identical factors as shown as follows:

2 22 4 2 2 2

3 3 3 3 x x

or

22 10

.3 9

x

Following that

2 1

3 3 x

0

or

2 10.

3 3 x

Therefore, there are two solutions to the quadratic equation, i.e.

2 10

3 3 x or

2 10.

3 3 x

This method is called "completing the square" method.

(c) Solving Quadratic Equations by Formula

The "completing the square" method, when applied to the general quadratic

equation ax2 bx c 0, will produce the formula to obtain the roots, that

is:

2 4.

2

b b ac x

a

(d) Nature of the Roots of Quadratic Equations

When we use the above formula to solve the quadratic equation ax2 bx c

0 we will get two roots, that is:

2 4

2

b b ac x

a

or

2 4.

2

b b ac x

a

Generally, a quadratic equation has two solutions (which is also known as

the roots of the equation). However, the nature of the solution can be

determined earlier by analysing the terms b2 4ac which is called the

discriminant.

If b2

4ac is positive,2

4b ac can be computed and the equation willhave two distinct real number roots.

If b2 4ac is zero, we will have only a single (repeated) root, which is

.2

b

a

If b2 4ac is negative,

2 4b ac does not have a real square root and

hence, the equation will have no real number roots.

Therefore, the conclusion is:

The equation ax2 bx c 0 will have

two distinct real roots when b2 4ac > 0,

two equal roots when b2 4ac 0,

and no real root when b2 4ac < 0.

Example 2.8Determine the nature of the roots of the equation 3 x

2– 5 x – 2 0 and then

find the roots.

Solution

Compare this equation to the standard quadratic equation ax2 bx c 0.

Therefore, for this equation we can recognise that a 3, b 5 and c 2.




These three values of a, b and c will enable us to compute the discriminant

for this equation which is equal to (5)2 4(3)(2) 49. Since the

discriminant is positive, we can conclude that this equation has two distinct

real number roots.

Next, we use the formula to compute the roots, that is:

2 2( 5) ( 5) 4(3)( 2) ( 5) ( 5) 4(3)( 2)

or 2(3) 2(3)

x x

Hence we have two distinct roots

x 2 or 1

3 x

for the quadratic equation 3 x2 – 5 x – 2 0.

Notice that the equation can also be factorised as follows:

3 x2

– 5 x – 2 (3 x 1)( x – 2) 0

which produces the same roots.

(e) Forming Quadratic Equation from its Roots

Let and be the roots of a quadratic equation. This means that in solvingthat quadratic equation we obtain two roots, i.e.

x or x .

But,

If x then x 0,

and

If x then x 0.




From the product rule, it means

( x )( x ) 0

or

2( ) x x 0 , A quadratic equation has resulted from the

roots and

or

x2

– (sum of roots) x (product of roots) 0

We can use the knowledge about a quadratic equation that relates to any

roots a and b in order to facilitate the process of finding its roots by using

factorisation. In trying to factorise the quadratic equation

20ax bx c ,

we can try numbers that are factors of c.

For example, consider the equation

2 6 7 x x 0 ,

We can try the numbers 1 or 7.

Observe that,

(i) x 1, we get 12 6(1) 7 0

(ii) x 1, we get (1)2 6(1) 7 0

(iii) x 7, we get (7)2 6(7) 7 0

(iv) x 7, we get (7)2 6(7) 7 0

It is obvious that 1 and 7 are the roots of the equation because these twovalues satisfy the original quadratic equation. Therefore the equation

can be factorised as:2 6 7 x x 0

26 7 ( 7)( 1) 0 x x x x

or

( 7)( 1) x x 0 .

2.4 INEQUALITIES

Inequalities in a single unknown are statements that involves two expressions

(at least one of these expressions contains the unknown) and both of these

expressions are separated by an inequality symbols of <, , >, or . The process of

solving an equality is the process of finding all values of the unknown that makesthe inequality true. All those values of the unknown is called the solution to the

inequality.

For example below are several inequalities in one unknown:

x – 3 > 5, 4 x 9 1,2

1 8, x 01

x

x

Before we can solve an equality, we need to know about severalproperties of an inequality:

(i) For any pair of real numbers a and b,

a 0.

(ii) For any real number a,

20.a

(iii) The following properties are true for any three real numbers a, b and c:

If a b and b > c, then a > c.

If a b then a c > b c.

(iv) The multiplication property of an inequality describes its property when it is

multiplied by a real number.

If a 0, then ac < bc.

If a bc.

If a > b and if c > 0, then ac > bc.

If a > b and if c < 0, then ac < bc.

(v) If a > 0, then1

0.a

If a < 0, then1

0.a

Two inequalities that have exactly the same set of solutions are called equivalent

inequalities.

Example 2.9

Solve the inequality 3 1 3 . x x

3 1 3

3 1 1 3

3 43 4

2 4

x x

1 x x

x x

x x

x

Add 1 to both sides of the inequality

Subtract x from both sides of the inequality

Divide both sides of the inequality with 2or x > 2

The solution set is { or, in interval notation, all the numbers in the interval

(2,∞) are solutions to the above inequality.

