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MODULE 1 Sub Module 1.2
ALGEBRA
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Contents EVALUATING SIMPLE ALGEBRAIC EXPRESSIONS .............................. 1 Addition and Subtraction of Algebraic Expressions ...................... 1
Sub Module 1.2 – Algebra
SIMULTANEOUS EQUATIONS ........................................................ 15 Elimination Method in Solving Simultaneous Equations............ 15
PRACTICE QUESTIONS ................................................................... 17
Multiplication and Division of Algebraic Expressions ................... 1
SECOND DEGREE EQUATIONS WITH ONE UNKNOWN (QUADRATIC
The Use of Brackets ...................................................................... 2
EQUATIONS) ................................................................................. 18
SIMPLE ALGEBRAIC FRACTIONS ....................................................... 4
Solution of Quadratic Equations by Factorization ...................... 18
PRACTICE QUESTIONS ..................................................................... 5 LINEAR EQUATIONS AND THEIR SOLUTION ..................................... 6
Solution of Quadratic Equations by Using the Quadratic Formula .................................................................................................... 19
PRACTICE QUESTIONS ................................................................... 20
Transposing Formulae................................................................... 6 Solution of Linear Equations ......................................................... 7
PRACTICE QUESTIONS ..................................................................... 8 INDICES AND POWERS .................................................................... 9 The Laws of Indices ..................................................................... 10 Substitution ................................................................................. 11
PRACTICE QUESTIONS ................................................................... 11 NUMBER SYSTEMS ........................................................................ 12 Binary Number System ............................................................... 12 Hexadecimal Number System ..................................................... 13
PRACTICE QUESTIONS ................................................................... 14
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EVALUATING SIMPLE A LGEBRA IC EXPRESSIONS
Multiplication and Division of Alg
Ad di ti on and Sub tr act io n o f A lg ebr aic Exp res si on s
The rules are exactly the same as those used with directed numbers:
Like terms are numerical multiplies of the same algebraic quantity. Thus 7x, 5x and -3x are three like terms.
(+ x)(+ y)
An expression consisting of like terms can be reduced to a
5x 3y
single term by summing (i.e. adding and/or subtracting) the numerical coefficients together. Thus:
(x)(-y) = - (xy) = - xy
7x - 5x + 3x = (7 - 5 + 3)x = 5x 2
2
3b + 7b
2
= (3 + 7)b
-3y - 5y q - 3q
= =
(-3 -5)y (1 - 3)q
= 10b
2
= -8y = -2q
Only like terms can be added or subtracted. Thus (7a + 3b - 2c) is an expression containing three unlike terms and it cannot be simplified any further. Similarly with (8a2b + 7ab3 + 6a2b2) which are all unlike terms. It is possible to have several sets of like terms in an expression and each set can then be simplified. 8x + 3y - 4z - 5x + 7z - 2y + 2z = (8 - 5)x + (3 - 2)y + (-4 + 7 + 2)z = 3x + y + 5z
= + (xy) = + xy = xy
3 x y = 15xy
= 5
(2x)(- 3y)
= - (2x)(3y) = -6xy
(- 4x)(2y)
= - (4x)(2y) = -8xy
(- 3x)(- 2y)
= + (3x)(2y) = 6xy
+x +y
= +
- 3x 2y
= -
3x 2y
- 5x - 6y
= +
5x 6y
-4x 3y
= - 4x 3y
