# The basics

## Summary and examples

To begin with there are some important rules for dealing with exponents. You will use them all the time and therefore you should learn them by heart.

#### Rules for exponents

$a^p.a^q=a^{p+q}$
Example: $x^3.x^2=x^5$

$\displaystyle\frac{a^p}{a^q}=a^{p-q}$
Example: $\displaystyle\frac{y^5}{y^3}=y^2$

$(ab)^p=a^pb^p$
Example: $(3z)^4=3^4.z^4=81z^4$

$(\displaystyle\frac{a}{b})^p=\displaystyle\frac{a^p}{b^p}$
Example: $(\displaystyle\frac{7}{x})^2=\displaystyle\frac{49}{x^2}$

$(a^p)^q=a^{pq}$
Example: $(x^4)^3=x^{12}$

$a^1=a$

$a^0=1$

$a^{-p}=\displaystyle\frac{1}{a^p}$

$\sqrt[q]{a^p}=a^{\frac{p}{q}}$
Example: $\sqrt{x}=\sqrt[2]{x^1}=x^{\frac{1}{2}}$

#### Roots

There is much to say about roots, but you should know the following characteristics anyway.

$\sqrt{49}=7$ because $7^2=49$

$\sqrt[3]{8}=2$ because $2^3=8$

$\sqrt{-25}$ does not exist, because there is no real number that has $-25$ as a square.

$\sqrt[3]{-125}=-5$ because $(-5)^3=-125$

$\sqrt[5]{32}=2$ because $2^5=32$

$\sqrt[144]{-2.38847}$ does not exist because there is no real number that has $144$ as a result when it is raised to an even power.

#### Fractions

We start with fractions containing only numbers. Fractions can be added or subtracted when the denominators of the fractions are equal. Note that a denominator is the expression below the vinculum ('divide by'); if the expression is not a denominator then it is a numerator.

$\displaystyle\frac{1}{7}+\displaystyle\frac{1}{3}=\displaystyle\frac{1.3}{7.3}+\displaystyle\frac{1.7}{3.7}=\displaystyle\frac{3}{21}+\displaystyle\frac{7}{21}=\displaystyle\frac{3+7}{21}=\displaystyle\frac{10}{21}$

We get equal denominators by multiplying the first denominator by $3$ and the second by $0$ (you could say by multiplying the denominators). This is allowed if the numerators are multiplied by $3$ and $7$ as well, respectively.
In the following example you would get equal denominators by multiplying by $4$ and $6$. However, it is better to take the lowest common multiple of $4$ and $6$ and that is $12$.

$\displaystyle\frac{5}{6}-\displaystyle\frac{3}{4}=\displaystyle\frac{5.2}{6.2}-\displaystyle\frac{3.3}{4.3}=\displaystyle\frac{10}{12}-\displaystyle\frac{9}{12}=\displaystyle\frac{1}{12}$

Usually we have to deal with fractions with letters. Look at the following example. We can make the denominators equal by multiplying the denominator of the first fraction by $x$ and the second denominator by $x-1$. Of course, again: when both the denominator and numerator of a fraction are multiplied or divided by the same number $(\neq0)$, the fraction retains the same value. See the following examples:

$\displaystyle\frac{1}{x-1}-\displaystyle\frac{1}{x}=\displaystyle\frac{x}{(x-1)x}-\displaystyle\frac{x-1}{x(x-1)}=\displaystyle\frac{x-(x-1)}{x(x-1)}=\displaystyle\frac{1}{x(x-1)}$

$\displaystyle\frac{1}{x(x-1)}+\displaystyle\frac{1}{x}=\displaystyle\frac{1}{x(x-1)}+\displaystyle\frac{x-1}{x(x-1)}=\displaystyle\frac{1+(x-1)}{x(x-1)}=$

$=\displaystyle\frac{x}{x(x-1)}$

Multiplying and dividing

Calculate the following product of fractions:

$\displaystyle\frac{5}{7}.\displaystyle\frac{4}{3}.\displaystyle\frac{21}{8}.\displaystyle\frac{13}{15}.\displaystyle\frac{27}{39}$

Most students think this is not possible without a calculator. But when the above-mentioned rule is used, namely, that a fraction retains its value, when both numerator and denominator are multiplied or divided by the same number $\neq0$, it is easy to carry out the calculation.

