# MU123 Unit 9 Notes

# 1 Number patterns and algebra

## 1.1 Arithmetic sequences

Sum of a sequence of natural numbers can be rewritten as a sum of pairs of numbers: $$ 1 + 2 + ... + 99 + 100 = (1 + 100) + (2 + 99) + ... + (50 + 51) $$ There are 50 pairs, each of them having a sum of $101$, so: $$ 1 + 2 + ... + 99 + 100 = 50 \times 101 = 5050 $$

In case of a sum of successive odd numbers the sum is always the square of how many odd numbers I add, e.g. $1 + 3 + 5 + 7 = 16 = 4^2$ . For $n$ numbers the formula is $n^2$.

Formula for $n$ successive natural numbers is:
$$
1 + 2 + ... + n = \frac{1}{2}n(n + 1)
$$
The numbers given by the formula are called **triangular numbers**.

Any list of numbers is called a **sequence**

- $1, 2, 3, ..., 100$ is
**finite**sequence - $1, 2, 3, ...$ is
**infinite**sequence

The numbers in a sequence are called the **terms** of the sequence.
**Arithmetic** sequence is a sequence where the **difference** between consequtive terms is constant, e.g.: $1, 2, 3, ...$ is a arithmetic sequence with difference of 1,
To specify an arithemtic sequence we can give the first term denoted by $a$, the difference denoted by $d$. If the sequence is finite the number of terms is denoted by $n$.

The $n$th term of an arithmetic sequence with first term $a$ and difference $d$ is given by the formula $$ n\text{th term} = a + (n - 1)d $$

The number of terms $n$ of a finite arithmetic sequence with first term $a$, last term $L$ and non-zero difference $d$ is given by the formula: $$ n = \frac{L - a}{d} + 1 $$

The sum of the finite arithmetic sequence with first term $a$, difference $d$ and numbers of term $n$ is given by the formula: $$ S = \frac{1}{2}n(2a + (n - 1)d) $$ An alternative formula involving the last term $L$: $$ S = \frac{1}{2}n(a + L) $$

# 2 Multiplying out pairs of brackets

## 2.1 Pairs of brackets

Strategy to multiply out two brackets

Multiply each term inside the first bracket by each term inside the second bracket, and add the resulting terms.

## 2.2 Squaring brackets

$$ \begin{aligned} (x + p)^2 = (x + p)(x + p) \\ = x^2 + xp + px + p^2 \\ = x^2 + 2px + p^2 \end{aligned} $$

Hence the general formula: $$ (x + p)^2 = x^2 + 2px + p^2 $$ and $$ (x - p)^2 = x^2 - 2px + p^2 $$

## 2.3 Differences of two squares

$$ (x - p)(x + p) = x^2 + xp - px - p^2 $$ Hence: $$ (x - p)(x + p) = x^2 - p^2 $$

# 3 Quadratic expressions and equations

## 3.1 Quadratic expressions

An expression of the form $ax^2 + bx + c$, where $a, b, c$ are numbers and $a \ne 0$ is called a **quadratic expression** in $x$. $a, b, c$ are called the **coefficients** of the quadratic.

## 3.2 Quadratic equations

An equation that can be expressed in the form:
$$
ax^2 + bx + c = 0
$$
is called a **quadratic equation** in $x$.

## 3.3 Solving simple quadratic equations

An equation of the form $x^2 = d$, where $d > 0$ has two solutions $x = \pm \sqrt{d}$.

Equations as $x^2 + 1 = 0$ have no solution among the real numbers.

## 3.4 Factorising quadratics of the form $x^2 + bx + c$

Fill in the gaps in the brackets on the right-hand side of the equation $$ x^2 + bx + c = (x...)(x...) $$ with two numbers whose product is $c$ and whose sum is $b$. I can search systematically by writing down all the factor pairs of $c$ and choosing (if possible) a pair whose sum is $b$.

## 3.5 Solving quadratic equations by factorisation

If the product of two or more numbers is 0, then at least one of the numbers must be 0. Hence in an equation as $(x - 2)(x - 3) = 0$ either $x - 2$ or $x - 3$ is 0. If $x - 2 = 0$ then $x = 2$, if $x - 3 = 0$ then $x - 3$ so the equation has two solutions.

### Strategy to solve $x^2 + bx + c = 0$ by factorisation

- Find a factorisation: $x^2 + bx + c = (x + p)(x + p)$
- Then $(x + p)(x + p) = 0$, so $x + p = 0$ or $x + q = 0$ and hence the solutions are $x = -p$ and $x = -q$.

When the two solutions are same it's said that equation has a **repeated solution**.

## 3.6 Factorising quadratics of the form $ax^2 + bx + c$

E.g. following quadratic expression; $$ 2x^2 - x - 6 $$ Where: $$ \begin{aligned} a &= 2\\ b &= -1\\ c &= -6 \end{aligned} $$

Find two numbers whose product is $ac$ and whose sum is $b$.

The numbers are $3$ and $-4$.

Rewrite the quadratic expression, splitting the term in $x$ using the above factor pair.

$$ 2x^2 - x - 6 = 2x^2 +3x -4x - 6 $$

Group the four terms in pairs and take out common factors to give the required factorisation.

$$ \begin{aligned} 2x^2 - x - 6 &= 2x^2 +3x -4x - 6 \\ &= x(2x + 3) -2(2x + 3) \\ &= (x - 2)(2x + 3) \end{aligned} $$

### Simplify a quadratic equation

- If the coefficient of $x^2$ is negative, then multiply the equation through by $-1$ to make this coefficient positive.
- If the coefficients have a common factor, then divide the equation through by this factor.
- If any of the coefficients are fractions, then multiply the equation through by a suitable number to clear them.

# 4 Manipulating algebraic fractions

$\frac{a}{b}$ is equivalent to $\frac{a(a + 1)}{b(b + 1)}$ and common factor can be cancelled out as usual.