x y
=
x y
=
5x 6y
When multiplying expressions containing the same symbols, indices are used:
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mm = m2
The Use of Brackets
3m 5m = 3 m 5 m = 15 m 2 (- m) m2 = (- m) m m = - m 5m2n 3mn3 = 5 m m n 3 m n nn = 15m3n4 3mn (-2n2) = 3 m n (- 2) n n= - 6mn3 When dividing algebraic expressions, cancellation between numerator and denominator is often possible. Cancelling is equivalent to dividing both numerator and denominator by the same quantity: pq p
p q
p
q
3p 2 q 3 p p q 3p p 6pq 2 6 p q q 6q 2q 2
18x 2 y z
Brackets are used for convenience in grouping terms together. When removing brackets each term within the bracket is multiplied by the quantity outside the bracket: 3(x + y) = 3x + 3y 5(2x + 3y) = 5 2x + 5 3y = 10x + 15y 4(a - 2b) = 4 a - 4 2b = 4a - 8b m(a + b) = ma + mb 3x(2p + 3q) = 3x 2p + 3x 3q = 6px + 9qx 4a(2a + b) = 4a 2a + 4a b = 8a2 + 4ab When a bracket has a minus sign in front of it, the signs of all the terms inside the bracket are changed when the bracket is removed. The reason for this rule may be seen from the following example: - 3(2x - 5y)
18 x x y y z 3xy
6xyz
6 x y z
= (- 3) 2x + (- 3) (- 5y) = - 6x + 15y
- (m + n) = - m - n - (p - q) = -p + q - 2(p + 3q) = - 2p - 6q
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When simplifying expressions containing brackets first remove the brackets and then add the like terms together: (3x + 7y) - (4x + 3y)
= 3x + 7y - 4x - 3y = - x + 4y
3(2x + 3y) - (x + 5y)
= 6x + 9y - x - 5y = 5x + 4y
2(5a + 3b) + 3(a - 2b)
= 10a + 6b + 3a - 6b = 13a
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SIMPLE ALGEBRA IC FRACTIONS
Example: Simplify
The addition for algebraic fractions is the same as for arithmetical fraction, that is:
Find the L.C.M. of the denominators. Express each fraction with the
m 2m + n m - 2n + . 12 4 3
The L.C.M. of 12, 4 and 3 is 12. m 2m + n m - 2n + 12 4 3 m + 3(2m + n ) - 4 (m - 2n)
common
denominators. Add or subtract the fractions.
=
12 m + 6m + 3n - 4m + 8n 12 3m + 11n = 12
a b c Example: Simplify + - . 2 3 4
=
The L.C.M. of 2,3 and 4 is 12. a b c + 2 3 4
Example: Simplify
6a 4b 3c + 12 12 12 6a + 4b - 3c = 12 =
2 3 4 + + . x 2x 3x
The L.C.M. of x, 2x and 3x is 6x. 2 3 4 12 + 9 + 8 + + = x 2x 3x 6x
The multiplication for algebraic fractions is also the same as for arithmetical fraction, that is, after using the cancellation rule of fraction multiplication, multiply all the numerators to get the numerator of the result and all the denominators to get the denominator of the result. x 7 2 Example: Simplify . 4 y x
x =
29 6x
The sign in front of a fraction applies to the fraction as a whole. The line which separates the numerator and denominator acts as a bracket.
7
y
4
x 2
4
2
x
7
y
x
7 2y
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Practice Questions
1. Simplify the following: 2. (a) 7x + 11x (b) 7x - 5x (c) - 8x 3x (d) 6b2 - 4b2 + 3b2 (- 2b) (e) 2x5y (f) (- 3a)
2
3
(g) 8mn- 5) (- 3m n ) (i) 2k(k (k) 3x2(x2 - 2xy + y2) (m) 5(2x - y) - 3(x + 2y) (o)
3 5 4 + y 3y 5y
+
(h) 3a) ÷+ (-4)3b) (j) -(-3y(3x (l) 3(x + 4) - (2x + 5) x x x (n) + + 3 4 5
3x 5 y 2 y 6 x
(p)
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Step 1: Since there are no roots get rid of the fraction by multiplying both sides of the equation by ( R r )
LINEAR EQUATIONS AND THEIR SOLUTION An equation is the relation of equality between two or more expressions.
V ( R r ) 2R
If the degree of the equation, i.e. the highest power among the variable of each term on both sides in the equation is 1, then the equation is said to be linear equation. For example, 2x + 5 = 0,
Step 2: Clear the bracket
x – 2y = 7 are linear equations.
Step 3: Collect the terms containing R on the LHS.
