note

1. 1
2.0 ALGEBRA
2.1 Solution of Simultaneous Equations with Three Unknowns
2.1.1 Introduction to Simultaneous Equations
Only one equation is necessary when finding the value of a single unknown quantity as with
simple equations. However, when an equation contains two unknown quantities it has an
infinite number of solutions. When two equations are available connecting the same two
unknown values then a unique solution is possible. Similarly, for three unknown quantities it is
necessary to have three equations in order to solve for a particular value of each of the unknown
quantities, and so on.
Equations that have to be solved together to find the unique values of the unknown quantities,
which are true for each of the equations, are called simultaneous equations.
Two methods of solving simultaneous equations analytically are:
(a) by substitution, and (b) by elimination.
Note:
Own your own read on the above two methods and matrices.
I. SOLUTION OF SIMULTANEOUS EQUATIONS USING MATRICES
The procedure for solving linear simultaneous equations in three unknowns using matrices is:
(i) write the equations in the form
a1x +b1 y +c1 z = d1
a2x +b2 y +c2 z = d2
a3x +b3 y +c3 z = d3
(ii) write the matrix equation corresponding to
these equations, i.e.
(
𝑎1 𝑏1 𝑐1
𝑎2 𝑏2 𝑐2
𝑎3 𝑏3 𝑐3
)×(
𝑥
𝑦
𝑧
)=(
𝑑1
𝑑2
𝑑3
)
(iii) determine the inverse matrix of
(
𝑎1 𝑏1 𝑐1
𝑎2 𝑏2 𝑐2
𝑎3 𝑏3 𝑐3
)
(iv) multiply each side of (ii) by the inverse
matrix, and
(v) solve for x, y and z by equating the corresponding

2. 2
elements.
Problem 1.
Use matrices to solve the simultaneous equations:
x + y +z −4 =0 (1)
2x −3y +4z −33 =0 (2)
3x −2y −2z −2 =0 (3)
(i) Writing the equations in the a1x +b1 y +c1 z = d1
form gives:
x + y +z = 4
2x −3y +4z = 33
3x −2y −2z = 2
(ii) The matrix equation is
(
1 1 1
2 −3 4
3 −2 −2
)×(
𝑥
𝑦
𝑧
)=(
4
33
2
)
(iii) The inverse matrix of
A =(
1 1 1
2 −3 4
3 −2 −2
)
is given by A−1
=
𝑎𝑑𝑗 𝐴
|𝐴|
The adjoint of A is the transpose of the matrix of the cofactors of the elements.
The matrix of cofactors is
(
14 16 5
0 −5 5
7 −2 −5
)
and the transpose of this matrix gives
adj A =(
14 0 7
16 −5 −2
5 5 −5
)
The determinant of A, i.e. the sum of the products of elements and their cofactors, using a first
row expansion is
1|
−3 4
−2 −2
|−1|
2 4
3 −2
|+1|
2 −3
3 −2
|= (1×14)−(1×(−16))+(1×5) = 35
Hence the inverse of A,
A−1
=
1
35
(
14 0 7
16 −5 −2
5 5 −2
)

