topic 2 equations and inequalities
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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.
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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
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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)
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TOPIC 2 EQUATIONS AND INEQUALITIES28
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.
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TOPIC 2 EQUATIONS AND INEQUALITIES 29
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.
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TOPIC 2 EQUATIONS AND INEQUALITIES30
(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.
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TOPIC 2 EQUATIONS AND INEQUALITIES 31
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
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TOPIC 2 EQUATIONS AND INEQUALITIES32
(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.
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TOPIC 2 EQUATIONS AND INEQUALITIES 33
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.
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TOPIC 2 EQUATIONS AND INEQUALITIES34
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
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TOPIC 2 EQUATIONS AND INEQUALITIES 35
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 < b or a b or b < a.
And if we found that b 0, then for any real number a,
a < 0 or a 0 or a > 0.
(ii) For any real number a,
20.a
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TOPIC 2 EQUATIONS AND INEQUALITIES36
(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 and b > c, then a > c.
If a < b then a c < b 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 < b and if c > 0, then ac < bc.
If a < b and if c < 0, then ac > 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
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TOPIC 2 EQUATIONS AND INEQUALITIES 37
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.
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TOPIC 2 EQUATIONS AND INEQUALITIES38
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
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TOPIC 2 EQUATIONS AND INEQUALITIES 39
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:
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TOPIC 2 EQUATIONS AND INEQUALITIES40
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
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TOPIC 2 EQUATIONS AND INEQUALITIES 41
(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.
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