VR Vr 2R
VR 2R Vr
Transposi ng Formulae The formula y ax b has y as is subject. By rearranging this formula we could make x the subject. The rules for transforming a formula are:
1) 2) 3) 4)
Remove square roots or other roots Get rid of brackets Clear brackets Collect together the terms containing the required subject 5) Factorize if necessary 6) Isolate the required subject
Step 4: Factorize the LHS.
R(V 2) Vr R Step 5: Isolate by dividing both sides of the equation by (V 2). R
Vr V 2
Although we used five steps to obtain the required subject, in very many cases far fewer steps are needed. Nevertheless, you should work through the steps in order given.
These steps should be performed in the order given. Example: (a) Transpose the formula V
2R
Rr
to make R the
subject.
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Solution of L inear E quations
−5x= −9
The solution of linear equation(s) depends upon the number of variable(s) used in the equation(s). The number of equations must be equal to or greater than the variables used in a system of equations.
and on division by −5, x= −9/-5 or,
x = 9/5
Example: Solve the following equations: (a). 3x – 4 = 6 − 2x; (b) 8 + 4(x − 1) − 5(x − 3) = 2(5 + 2x). (a) For this equation, all we need to do is to collect all terms involving the unknown x on to the left-hand side of the equation, simply by using our rules for transposition of formula. Then:
3x + 2x − 4 = 6
so,
3x + 2x = 6 + 4
or,
5x = 10
and so
x=2
(b) In this equation first we need to multiply out the brackets, then collect all terms involving the unknown x onto one side of the equation and the numbers onto the other side, then divide out to obtain the solution. So: 8 + 4(x − 1) − 5(x − 3) = 2(5 + 2x) 8 + 4x − 4 − 5x + 15 = 10 + 4x 4x − 5x − 4x = 10 + 4 − 8 − 15 ISO 9001:2008 Certified
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Practice Questions
1. Transpose the equation V Ah for h . 2. The formula P
RT V
is used in connection with the
expansion of gases. Transpose it to find the value of R . 3. Solve the following equations: (a) 2x 5 7 (b) 3x 1 6 x 10 (c)
1
x
3 2x
5 2
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INDICES AND POWERS When a number is the product of the same factor multiplied by itself, this number is called a power of the factor. For example, we know that 3 × 3 = 9. Therefore, we can say that 9 is a power of 3. To be precise, it is the second power of 3, because two 3s are multiplied together to produce 9. Similarly, 16 is the second power of 4. We may use literal terminology to generalize the
Now, providing the base of two or more numbers expressed in index (exponent) form are the same, we can perform multiplication and division on these numbers, by adding or subtracting the indices accordingly. Consider the following literal numbers in index form:
x2 x2 x x x x x4
relationship between powers and factors. So the second power of a means a × a or ( a · a ), this is written 2
as a , where a is known as the base (factor) and 2 is the exponent (or index). Thus writing the number 9 in exponent form we get 9 = 32 where; 9 is the second power, 3 is the base (factor) and 2 is the exponent (index). The above idea can be extended to write arithmetic numbers in exponent or index form. For example, 52 = 25, 92 = 81 and 33 = 27. Notice that the second power of 5 gives the number 25 or 5 × 5 = 25; similarly 33 means the third power of 3, literally 3 × 3 × 3 = 27. The idea of powers and exponents (indices) can be extended to literal numbers. For example: a a a a a or
a 5 or in general a m where a is the base (factor) and the m exponent m (or index) is any positive integer. a means a used
x 2 x 4 ( x x) ( x x x x)
x x x x x x x6 x2 x2 x2 x4
x x x x
x0 1
x x x x x x
1
x x
x 2
For multiplication of numbers with the same base, we add the indices and for division of numbers with the same base, we subtract the indices in the denominator from those in the numerator.
as a factor m times and is read as the “mth power of a”. Note
We will now generalize our observations and so formulate the
that since any number used as a factor once would simply be the number itself, the index (exponent) is not usually written; in
laws of indices.
1
other words a means a .