3. 3
(iv) Multiplying each side of (ii) by (iii), and remembering that A× A−1
= I , the unit matrix,
gives
(
1 0 0
0 1 0
0 0 1
)×(
𝑥
𝑦
𝑧
)=
1
35
(
14 0 7
16 −5 −2
5 5 −5
)×(
4
33
2
)
(
𝑥
𝑦
𝑧
) =
1
35
(
(14 × 4) + (0 × 33) + (7 × 2)
(16 × 4) + ((−5) × 33) + ((−2) × 2
(5 × 4) + (5 × 33) + ((−5) × 2)
)=
1
35
(
70
−105
175
) =(
2
−3
5
)
(v) By comparing corresponding elements, x = 2, y=−3, z = 5, which can be checked in the
original equations.
ASSIGNMENT 1
In Problems 1 to 3 use matrices to solve the simultaneous equations given.
1. x +2y +3z = 5
2x −3y −z = 3
−3x +4y +5z = 3
2. 3a +4b−3c = 2
−2a +2b +2c = 15
7a −5b+4c = 26
3. p +2q +3r +7.8 = 0
2 p +5q −r −1.4 = 0
5 p −q +7r −3.5 = 0
4. The relationship between the displacement, s, velocity, v, and acceleration, a, of a piston
is given by the equations:
s +2v +2a = 4
3s −v +4a = 25
3s +2v −a =−4
Use matrices to determine the values of s, v and a.
5. In a mechanical system, acceleration 𝑥̈, velocity 𝑥̇ and distance x are related by the
simultaneous equations:
3.4𝑥̈ +7.0𝑥̇ −13.2x =−11.39
−6.0𝑥̇ +4.0𝑥̇ +3.5x = 4.98
2.7𝑥̈ +6.0𝑥̇ +7.1x = 15.91
Use matrices to find the values of ¨ x, ˙ x and x
II. SOLUTION OF SIMULTANEOUS EQUATIONS USING DETERMINANTS
When solving simultaneous equations in three unknowns using determinants:

4. 4
(i) Write the equations in the form
a1x +b1 y +c1z +d1 = 0
a2x +b2 y +c2 z +d2 = 0
a3x +b3 y +c3z +d3 = 0
and then
(ii) the solution is given by
𝑥
𝐷𝑥
=
−𝑦
𝐷𝑦
=
𝑧
𝐷𝑧
=
−1
𝐷
where D = x|
𝑏1 𝑐1 𝑑1
𝑏2 𝑐2 𝑑2
𝑏3 𝑐3 𝑑3
|
i.e. the determinant of the coefficients obtained by covering up the x column.
Dy =|
𝑎1 𝑐1 𝑑1
𝑎2 𝑐2 𝑑2
𝑎3 𝑐3 𝑑3
|
i.e., the determinant of the coefficients obtained by covering up the y column.
Dz =|
𝑎1 𝑏1 𝑑1
𝑎2 𝑏2 𝑑2
𝑎3 𝑏3 𝑑3
|
i.e. the determinant of the coefficients obtained by covering up the z column.
and
D =|
𝑎1 𝑏1 𝑐1
𝑎2 𝑏2 𝑐2
𝑎3 𝑏3 𝑐3
|
i.e. the determinant of the coefficients obtained by covering up the constants column.
Problem 1.
A d.c. circuit comprises three closed loops. Applying Kirchhoff’s laws to the closed loops gives
the following equations for current flow in milliamperes:
2I1 +3I2 −4I3 = 26
I1 −5I2 −3I3 =−87
−7I1 +2I2 +6I3 = 12
Use determinants to solve for I1, I2 and I3
(i) Writing the equations in the a1x +b1 y +c1 z +d1 = 0 form gives:
2I1 +3I2 −4I3 −26 = 0
I1 −5I2 −3I3 +87 = 0
−7I1 +2I2 +6I3 −12 = 0
(ii) The solution is given by

5. 5
𝐼1
𝐷𝐼1
=
−𝐼2
𝐷𝐼2
=
𝐼3
𝐷𝐼3
=
−1
𝐷
where DI1 is the determinant of coefficients obtained by covering up the I1 column, i.e.
DI1=|
3 −4 −26
−5 −3 87
2 6 −12
|
= (3)|
−3 87
6 −12
|−(−4)|
−5 87
2 −12
|+(−26)|
−5 −3
2 6
|
= 3(−486)+4(−114)−26(−24)
= −1290
DI2=|
2 −4 −26
1 −3 87
−7 6 −12
|
= (2)(36−522)−(−4)(−12+609)+(−26)(6−21)
=−972+2388+390
= 1806
DI3=|
2 3 −26
1 −5 87
−7 2 −12
|
= (2)(60 −174)−(3)(−12+609)+(−26)(2−35)
=−228−1791+858 = −1161
and D =|
2 3 −4
1 −5 −3
−7 2 6
|
= (2)(−30+6)−(3)(6−21)+(−4)(2−35)
=−48+45+132 = 129
Thus
𝐼1
−1290
=
−𝐼2
1806
=
𝐼3
−1161
=
−1
129
giving
I1 =
−1290
−129
=10mA,
I2 =
1806
129
= 14mA
and I3 =
1161
129
= 9mA

