A-level Mathematics/MEI/C1/Algebra

< A-level Mathematics‎ | MEI‎ | C1

Algebra is the branch of mathematics that deals with the relation of quantities. In an equation, both sides are equal, and in an inequality, one side is usually greater than another.

An equation consists of two expressions joined by the equals sign ($=$). Everything on the left-hand side is equal to everything on the right-hand side, for example $2+3=4+1$. Some equations contain a variable, usually denoted by $x$, $y$ or $z$. An equation with a variable will only hold true for certain values of that variable. For example $2 + x = 5$ is only true for $x=3$. The values that the variables have when the equation is true are called the solutions of the equation. Therefore $x=3$ is the solution of the equation $2 + x = 5$.

The language of algebra

There are several terms that you need to be familiar with before you begin to work with algebra.

Variable

A variable is a quantity, whose value is usually unknown. Usually, a variable is given the symbol $x$, although any letter can be used.

Constant

A constant is usually a known quantity, which does not involve the variable. Later you will come across unknown constants, which are usually given the symbol $c$.

Index

Generally, an index is anything that is written superscript to (slightly above) a symbol. Often indicies are used to indicate something raised to a power, such as $x^3$ (read as x-cubed), which is the same as $x \times x \times x$.

Expression

An expression is a group of symbols which form a mathematical statement. $2 + 2$ is an example of an expression.

Term

A term is any variable, constant or a product of variables or constants which are separated by a $+$ or a $-$ sign. In the expression $3x+4xy-2y$, the separate terms are $3x$, $4xy$, and $2y$.

Coefficient

A coefficient is the constant part of a term which multiplies the variable part of the term. For example, in $2x^3$, the coefficient of $x^3$ is $2$.

Equation

An equation is a mathematical statement that two things are equal. There is an equals sign ($=$) in between two expressions. $2+2=4$ is an example of an equation, as well as $3+x=7$.

Identity

An identity is an equation with an unknown, and is true for all values of that unknown. The identity symbol ( ≡ ) is used in place of the equals sign. As an example, $x - x = 0$ is always correct no matter what the value of $x$ is. It is usual to write $x - x$$0$ when you want to emphasize the fact that it is an identity and not just an equation.

Function

A function relates one input value to one output value. A function of $x$ is usually noted as $f(x)$. $f(x)=x^2$ is an example of a function. It is then possible to write $f(3)=3^2=9$ once the function has been defined. $f(x)= \sqrt x$ is not a function, because there are two output values (positive and negative) for each input.

Manipulating expressions

Sometimes, expressions will be messier than they need to be, and they can be represented in an easier to understand form. The skills here are essential to the rest of the A-level course, although it is very likely that you have already covered them at GCSE.

Collecting like terms

When collecting like terms, you simply add all the terms in $x$ together, all the terms in $y$ together, and all the terms in $z$ together. The same applies for any other letter that represents a variable.

For example, $2x+4y+8z-3x-7y-2z+4x$ becomes:

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

$4y-7y = -3y$

$8z-2z = 6z$

So, by adding all the answers, $2x+4y+8z-3x-7y-2z+4x$ simplified is $3x-3y+6z$.

Multiplication

Multiplication of different variables such as $a \times b$ becomes $ab$. Single variables become indices, so $x \times x$ is $x^2$.

Like addition and subtraction, you keep like terms together. So, for example:

$2x^2z \times 3yz^2 \times 4xy^3$ becomes:

$2 \times 3 \times 4 \times x^2 \times x \times y \times y^3 \times z \times z^2$

which can finally be simplified as:

$24{x^3}{y^4}{z^3}$

(Shouldn't this be 234{x^3}{y^4}{z^3})

Expansion of brackets

The skill of expanding brackets is illustrated by the following example.

$(8x+5y)(3x-6y)$

$=(8x+5y)3x-(8x+5y)6y=8x(3x-6y)+5y(3x-6y)$ use FOIL

F=> first term in each bracket 8x . 3x =24x^2 O=>outside terms(1st in 1st x 2nd in 2nd) 8x . -6y = -48xy I=> inside terms (2nd in 1st x 1st in 2nd) 5y . 3x =15xy L=> last term in each bracket 5y . -6y = -30 y^2

and terms together(usually only the middle 2) 24x^2 -48xy +15xy -30y^2 answer 24x^2-33xy-30y^2

Factorising

Sometimes, expressions can be re-written as the product of their factors. You divide an expression by a factor common to all of the terms in the expression. This is essentially the opposite of expanding brackets, since most of the time the common factor is placed outside of brackets. For example:

To factorise $10x+15y$ you must first find a common factor of $10x$ and $15y$. $5$ is easily spotted as a factor. Now you divide the whole expression by $5$ leaving you with $2x+3y$. Place $2x+3y$ within brackets, and then put the $5$ outside the bracket. The factorised expression is now $5(2x+3y)$, and you can multiply out the expression to make sure you get the original expression.