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The Laws of Indices In the following laws a is the common base, m and n are the indices (exponents). Each law has an example of its use alongside:
1. a a a m
am
2.
an
n
mn
a m n
22 24 224 26 64 34 32
342 32 9
m n mn 3. (a ) a
(2 2 ) 3 2 23 2 6 64
4. a 1
5 1
0
0
m
4
m n n 5. a a
a n
6.
1
an
27 3 3 27 4 34 81 6 2
1 62
Law 2: We have again used, when dividing numbers with a common base in this case, the base is 3. Note that since division is the opposite arithmetic operation to multiplication. It follows that we should perform the opposite arithmetic operation on the indices, that of subtraction. Remember we always subtract the index in the denominator from the index in the numerator. Law 3: It is concerned with raising the powers of numbers. Do not mix this law up with law 1. When raising powers of numbers in index form, we multiple the indices. Law 4: As you have also met, this law simply states that any number raised to the power 0 is always 1. Knowing that any number divided by itself is also 1, we can use this fact to show that a number raised to the power 0 is also 1. What we need to do is use the second law concerning the division of numbers in index form.
1
We know that
36
9
We need to study these laws carefully in order to understand the significance of each. Law 1: As youinhave met, it a enables usbase. to multiply numbers given indexalready form that have common In the example the common base is 2, the first number raises this base (factor) to the power 2 and the second raises the same base to the power 3. In order to find the result we simply add the indices.
9
1
32 or 3
2
0
which shows that 3
1 and in fact because we have used the
second law of indices, this must be true in all cases. Law 5: This, rather complicated looking, law simply enables us to find the decimal equivalent of a number in index form; where the index is a fraction. All that you need to remember is that the
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index number above the fraction line is raised to that power and the index number below the fraction line has that number root.
These missed multiplication signs must reappear when the numbers are substituted for the symbols.
2
So for the number 8 3 , we raise 8 to the power 2 and then take the cube root of the result. It does not matter in which order we perform these operations. So we could have just as easily taken the cube root of 8 and
(a)
2y + 4 = 2 4 + 4 = 8 + 4 = 12
(b)
3y + 5z = 3 4 + 5 5
(c)
8 - x = 8 - 3 = 5
(d)
y x
(e)
3y + 2z x + z
then raised it to the power 2. Law 6: This is a very useful law, when you wish to convert the division of a number to multiplication. In other words, bring a number from underneath the division line to the top of the division line. As the number crosses the line we change the sign of its index. This is illustrated in the example, which accompanies this law.
The process of finding the numerical value of an algebraic expression for given values of the symbols that appear in it is called substitution.
1 = 1 3 3 4 + 2 5 3 + 5 12 + 10 22 = = 8 8
=
3 = 2 4
(b) 3y + 5z (e)
1. Simplify: 3
(a) 3
Example: If x = 3, y = 4 and z = 5, find the values of:
1 23 8
b b b (c) 8
3y + 2z x + z
Note that multiplication signs are often missed out when writing algebraic expressions so that, for instance, 2y means 2 y.