6. 6
ASSIGNMENT 2
In Problems 1 to 2 use determinants to solve the simultaneous equations given.
1. 3x +4y +z = 10
2x −3y +5z +9 = 0
x +2y −z = 6
2. 1.2 p −2.3q −3.1r +10.1 = 0
4.7 p +3.8q −5.3r −21.5 = 0
3.7 p −8.3q +7.4r +28.1 = 0
8. Kirchhoff’s laws are used to determine the current equations in an electrical network and
show that
i1 +8i2 +3i3 =−31
3i1 −2i2 +i3 =−5
2i1 −3i2 +2i3 = 6
Use determinants to solve for i1, i2 and i3.
III. SOLUTION OF SIMULTANEOUS EQUATIONS USING CRAMERS RULE
Cramers rule states that if
a11x +a12 y +a13z = b1
a21x +a22 y +a23z = b2
a31x +a32 y +a33z = b3
then x =
𝑫𝒙
𝑫
, y =
𝑫𝒚
𝑫
and z =
𝑫𝒛
𝑫
where D =|
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
|
Dx =|
𝑏1 𝑎12 𝑎13
𝑏2 𝑎22 𝑎23
𝑏3 𝑎32 𝑎33
|
i.e. the x-column has been replaced by the R.H.S. b column
Dy =|
𝑎11 𝑏1 𝑎13
𝑎21 𝑏2 𝑎23
𝑎31 𝑏3 𝑎33
|
i.e. the y-column has been replaced by the R.H.S. b column
Dz =|
𝑎11 𝑎12 𝑏1
𝑎21 𝑎22 𝑏2
𝑎31 𝑎32 𝑏3
|
i.e. the z-column has been replaced by the R.H.S. b column.

7. 7
Problem 1.
Solve the following simultaneous equations using Cramers rule
x + y +z = 4
2x −3y +4z = 33
3x −2y −2z = 2
D =|
1 1 1
2 −3 4
3 −2 −2
|
= 1(6−(−8))−1((−4)−12)+1((−4)−(−9)) = 14+16+5 = 35
Dx =|
4 1 1
3 −3 4
2 −2 −2
|
= 4(6−(−8))−1((−66)−8)+1((−66)−(−6)) = 56+74−60 = 70
Dy =|
1 4 1
2 33 4
3 2 −2
|
= 1((−66)−8)−4((−4)−12)+1(4 −99)=−74+64−95 = −105
Dz =|
1 1 4
2 −3 33
3 −2 2
|
= 1((−6)−(−66))−1(4 −99)+4((−4)−(−9)) = 60+95+20 = 175
Hence
x =
𝐷𝑥
𝐷
=
70
35
= 2, y =
𝐷𝑦
𝐷
=
−105
35
= −3 and z =
𝐷𝑧
𝐷
=
175
35
= 5
ASSIGNMENT 3
In Problems 1 to 2 use Cramers rule to solve the simultaneous equations given.
1. x +2y +3z = 5
2x −3y −z = 3
−3x +4y +5z = 3
2. 3a +4b−3c = 2
−2a +2b +2c = 15
7a −5b+4c = 26