Another expression, $x^2-xy^2+3xz$ has a common factor $x$. The factorised form of this expression is $x(x-y^2+3z)$.

Fractions

When working with fractions, the rule is to make all of the denominators equal, and then write the expression as one fraction. You need to multiply both the top and bottom by the same amount to keep the meaning of the fraction the same.

For example, for $\frac {3x}{2} + \frac {2y}{5} - \frac {z}{10}$, the common denominator is $10$.

Multiply both parts by $5$: $\frac {15x}{10}$

Multiply both parts by $2$: $\frac {4y}{10}$

Leave this as it is: $\frac {z}{10}$

You now have $\frac {15x}{10} + \frac{ 4y}{10} - \frac {z}{10}$, which becomes $\frac {15x+4y-z}{10}$.

Solving equations

Often, to solve an equation you must rearrange it so that the unknown term is on its own side of the equals sign. By rearranging $2 + x = 5$ to $x = 5 - 2$, $x$ has been made the subject of the equation. Now by simplifying the equation, you can find that the solution is $x=3$.

Changing the subject of an equation

You will usually be given equations that are more complex than the example above. To move a term from one side of the equals sign to the other, you have to do the same thing on both sides of the equals sign. For example, to make $x$ the subject of $y = \frac{4 a (x^2 +b)}{3}$:

 Multiply both sides by $3$ $3y = 4 a (x^2 + b)$ Divide both sides by $4a$ $\frac {3y}{4a} = x^2 + b$ Subtract $b$ from both sides $\frac {3y}{4a} -b = x^2$ Square root both sides $\pm\sqrt{\frac {3y}{4a} -b } = x$

Quadratic equations are equations where the variable is raised to the power of 2 and, unlike linear equations, there are a maximum of two roots. A root is one value of the variable where the equation is true, and to fully solve an equation, you must find all of the roots. For a quadratic equation you can factorise it and then easily find which values make the equation valid. The example above is quite a simple case. You will usually be given a more complicated equation such as $2x^2 + 5x + 3 = 0$. If the equation isn't already in the form $ax^2 + bx + c = 0$, rearrange it so that it is. The steps needed to factorise $2x^2 + 5x + 3 = 0$ are:

 Multiply $2$ by $3$ (coefficient of $x^2$ multiplied by the constant term) $2 \times 3 = 6$ Find two numbers that add to give $5$ (coefficient of $x$) and multiply to give $6$ (answer from previous step) $2 \times 3 = 6$, $2 + 3 = 5$ Split $5x$ to $2x + 3x$ (from the results of the previous step) $2x^2 + 2x + 3x + 3 = 0$ Simplify $2x(x + 1) + 3(x + 1) = 0$ Simplify further $(2x + 3)(x + 1) = 0$

So $(2x + 3)(x + 1) = 0$ is $2x^2 + 5x + 3 = 0$ in factorised form. You can now use the fact that any number multiplied by $0$ is $0$ to find the roots of the equation. The numbers that make one bracket equal to $0$ are the roots of the equation. In the example, the roots are $-1.5$ and $-1$.

It is also possible to solve a quadratic equation using the quadratic formula or by completing the square.

Simultaneous equations

Simultaneous equations are useful in solving two or more variables at once. Basic simultaneous equations consist of two linear expressions and can be solved by three different methods: elimination, substitution or by plotting the graph.

Elimination method

The basic principle of the elimination method is to manipulate one or more of the expressions in order to cancel out one of the variables, and then solve for the correct solution.

An example of this:

 $2x + 3y = 10$ (1) (Assigning the number (1) to this equation) $2x + 6y = 6$ (2) (Assigning the number (2) to this equation)

From this, we can see that by multiplying equation (1) by a factor of 2 and then subtracting this new equation from (2), the $y$-variable will be eliminated.

(1) $\times 2 \rightarrow 4x + 6y = 20$ (1a) (Assigning the number (1a) to this equation)

Now subtracting (2) from (1a):

 $4x + 6y = 20$ (1a) $-$ $2x + 6y = 6$ (2) $=$ $2x + 0y = 14$

Now that we have $2x = 14$, we can solve for $x$, which in this case is $7$.

$x = 7$.

Substitute the newly found $x$ into (1):

$2 \times 7 + 3y = 10$

$14 + 3y = 10$

And we find that $y = - 4/3$

So, the solution to the two equations (1) and (2) are:

$x = 7$

$y = -4/3$

Substitution method

The substitution method relies on being able to rearrange the expressions to isolate a single variable, in the form variable = expression. From this result this new expression can then be substituted for the variable itself, and the solutions evaluated.

An example of this:

 $2x + 3y = 12$ (1) (Assigning the number (1) to this equation) $x + y = 6$ (2) (Assigning the number (2) to this equation)

From this expression, it is possible to see that (2) is the most simplistic expression, and thus will be the better choice to rearrange.