First we may write the fraction as follows:

$\displaystyle\frac{5.4.21.13.27}{7.3.8.15.39}$

Next we divide both numerator and denominator by common factors. As an example, if we divide both numerator and denominator by $13$, then we get $1$ in the numerator instead of $13$ and in the denominator $3$ instead of $39$. Continuing in this way we get:

$\displaystyle\frac{3}{2}$.

Next we want to simplify the following fraction:

$\displaystyle\frac{a^3\sqrt{a^5}}{a^2\sqrt[4]{a}}$

We may write (see the rules concerning exponents):

$\sqrt{a^5}=a^{\frac{5}{2}}$ en $\sqrt[4]{a}=a^{\frac{1}{4}}$

resulting in:

$\displaystyle\frac{a^3\sqrt{a^5}}{a^2\sqrt[4]{a}}=\displaystyle\frac{a^{\frac{11}{2}}}{a^{\frac{9}{4}}}=a^{\frac{13}{4}}=a^3\sqrt[4]{a}$

#### Factorization and removing brackets

The following is an example of factorization.

$a^2b^3+ab^2+a^2bc=ab(ab^2+b+ac)$

The three terms in the left-hand side have the factors $a$ and $b$ in common and we want to use them as separate factors.

Sometimes we want to remove brackets. The following two formulas are used frequently and need to be learned by heart.

$a(b+c)=ab+bc$

$(a+b)(c+d)=ac+ad+bc+bd$

There are a few cases that occur frequently and therefore must be learned by heart. They are known as special products. They are special because a product of two terms in generally yields four terms, but not in these cases. The result can be verified with the above formulas.

$(a+b)^2=a^2+2ab+b^2$

$(a-b)^2=a^2-2ab+b^2$

$(a+b)(a-b)=a^2-b^2$

#### Linear functions and equations

A linear function (also called a first-degree function) has the following form:

$y=ax+b$

The graph of this function is a straight line.

The parameters $a$ en $b$ have a special meaning.
The parameter $a$ is the slope of the line. When $a>0$ then the line is ascending (from bottom left to top right), in the case $a<0$ the line descending (from top left to bottom right), when $a=0$ we are dealing with a horizontal line.
The parameter $b$ indicates the point where the line crosses the $Y$-axis, namely in the point $(0,b)$, because all points on the $Y$-axis have the $x$-coördinate $x=0$.
When we want to find the point where the line crosses the $X$-axis, we must choose $y=0$. After all, points on the $X$-axis have a $y$-coördinate $0$.

$ax+b=0$,  $(a\neq0)$, so $x=-\displaystyle\frac{b}{a}$

Solve the following linear equation, that is, find the value of $x$ for which the left and right side are equal.

$3(2x+2)-4(x+1)+x=2(2x-4)-4$

Take the following steps.

step 1: remove left and right the brackets;

step 2: take left and right the $x$-terms together; do the same with the numbers;

step 3: place the $x$-variable in the left side and the number in the right side;

step 4: make sure that the result looks like this: $x=number$.

##### Example 1

step 1: $6x+6-4x-4+x=4x-8-4$

step 2: $3x+2=4x-12$

step 3: $-x=-14$

step 4: $x=14$

##### Example 2

Sometimes a linear equation is disguised as a higher degree equation.

Look at the following example:

$x^2(x-1)+x^2+3x-4=x(x^2+5)-x+6$

In simplifying the third power and the square disappear and a linear equation remains:

$x^3-x^2+x^2+3x-4=x^3+5x-x+6$

$3x-4=4x+6$

$x=-10$

#### Liner inequalities

It is important to remember that the inequality sign of an inequality turns around as left and right side are multiplied by the same negative number.