2 7
1 2 5
2
16 4 (b) 81
4
(c)
2
b 0 b 5 2. Simplify:
(a) (2 2 ) 6 3 24
(b) 1 2 12 11 2
3
3
3. If a = 2, b = 3 and c = 5, find the values of the following:
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Practice Questions
Substitution
(a) 2y + 4 - x y (d) x
=
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(a) a + 2b + 5c
Sub Modul e 1.2 – Algebra
abc (b) 6
(c)
5a + 9b + 8c a+b+c NUMBER SYSTEMS The decimal system of numbers we have been studying up till now use the integers 0 – 9. There are in fact 10 integers and for this reason we often refer to the decimal system as the denary (ten) system. Thus, for example, the denary number 245.5 is equivalent to: (2 × 102) + (4 × 101) + (5 × 100) + (5 × 10−1) This arrangement of the number consists of an integer ≥1.0 and ≤10.0 multiplied by the base raised to the power. Binary Number System In the binary system of numbers, the base is 2 and so, for example, the denary number 43 to the base 10, written as 4310 is equivalent to the number: 25 + 23 + 21 + 20 = 3210 + 810 + 210 + 110 As a reminder and source of reference the binary and denary equivalents for some important numbers related to computing are detailed below:
To convert denary to binary, we repeatedly divide by 2 and note the remainder at each stage. For example, to convert the number 2510 to binary, we proceed as follows: 25/2 = 12 ----12/2 = 6 -----6/2 = 3 -------3/2 = 1 -------1/2 = 0 --------
Least significant digit (LSD)
Most significant digit (MSD)
The binary2 equivalent of 2510 is 110012. Note the order in which the digits of the binary number are laid out from the MSD to the LSD; i.e. in reverse order to the successive division. To convert binary to denary, we lay out the number in successive powers. For example, to convert binary number 11012 into denary, we proceed as follows: 11012 = (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20) = (1 × 8) + (1 × 4) + (0 × 2) + (1 × 1) = 8 + 4 + 0 + 1 = 1310
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When numbers are placed in binary form, we can see from above that they consist of a number of ones (1) and noughts (0). If in electronic logic circuits we allow the binary digit “1” to represent “ON” and the binary digit “0” to represent “OFF”. We can apply this binary code to electronic logic systems. It is this powerful application of binary numbers that makes their study important.
Hexadecimal Number System In order to get more digital information down computer communication lines, we can use another number system that allows us to send 16 individual pieces of information (bytes) down parallel lines, all at the same time. This type of communication may be coded using hexadecimal representation. Thus, for hexadecimal numbers their base is 16. However, because in our decimal number counting system we only have 10 digits (0 – 9), we make up for this in the hexadecimal system by allocating capital letters to the remaining decimal numbers 10 – 15 (remembering that decimal zero is counted as part of the 16 digit base). Hexadecimal representation, together with their denary and binary equivalents, are shown in the table below.
Sub Module 1.2 – Algebra
Denary10
Binary2
Hexadecimal16
0
0000
0
1
0001
1
2
0010
2
3
0011
3
4
0100
4
5
0101
5
6
0110
6
7
0111
7
8
1000
8
9
1001
9
10
1010
A
11
1011
B
12
1100
C
13
1101
D
14
1110
E
15
1111
F
Thus in a similar manner to before, the denary number 54210 may be represented as: 54210 = (5 × 102) + (4 × 101) + (2 × 100) which is equivalent to ISO 9001:2008 Certified
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21E16 = (2 × 162) + (1 × 161) + (E × 160).
The denary10 equivalent of hexadecimal number BA4516 is 4768510.
To convert denary to hexadecimal, we repeatedly divide by 16 in a similar manner to the way in which we converted denary to binary. To convert the denary number 513610 to hexadecimal, we proceed as follows:
Practice Questions
1. Convert the following denary numbers into binary: (a) 17 (b) 23 (c) 40 2. Convert the following binary numbers into denary:
5136/16 = 321 -----321/16 = 20 --------20/16 = 1 -----------1/16 = 0 --------------
remainder 0 remainder 1 remainder 4 remainder 1
LSD
(a) 1011 MSD
So the hexadecimal16 equivalent of 513610 is 141016. Similarly, to convert the number 9410 to hexadecimal, we proceed as follows: 94/16 = 5 ------------- remainder 14 (=E16) 5/16 = 0 -------------- remainder 5
3. Convert the hexadecimal: (a) 5890
(b) 11111 following
(c) 1010101
denary
numbers
into
(b) 16892
4. Convert the following hexadecimal numbers into denary: (a) 6E (b) CF18
So the hexadecimal16 equivalent of 9410 is 5E16. To convert hexadecimal to denary, we proceed in a similar manner as for binary to denary. For example, to convert BA4516 to denary, we proceed as follows: BA4516=(B × 163) + (A × 162) + (4 × 16 1) + (5 × 160) = (11 × 4096) + (10 × 256) + (4 × 16) + (5 × 1) = (45056) + (2560) + (64) + (5) = 4768510 ISO 9001:2008 Certified
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We can now eliminate equation [3].