8. 8
2.2 Solution of Quadratic Equations
2.2.1 Introduction to Quadratic Equations
An equation is a statement that two quantities are equal and to ‘solve an equation’ means ‘to
find the value of the unknown’. The value of the unknown is called the root of the equation.
A quadratic equation is one in which the highest power of the unknown quantity is 2. For
example, x2
−3x +1=0 is a quadratic equation.
2.3 Solution of quadratic equations.
There are four methods of solving quadratic equations.
These are: (i) by factorisation (where possible)
(ii) by ‘completing the square’
(iii) by using the ‘quadratic formula’
or (iv) graphically
I. Solution of quadratic equations by factorization
Multiplying out (2x+1)(x −3) gives 2x2
−6x +x −3, i.e. 2x2
−5x −3. The reverse process of moving
from 2x2
−5x −3 to (2x+1)(x−3) is called factorising.
If the quadratic expression can be factorised this provides the simplest method of solving a
quadratic equation.
For example, if 2x2
−5x −3 = 0, then, by factorising: (2x +1)(x −3) = 0
Hence either (2x +1) =0 i.e. x = −
1
2
or (x −3) =0 i.e. x = 3
The technique of factorising is often one of ‘trial and error’.
Problem 1.
Solve the equations:
(a) x2
+2x −8=0 (b) 3x2
−11x −4=0 by factorization
(a) x2
+2x −8=0.
The factors of x2
are x and x. These are placed in brackets thus: (x )(x )
The factors of−8 are+8 and−1, or−8 and+1, or +4 and −2, or −4 and +2. The only combination
to given a middle term of +2x is +4 and −2, i.e. x2
+2x −8 = (x +4)(x −2)
(Note that the product of the two inner terms added
to the product of the two outer terms must equal
to the middle term, +2x in this case.)
The quadratic equation x2
+2x −8=0 thus becomes (x +4)(x−2)=0.
Since the only way that this can be true is for either the first or the second, or both factors to be
zero, then
either (x +4) =0 i.e. x =−4

9. 9
or (x −2) =0 i.e. x = 2
Hence the roots of x2 + 2x − 8=0 are x=−4 and 2
(b) 3x2
−11x−4=0
The factors of 3x2
are 3x and x. These are placed in brackets thus: (3x )(x )
The factors of −4 are −4 and +1, or +4 and −1, or −2 and 2.
Remembering that the product of the two inner terms added to the product of the two outer terms
must equal −11x, the only combination to give this is +1 and −4, i.e.
3x2
−11x −4 = (3x +1)(x −4)
The quadratic equation 3x2
−11x−4=0 thus becomes (3x +1)(x−4)=0.
Hence, either (3x +1) =0 i.e. x=−
𝟏
𝟑
or (x −4) =0 i.e. x = 4
and both solutions may be checked in the original equation.
Problem 2.
Determine the roots of:
(a) x2
−6x +9=0, and (b) 4x2
−25=0, by factorization
(a) x2
−6x+9=0.
Hence (x −3) (x −3)=0, i.e. (x−3)2=0 (the left-hand side is known as a perfect square). Hence
x=3 is the only root of the equation x2
−6x +9=0.
(b) 4x2
−25=0
(the left-hand side is the difference of two squares, (2x)2
and (5)2
). Thus
(2x+5)(2x−5)=0. Hence either (2x +5) =0 i.e. x=−
𝟓
𝟐
or (2x −5) =0 i.e. x =
𝟓
𝟐
Problem 3.
The roots of quadratic equation are
1
3
and −2. Determine the equation.
If the roots of a quadratic equation are α and β then (x −α)(x −β)=0.
Hence if α=
1
3
and β=−2, then
(𝑥 −
1
3
) − (𝑥 − (−2)) = 0
(𝑥 −
1
3
) (𝑥 + 2) = 0
𝑥2
−
1
3
𝑥 + 2𝑥 −
2
3
= 0
𝑥2
+
5
3
𝑥 −
2
3
= 0