2} x x

Example 2.10

Sometimes we are faced with a combination of inequalities like the following:

–3 < 2 x – 1 < 1

We can still use the same approach as before. However, we must remember that

any operation chosen must be applied to the entire inequality.

–3 < 2 x – 1 < 1Add 1 to each side of the inequality

–3 1 < 2 x – 1 1 < 1 1

–2 < 2 x < 2Divide each side of the inequality with 2

or –1 < x

< 1.

To solve inequalities that contain polynomial of the second degree or higher, and

inequalities that contain irrational expression, we need to re-arrange the

inequality so that the polynomial or the irrational expression is on the left-hand

side and zero is on the right-hand side of the inequality. The following example

shows why we need to take such a step.

Example 2.11

x

2

– 7x > 8.

Solution

The first step is to make the right-hand side equals to zero. Subtract 8 from both

sides of the inequality and we have:

x2 – 7 x > 8 > 0.

Observe that the left-hand side of the inequality can be factorised as

( x – 8)( x 1) > 0.

Now we can use the real-number line to form a graph based on the solution for the

equation

x2

– 7 x – 8 ( x – 8)( x 1) 0.

i.e. x 8 and x 1. These two numbers divide the real-number line into threeintervals:

| |

1 8

–1, 1 8, 8. x x

x

Next, construct a table as follows:

entries for the first row are the three intervals;

entries for the first column are the factors to the left of the inequality and their

products; and

other entries are the sign of each of the factor and the sign of their products for

each of the intervals.

x < –1 –1 < x < 8 x > 8

(x – 8) – –

(x 1) –

(x – 8)(x 1) –

For example, the second row is read as ( x 8) is negative when x < 1 and

1 < x < 8 and ( x 8) is positive when x > 8.

The third row is read as the product of ( x 8) (which is negative when 1 < x < 8)

with the factor ( x 1) (which is positive when 1 < x < 8) is a negative number.

Now, to solve the original inequality, we need ( x 8)( x 1) > 0.

From the table, this is given in the second and fourth column. Therefore, the

solution is obtained when x < 1 or x > 8 and can be written in set notation as

follows:

{ x | x < 1 or x > 8}.

Example 2.12

The same approach can be used to solve the inequality

1 3

2 1 x x




Solution

Subtract3

1 x

from both sides of the equation so that the unknown x are on the

left-hand side,

1 3 3 3

2 1 1 1 x x x x

or 1 3

02 1 x x

common factor of the denominator

eliminate the bracket

( 1) 3( 2)0

( 2)( 1)

x x

x x

1 3 60

( 2)( 1)

x x

x x

or (2 5)

0( 2)( 1)

x

x x

Now, multiply the entire inequality with 1, and from property (iv) above,multiplication with a negative number will affect the sign of the inequality. The

result is

(2 5)0.

( 2)( 1)

x

x x

When we have reached this stage, we can construct a table similar to Example

2.11 above to decide when this division will yield negative values. This sign will

depend on the sign of the numerator and the denominator of the inequality.

Therefore, in this case, we divide the number line into intervals using the numbers

obtained when the numerator and the denominator is set to zero. In this example,

the numerator is zero when 5 ,2

x and the denominator is zero when x 2 or

x 1. These values divide the number line as follows:

5

2 x

52

2 x

2 1 x 1 x

(2x 5) – (x 2) – –

(x 1) – – –

(2 5)

( 2)( 1)

x

x x

– –

It is clear from the table that

(2 5)

( 2)( 1)

x

x x

when

5

2 x

or

2 < x <

1. This

solution can be written in set notation as follows:

5{ or 2

2 x x x

1}

(a) Quadratic Equations

(i) Any equation of the form 2 0ax bx c

where a, b, and c are real numbers and a 0 is called a quadratic

equation.

(b) Solving Quadratic Equations

(i) There are three methods that can be used to solve quadratic equations:

by factoring When a quadratic equation is written in the standard form

2 0ax bx c , the expression on the left hand side may be

factorised into product of two polynomials of the first degree.

by completing the square method.

by using formula,2

4.

2

b b ac x

a

(ii) Nature of the Roots of Quadratic Equations

The equation ax

2

bx c 0 will have

two distinct real roots when b2 4ac > 0,

two equal roots when b2 4ac 0,

and no real root when b2 4ac < 0.

(iii) Forming Quadratic Equation from its Roots

A quadratic equation resulted from the roots and

( x ) ( x ) 0 or 2

( ) x x 0, or

x2

– (sum of roots) x (product of roots) 0

Discriminat

Inequalities

Linear equations

Quadractic equations

Roots

1. Solve( 1)

0.

( 2)( 3)

x

x x

Hence write the solution to1

0.( 2)( 3)

x

x x

2. Let and be roots of a quadratic function2

.ax bx c Find a, b and c if

4 and 2.

topic 2 equations and inequalities

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