x
3x + 21y 3x + 4y 17y
45 11 34 2
by subtracting equation [1] from
SIMULTANEOUS EQUATIONS Consider the two equations: 2x + 3y = 13 3x + 2y = 12
[1] [2]
Each equation contains the unknown quantities x and y. The solutions of the equations are the value if x and y which satisfy both equations. Equations such as these are called simultaneous equations.
y
= = = =
[3] [1]
To find x we substitute for y = 2 in either of the srcinal equations. Thus, substituting for y = 2 in equation [1], 3x + 4 2 3x + 8
There are various methods of solving simultaneous equations. We will only discuss one method, that is elimination method.
= = 3x = 3x = x=
11 11 11 - 8 3 1
Elimination Method in Solving Simultaneous Equations Hence the solutions are: The method will be shown by considering the following examples.
Hence the solutions are correct since the L.H.S. and R.H.S. are equal.
Example: Solve the equations: 3x + 4y x + 7y
= 11 = 15
[1] [2]
If we multiply equation [2] by 3 we shall have the same coefficient of x in both equations: 3x + 21y
= 45
x = 1 and y = 2
[3]
Example: Solve the equations: 5x + 3y 4x + 7y
[1] [2]
The same coefficient of x can be obtained in both equations if equation [1] is multiplied by 4 (the coefficient of x in equation
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[2]) and equation [2] is multiplied by 5 (the coefficient of x in equation [1]).
Multiply equation [1] by 4, 20x + 12y
= 116
[3]
Multiply equation [2] by 5, 20x + 35y
= 185
7x + 4y= 41
[1]
4x - 2y = 2
[2]
In these equations it is easier to eliminate y because the same coefficient of y can be obtained in both equations by multiplying equation [2] by 2.
[4]
Subtracting equation [3] from equation [4], 23y y
Example: Solve the equations:
= 69 = 3
Multiplying equation [2] by 2, 8x - 4y
= 4
Adding equation [1] and [3], 15x
Substituting for y = 3 in equation [1], 5x + 3 3 = 29 5x + 9 = 29 5x = 20 x = 4 Hence the solutions are: y = 3 and x = 4
[3]
x
= 45 = 3
Substituting for x = 3 in equation [1], 7 3 + 4y = 41 21 + 4y = 41 4y = 20 y = 5 Hence the solutions are: x = 3 and y = 5
Check in equation [2], L.H.S. = 4 4 + 7 3 = 16 + 21 = 37 = R.H.S.
Check in equation [2], L.H.S. = 4 3 - 2 5 = 12 - 10 = 2 = R.H.S.
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x
= 2
Example: Solve the equations: Substituting for x = 2 in equation [3], 2x y 7 = 3 4 12 3x 2y 3 = 4 5 10
[1] [2]
8 2 - 3y = 7 16 - 3y = 7 - 3y = -9 y = 3
It is best to clear each equation of fractions before attempting to solve.
Hence the solutions are: x = 2 and y = 3
In equation [1] the L.C.M. of the denominators is 12. Hence by multiplying equation [1]by 12, Practice Questions 8x - 3y
= 7
[3]
In equation [2] the L.C.M. of the denominators is 20. Hence by multiplying equation [2] by 20, 15x - 8y
= 6
[4]
1. Solve the following equations for x and y and check the solutions: (a) 3x + 2y = 7 ; x+y =3 (b) x - 3y = 1 ; x + 3y = 19 (c) x + 3y = 7 ; 2x - 2y = 6
We now proceed in the usual way. Multiplying equation [3] by 8, 64x - 24y
= 56
[5]
Multiplying equation [4] by 3, 45x - 24y
= 18
[6]
Subtracting equation [6] from equation [5], 19x
= 38
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Sub Module 1.2 – Algebra
SECOND DEGREE EQUATIONS WITH ONE UNKNOWN (QUADRATIC EQUATIONS) A quadratic equation is one in which the unknown variable is raised to the second power (or degree). For example, the equation x2 =4 is perhaps one of the simplest of quadratic equations. We can solve this equation by taking the square root of both sides, as:
or
x2 4 x 2
term a = 1. What about the constant b? Well there is no x term in our equation so b = 0. What about the constant c? Our equation is not in standard form, because the equation should be equated to zero. Then in standard form our equation becomes x2 – 4 = 0 by simple transposition! So now we know that for our equation the constant term c = −4. There are several ways in which quadratic equations may be solved, that is finding the values of the unknown variable. We shall concentrate on just two methods of solution; factorization, and using the quadratic formula. Solution of Quadratic Equations by
Factori zation
Note that even for this simple equation there are two possible solutions, either x 2 or x 2 , remembering the laws of signs!