10. 10
Hence 3x2+5x−2 = 0
ASSIGNMENT
In Problems 1 to 4, solve the given equations by factorisation.
1. x2
+4x −32=0
2. x2
−16=0
3. (x +2)2
=16
4. 2x2
−x −3=0
In Problems 5 to 6, determine the quadratic equations in x whose roots are:
5. 3 and 1
6. 2 and −5
II. Solution of quadratic equations by ‘completing the square’
An expression such as x2
or (x+2)2
or (x −3)2
is called a perfect square.
If x2
=3 then x=±√3
If (x+2)2
=5 then x+2=±√5 and x=−2±√5
If (x−3)2
=8 then x−3=±√8 and x =3±√8
Hence if a quadratic equation can be rearranged so that one side of the equation is a perfect
square and the other side of the equation is a number, then the solution of the equation is readily
obtained by taking the square roots of each side as in the above examples. The process of
rearranging one side of a quadratic equation into a perfect square before solving is called
‘completing the square’.
(x +a)2
= x2
+2ax +a2
Thus in order to make the quadratic expression x2
+2ax into a perfect square it is necessary to add
(half the coefficient of x)2
i.e. (
2𝑎
2
)
2
or a2
For example, x2
+3x becomes a perfect square by adding
(
3
2
)
2
, i.e. x2
+3x +(
3
2
)
2
=(𝑥 +
3
2
)
2
The method is demonstrated in the following worked problems.
Problem 1.
Solve 2x2
+5x =3 by ‘completing the square’
The procedure is as follows:
1. Rearrange the equations so that all terms are on the same side of the equals sign (and the
coefficient of the x2
term is positive).
Hence 2x2
+5x −3=0
2. Make the coefficient of the x2
term unity. In this case this is achieved by dividing throughout
by 2. Hence

11. 11
2𝑥2
2
+
5𝑥
2
−
3
2
= 0
i.e. x2
+
5𝑥
2
−
3
2
= 0
3. Rearrange the equations so that the x2
and x terms are on one side of the equals sign and the
constant is on the other side, Hence
𝑥2
+
5𝑥
2
=
3
2
4. Add to both sides of the equation (half the coefficient of x)2
. In this case the coefficient of x is
5
2
.
Half the coefficient squared is therefore (
5
4
)
2
.
Thus, x2
+
5𝑥
2
+(
5
4
)
2
=
3
2
+(
5
4
)
2
The LHS is now a perfect square, i.e.
(𝑥 +
5
4
)
2
=
3
2
+(
5
4
)
2
5. Evaluate the RHS. Thus
(𝑥 +
5
4
)
2
=
3
2
+
25
16
=
24+25
16
=
49
16
6. Taking the square root of both sides of the equation (remembering that the square root of a
number gives a ± answer). Thus
√(𝑥 +
5
4
)
2
=√
49
16
i.e. 𝑥 +
5
4
=±
7
4
7. Solve the simple equation. Thus
x =−
5
4
±
7
4
i.e. x =−
5
4
+
7
4
=
2
4
=
1
2
and x =−
5
4
−
7
4
= −
12
4
= −3
Hence x=
𝟏
𝟐
or −3 are the roots of the equation 2x2
+5x =3
Problem 7.
Solve 2x2
+9x +8=0, correct to 3 significant figures, by ‘completing the square’

12. 12
Making the coefficient of x2
unity gives:
x2
+
9
2
𝑥 + 4 = 0
and rearranging gives: x2
+
9
2
𝑥 = −4
Adding to both sides (half the coefficient of x)2 gives:
x2
+
9
2
𝑥+(
9
4
)
2
= (
9
4
)
2
− 4
The LHS is now a perfect square, thus:
(𝑥 +
9
4
)
2
=
81
16
− 4 =
17
16
Taking the square root of both sides gives:
𝑥 +
9
4
= √
17
16
=±1.031
Hence x =−
9
4
±1.031
i.e. x=−1.22 or −3.28, correct to 3 significant figures.
ASSIGNMENT
Solve the following equations by completing the square, each correct to 3 decimal places.
1. x2
+4x +1=0
2. 2x2
+5x −4=0
3. 3x2
−x −5=0
III. Solution of quadratic equations by ‘quadratic formula’
Let the general form of a quadratic equation be given by:
ax2
+bx +c = 0
where a, b and c are constants.
Dividing ax2
+bx+c=0 by a gives:
x2
+
𝑏
𝑎
𝑥 +
𝑐
𝑎
= 0
Rearranging gives:
x2
+
𝑏
𝑎
𝑥 = −
𝑐
𝑎
Adding to each side of the equation the square of half the coefficient of the terms in x to make
the LHS a perfect square gives:
x2
+
𝑏
𝑎
𝑥 + (
𝑏
2𝑎
)
2
= (
𝑏
2𝑎
)
2
−
𝑐
𝑎
Rearranging gives:
(x +
𝑏
𝑎
)
2
=
𝑏2
4𝑎2 −
𝑐
𝑎
=
𝑏2−4𝑎𝑐
4𝑎2

