1
ALGEBRA
Indices - - - - - 2
Algebraic Identities - - - - - 6
Perfect Squares - - - - - 10
Absolute Value - - - - - 15
Quadratic Equations - - - - - 19
Complex Numbers - - - - - 25
Inequalities - - - - - 27
INDICES
LAWS OF INDICES
Problem 1
Solution
Since the bases are the same, we equate the powers
Problem 2
Simplify
Solution
Problem 3
Solution
Equating powers we have
Problem 4
Solution
ALGEBRAIC INDENTITES
• Difference of two squares:
• Difference of two cubes:
• Sum of two cubes:
• When the sum of three quantities equals zero:
Problem 1
Simplify
Solution
Problem 2
Simplify
Solution
Problem 3
Factorise:
Solution
Problem 4
Evaluate the expression
Solution
Problem 5
Given
Solution
PERFECT SQUARES
• A perfect square is an expression that can be expressed as the square of another expression.
• A perfect square number is a number that can be expressed as the square of another number. Examples include 4 ( and 9(=
•
• is a perfect square because it is equivalent to
• is also a perfect square because it is equivalent to
• It is important to note the following
• Also,
ILLUSTRATION
•
•
Problem 1
Evaluate (Nigeria, 2009)
Solution
Problem 2
Find the smallest positive integer n such that the number is a perfect square? (Kenya, 2019)
Solution
We start by observing that the average of 2011, 2013, 2017 and 2019 is 2015.
So let 2015 be
Thus, the smallest positive integer n such that the number is a perfect square is 36.
Problem 3
Without using a calculator: what integer equals (Uganda, 2017)
Solution
Problem 4
Find the sum of the digits of the number above.
Solution
Required sum
ABSOLUTE VALUE
• Given a real number its absolute value or modulus is defined as
.
• The absolute value of any real number is always a positive number.
• Absolute value of any real number is the positive difference of that number and zero.
Properties of Absolute Value
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Illustrations
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Note:
Problem 1
Simplify the following expression
(MATHEMATICS WITHOUT BORDERS, 2019)
Solution
Problem 2
If (American Olympiad, 2000)
Solution
since ,
Subtract from both sides
Problem 3
If then the value of
is:
(MATHEMATICS WITHOUT BORDERS, 2015)
Solution
Problem 4
If are non-zero real numbers, find all possible values of the expression
Solution
Due to the symmetry in the question, we need consider only 4 cases:
Case 1: when are all positive
Case 2: when are all negative
Case 3: when two of are negative, and the third is positive. Without loss of generality, let the only positive be
Case 4: when two of are positive and the third is negative. Without loss of generality, let the negative be
QUADRATIC EQUATIONS
• A quadratic equation is an equation of the form where
are constants, and
is the unknown to be solved for.
• The results obtained after solving a quadratic equation are called solutions, roots or zeros of the equation.
• The greatest power of the unknown variable in a quadratic equation is always 2. Consequently, when solved it always yields two solutions.
• The two solutions of a quadratic equation must be one of the following: equal real roots, distinct real roots or conjugate complex roots.
• Solutions of a quadratic equation are called zeros of the equation because they will produced a zero value when plug into the equation to replace the unknown variable
• Given a quadratic equations with roots it follows that
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•
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• Every quadratic equations of the form can be expressed in terms of its roots as:
where
are the roots.
• If are roots of a quadratic equation, then
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• Given that where
are constants,
we can obtain the roots of the equations by Quadratic formula method:
• Let and
be the roots. WOLOG let
•
•
• Thus, given the quadratic equation sum of roots is
while the product of roots is
• Given that where
are constants,
•
• is either less than zero, equal to zero or greater than zero.
• if and only if we have equal real roots.
• if and only if we have distinct real roots. If
is a perfect square, the roots will be rational numbers; otherwise, the roots will be irrational numbers.
• if and only if we have complex roots.
Problem 1
What is the sum of all possible values of k for which the polynomials and
have a root in common? (Question 5, AMC 12A, 2018)
Solution
We first solve and obtain the roots of
2 and 1 are the roots/zeros of , thus must be the roots/zeros of
as well. To obtain the possible values of k, we plug
into the equation.