##### Example 3

$5x-6>7x-8$

$-2x>-2$

$x<1$

##### Example 4

$3(2x+2)-4(x+1)+x\geq2(2x-4)-4$

This inequality can be rewritten as:

$-x\geq-14$

Multiply left and right side by $-1$ and the inequality sign will be $\leq$ instead of $\geq,$ so:

$x\leq14$

#### Quadratic or second degree equations

The general formula is:

$y=ax^2+bx+c$

($a\neq0$, if $a=0$ then we do not have a quadratic function).

The graph is a parabola; if $a>0$: $\cup$ (opens up); $a<0$: $\cap$ (opens down).

The number $+c$ determines the intersection point of the parabola with the $Y$-as. After all, all points on the $Y$-axis have the coordinate $x=0$ and thus $(0,c)$ is the intersecion point of the graph with the $Y$-axis.

To find the intersection points with the $X$-axis (i.e., $y=0$) you have to solve the equation: $ax^2+bx+c=0$.

This equation has not always (real) solutions. An 'opens up' parabola with a minimum above the $X$-axis and an 'opens down' parabola with a maximum below the $X$-axis do not have intersection points with the $X$-axis. Whether the graph of a parabola will cross the $X$-axis, depends on the discriminant $D=b^2-4ac$, see below.

How do we solve a quadratic equation $ax^2+bx+c=0$?

We distinguish the following methods:

• Factorization (sometimes that is easy to do and then this method is preferred, but this is not always the case);
• The $abc$-formula (always delivers a solution if one exists);
• Special cases.

The quadratic formula definitely delivers a solution if one exists, but the other methods are sometimes faster to calculate.

Factorization
Try the expression $ax^2+bx+c$ to write in the form $(x+p)(x+q)$. where $p$ and $q$ are whole numbers which have $+c$ as a product and $+b$ as a sum (the so-called sum-product method).

##### Example 5

$x^2+5x+6=0$ can be written as $(x+3)(x+2)=0$ for $3*2=6$ and $3+2=5$

$x^2-2x-8=0$ can be written as $(x+(-4))(x+2)=0$ for $-4*2=-8$ and $-4+2=-2$

This method is only easy to use if  $a=1$. If $a\neq1$ left and right side are divided by $a$. However, when because of this operation one of the coefficients will become a fraction, the sum-product method will not be used.

abc-formula
The solution of the quadratic equation $ax^2+bx+c=0$ can be found by using the following formula.

$x_{1,2}=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

This formula is known as the $abc$-formula. $D$ is the discriminant: $D=b^2-4ac$.

There are three possibilities:

• $D>0$: 2 different solutions $x_1$ or $x_2$
• $D=0$: 1 solution (actually 2 equal solutions $x_1=x_2$)
• $D<0$: no (real) solutions
##### Example 6

$x^2+5x+6=0$

so:

$x_{1,2}=\displaystyle\frac{-5\pm\sqrt{5^2-4.1.6}}{2.1}=\frac{-5\pm\sqrt{1}}{2}$

So the solutions are: $x=-3$ of $x=-2$.

This equation can also be solved with the sum-product method. Try!

$2x^2+x-1=0$

so

$x_{1,2}=\displaystyle\frac{-1\pm\sqrt{1^2-4.2.-1}}{2.2}=\frac{-1\pm\sqrt{9}}{4}$

So the solutions are $x=-1$ of $x=\displaystyle\frac{1}{2}$

Special cases

The first case is when $b=0$.

$x^2-16=0$, so $x^2=16$, and so $x=4$ of $x=-4$ (do not forget the minus solution).

In the other case we have $c=0$.

Then we just get $x$ outside brackets.

$x^2-5x=0$, so $x(x-5)$ and so $x=0$ of $x=5$.

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