13. 13
Taking the square root of both sides gives:
x +
𝑏
2𝑎
=√
𝑏2−4𝑎𝑐
4𝑎2 =
±√𝑏2−4𝑎𝑐
2𝑎
Hence x =−
𝑏
2𝑎
±√𝑏2−4𝑎𝑐
2𝑎
i.e. the quadratic formula is: x =
−𝑏±√𝑏2−4𝑎𝑐
2𝑎
(This method of solution is ‘completing the square’.
Summarising:
if ax2
+bx +c = 0
then x =
−𝑏±√𝑏2−4𝑎𝑐
2𝑎
This is known as the quadratic formula.
Problem 9.
Solve (a) x2
+2x −8=0 and (b) 3x2
−11x−4=0 by using the quadratic formula
(a) Comparing x2
+2x −8=0 with ax2
+bx+c=0 gives a=1, b=2 and c=−8.
Substituting these values into the quadratic formula
x =
−𝑏±√𝑏2−4𝑎𝑐
2𝑎
gives
x =
−2±√22−4(1)−(8)
2(1)
=
−2±√4+32
2
=
−2±√36
2
=
−2±6
2
=
−2+6
2
or
−2−6
2
Hence x=
4
2
=2 or
−8
2
=−4
(b) Comparing 3x2
−11x−4=0with ax2
+bx+c=0 gives a=3, b=−11 and c=−4. Hence,
x =
−(−11)±√(−11)2−4(3)(−4)
2(3)
=
−(−11)±√121+48
6
=
11±√169
6
=
11±13
6
=
11+13
6
or
11−13
6
Hence x=
24
6
=4 or
−2
6
= −
1
3
ASSIGNMENT
Solve the following equations by using the quadratic formula, correct to 3 decimal places.
1. 2x2+5x −4=0
2. 5.76x2+2.86x −1.35=0

14. 14
PRACTICAL PROBLEMS INVOLVING QUADRATIC EQUATIONS
There are many practical problems where a quadratic equation has first to be obtained, from
given information, before it is solved.
Problem 1.
Calculate the diameter of a solid cylinder which has a height of 82.0cm and a total surface area
of 2.0m2
Total surface area of a cylinder = curved surface area + 2 circular ends = 2πrh +2πr2
(where r =radius and h=height)
Since the total surface area = 2.0m2
and the height h=82cm or 0.82m, then
2.0 = 2πr(0.82)+2πr2
i.e. 2πr2
+2πr(0.82)−2.0=0
Dividing throughout by 2π gives:
r2 +0.82r −
1
𝜋
= 0
Using the quadratic formula:
r =
−0.82±√(0.82)2−4(1)(
1
𝜋
)
2(1)
=
−0.82±√1.9456
2
=
−0.82±1.3948
2
= 0.2874 or −1.1074
Thus the radius r of the cylinder is 0.2874m (the negative solution being neglected).
Hence the diameter of the cylinder = 2×0.2874= 0.5748m or 57.5 cm correct to 3 significant
figures.
Problem 12.
The height s metres of a mass projected vertically upward at time t seconds is s=ut −
1
2
gt2
.
Determine how long the mass will take after being projected to reach a height of 16m
(a) on the ascent and (b) on the descent, when u=30m/s and g=9.81m/s2
.
When height s=16m, 16=30t −
1
2
(9.81)t2
i.e. 4.905t2
−30t +16 = 0
Using the quadratic formula:
t =
−(−30)±√(−30)2−4(4.905)(16)
2(4.905)
=
30±√586.1
9.81
=
30±24.21
9.81
= 5.53 or 0.59
Hence the mass will reach a height of 16 m after 0.59 s
on the ascent and after 5.53 s on the descent.

15. 15