When
When
Our required sum
Problem 2
Both roots of the quadratic equation are prime numbers. The number of possible value of
is? (Question 12, AMC 12A, 2002)
Solution
Let the roots of the above equation be , the resulting equation is:
By equating coefficient we have:
We know that the sum of two odd prime must be even. The RHS of our equation above is odd; it follows that one of the prime must be even, that is 2.
The only possibility is
Thus, there is only one possible value of which is 122(=2x61).
Our required answer is 1.
Problem 3
Suppose that are non-zero real numbers, and that the equation
has solutions
Then the pair
is?
(Question 6, AMC 12B, 2002)
Solution
We are given:
Also, and
are the solutions of
……(1)
The resulting quadratic equation with roots and
is
…..(2)
Equations (1) and (2) are equivalent. Thus equating coefficients we have
From equation (4):
Plugging into equation (3) gives
Thus, our required pair
Problem 4
Given that a and b are distinct prime numbers, and
find the value of
( Question 6, 2011-12 HKMO)
Solution
and
are distinct prime numbers,
Subtracting equation (2) from equation (1), we have
The sum of two distinct prime is odd if and only if one of the primes is the only even prime 2.
This implies that satisfies equation (3) are
Problem 5
The quadratic equation has roots twice those of
and none of
and
is zero. What is the value of
(Problem 12, AMC 12B, 2005).
Solution
Let the roots of be
and
has roots twice those of
.
This means and
are the roots of
COMPLEX NUMBERS
• Complex numbers are numbers of the form where
and
are real numbers.
• is the real part of the complex number, while
(the coefficient of
is the imaginary part of the complex number.
• Two complex numbers are equal to each other if and only if both their real part and imaginary part are equivalent.
• What makes a number a complex number is the existence of
•
•
•
• where
is an integer, is an element of
•
Problem 1
There is a complex number with imaginary part 164 and a positive integer n such that
Find
Solution
Let the real part of be
The complex number on the LHS is equal to the complex number on the RHS
INEQUALITIES
• Means are averages. A Mean is a representative value of a given set of observations. This representative value always lie between the smallest and largest observations; thus, Means are also known as measures of central tendency.
• Some types of Means are Quadratic Mean (also known as Root-Mean-Square), Arithmetic Mean, Geometric Mean and Harmonic Mean
• Quadratic Mean (QM) is the square root of “the sum of the squares of the given observations divided by the number of observation”.
• Arithmetic Mean (AM) of a given set of observations is their sum divided by the number of observations.
• The Geometric Mean (GM) of a set of observations is the
root of their product.
• Harmonic Mean (HM) is the reciprocal of the arithmetic mean of the reciprocals of the given observations.
Given a set of observations:
• QM
• AM
• GM
• HM
QM-AM-GM-HM INEQUALITY THEOREM
Given a set of observations:
Equality holds when
HINTS TO CONSIDER WHEN PROVING INEQUALITIES
• If sum of the variables occurs, example consider an inequality theorem involving Arithmetic mean.
• If sum of squares of the variables occurs, exampleconsider an inequality theorem involving Quadratic mean.
• If product of the variables occurs, example consider an inequality theorem involving Geometric mean.
• If sum of the inverses of the variables occurs, example consider an inequality theorem involving Harmonic mean.
Problem 1
Given that are real numbers and
Find
Solution
The least value of a perfect square is zero. The RHS is equal to zero; every term on the LHS is a perfect square, thus, each term on the LHS must be zero.
Problem 2
Given that are positive real numbers, prove
PROOF
We know that for any given set of positive real numbers, the Arithmetic mean is at least the Harmonic mean.
Thus,
Multiplying both sides by
Problem 3
Given that are nonnegative real numbers. Prove that
PROOF
From AM-GM inequality, we know that
Similarly,
Adding the above three inequalities, we have
The above inequality is equivalent to
Problem 4
Given that are positive real numbers and
. Find the minimum value of
Solution
(AM-HM Inequality)
The minimum value of is
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