Example: Solve the equation 3x
In general, a quadratic equation is of the type
A quadratic equation can not be solved until it is converted into the standard form. So first we have to convert the given equation in standard form as:
ax2 bx c 0 ,
3x 2 5 2x 4
where the constants a, b and c can take any numerical value, positive or negative, decimal or fraction.
or
Like linear equations, quadratic equations do not always appear in standard form, i.e. they are not always arranged in exactly the 2 same order as their qualifying equation, ax bx c 0 . How is our simple equation x2 = 4 related to its qualifying equation? Well the coefficient of x2 that is the number multiplying the x2
5 2x 4 .
3x2 5 2x 4 0
3x 2x 1 0 or The second step is to factorize the L.H.S. of the equation, as: 2
2
3x
or
3x x 1 0
3x( x 1) 1( x 1) 0
or
( x 1)(3x 1) 0
So,
either x 1 0 , giving x 1
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Sub Module 1.2 – Algebra
or 3x 1 0 , giving x
1
Solution of Quadratic Equations by Using the Quadratic Formula
.
3
Thus, the solution of the given equation is either x 1 or 1 x 3 Example: Solve the equation x
2
4x 4 0 .
It is not always possible to solve quadratic equations by factorization. When we cannot factorize a quadratic expression, we may resort to use of the standard formula. Now we know that the standard form of the quadratic equation is
ax2 bx c 0 and it can be shown that the solution of this equation is:
The given equation is already in the standard form; so we will factorize it, as:
x 2 4x 4 0 or or or
x 2 2 x 2x 4 0 x( x 2) 2( x 2) 0 ( x 2)(x 2) 0
If we get the same factors, then we combine them as a square power of single factor and solve it, as:
( x 2) 2 0
x
b b 2 4 ac 2a
The above formula, commonly known as ‘Quadratic Formula, may look complicated but it is relatively simple to use. The coefficients a, b and c are the same coefficients, as in the standard form of the quadratic. So in finding a solution for the variable x, all we need to do is substitute the coefficients into the above formulae, for the quadratic equation we are considering. All we need to remember is that, before using the above formula, always put the equation to be solved, into standard form. Example: Solve the equation 5x( x 1) 2x(2x 1) 20 .
( x 2) 2 0
or or
x20
or
x
2
A quadratic equation can not be solved until it is converted into the standard form. So first we have to convert the given equation in standard form as:
5x( x 1) 2x(2x 1) 20
Thus, the given equation has only one solution as x 2 .
5x 2 5x 4x2 2x 20
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Module 1 – MATHEMATICS
Category – A/B 1/B2
Sub Module 1.2 – Algebra
or
x 2 7 x 20 0
Here,
a 1 , b 7 and c 20
Practice Questions
1. Solve the following equations: (a) 6x x 2 0 2
b b 4 ac 2
x
2a
Since
(b) 2x 20x 32 2
2
or
x 7 (7) 4(1)(20) 2(1) x
7 49 80
x or
1
x2
2 3
0
2 2
7 11 .358 2
or or
7 11 .358
or
x
1
x 1
7 129
or
x
(d)
2
or
x
(c) x 1 3 x
4.358 2
x or
7 11.358 x 2 or 18 .358 6
x 2.18 or x 9.18
Thus, the given equation has only one solution as x 2.18 or x 9.18 .
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