Taking (2), and rearranging this into $x = 6 - y$. (2a)

Subbing (2a) into (1) we get

$2(6 - y) + 3y = 12$

Solving this, we get that $y = 0$

Again we can sub this result into one of the original equations to solve for $x$. In this case $x = 6$.

Note that for situations in which one of the equations is non-linear, you must isolate one variable in the linear equation and substitute it into the non-linear one. Then you can solve the quadratic equation with one of the methods above.

Another form of substitution is if you've got a similar expression in both equations, like in this case:

 $2x + 3y = 10$ (1) (Assigning the number (1) to this equation) $2x + 6y = 6$ (2) (Assigning the number (2) to this equation)

Here, $2x$ is found in both equations, so:

 $2x = 10 - 3y$ (1) $2x = 6 - 6y$ (2)

And since $2x = 2x$, you could do:

$10 - 3y = 6 - 6y$

$6y - 3y = 6 - 10$

$3y = -4$

$y = -4/3$

Now you've got $y$, and finding $x$ will be the same as above.

Graphical method

By plotting the lines of the two equations, you can solve them by seeing where the lines intersect. If the intersection is at the point (a,b), then the solution is $x=a$ and $y=b$.

Solving problems with simultaneous equations

Often, you will be given problems which you must be able to write out as a pair of simultaneous equations. You will need to recognise such problems, and write them out correctly before solving them. Most problems will be similar to these examples with some differences.

Example 1

At a record store, 2 albums and 1 single costs £10. 1 album and 2 singles cost £8. Find the cost of an album and the cost of a single.

Taking an album as $a$ and a single as $s$, the two equations would be:

$2a+s=10$

$a+2s=8$

You can now solve the equations and find the individual costs.

Example 2

Tom has a budget of £10 to spend on party food. He can buy 5 packets of crisps and 8 bottles of drink, or he can buy 10 packets of crisps and 6 bottles of drink.

Taking a packet of crisps as $c$ and a bottle of drink as $d$, the two equations would be:

$5c+8d=10$

$10c+6d=10$

Now you can solve the equations to find the cost of each item.

Example 3

At a sweetshop, a gobstopper costs 5p more than a gummi bear. 8 gummi bears and nine gobstoppers cost £1.64.

Taking a gobstopper as $g$ and a gummi bear as $b$, the two equations would be:

$b+5=g$

$8b+9g=164$

The problem can now be solved by using one of the methods above.

A quadratic is a Polynomial of degree 2, in the form $f(x)=ax^2+bx+c$.

Graph

A quadratic graph is one that can be written in the form $y=ax^2+bx+c$. The graph of $y=2x^2 + 8x + 2$ is shown on the right, and as you can see it has the same characteristic "bucket" shape that all quadratics have, called a parabola. The line of symmetry and the vertex (maximum or minimum point) of the graph can be found by plotting the curve point by point, and the root can be found by factorising the quadratic.

However, these properties can be more easily deduced from its completed square form:

$y=2(x + 2)^2 - 6$.

Completing the square

Completing the square is the process of changing a quadratic from the form $ax^2 + bx +c$ to the equivalent form $a(x+d)^2 + e$, where $a$, $d$ and $e$ are constants. For example, the quadratic $2x^2 + 8x +2$ would become $2(x+2)^2 - 6$.

Changing a quadratic to completed square form makes it easy to find several things, such as the roots of the quadratic and the vertex of the quadratic, without even requiring a sketch.

Here are the steps for completing the square. Don't worry, it's easier than it looks.

Step Action Example General case
1. Ensure the quadratic is in the conventional form: $ax^2 + bx + c$. $2x^2 + 8x + 2$ $ax^2 + bx + c$
2. Unless $a=1$, "pull/factor out a", that is, divide the entire quadratic by $a$ and put it outside a bracket.

Note: If the quadratic is part of an equation you can divide each side by $a$, for example $2x^2 + 8x + 2=0$ simply becomes $x^2 + 4x + 1=0$.

$2\left(x^2 + 4x + 1\right)$ $a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)$
3. Replace the $x^2 + kx$ part with $\left(x + \frac{k}{2}\right)^2$.

It is important to realise that $\left(x + \frac{k}{2}\right)^2=x^2 + kx + \frac{k^2}{4}$ which is close to what it was there before but not equal. This will be corrected in the next step. To prevent writing something that isn't actually equal it is a good idea to do both these steps at once in your working once you have got used to the method.

$2\left((x+2)^2 + 1\right)$ $a\left(\left(x + \frac{b}{2a}\right)^2 + \frac{c}{a}\right)$
4. Correct the error introduced in the previous step by inserting the subtraction of a suitable number. This suitable number can be found in two ways:
1. By expanding the term inserted in the previous step and comparing it to the original; or
2. By remembering that the error is always $\frac{k^2}{4}$.

This step is known as "completing the square" and gives the method its name.

1. $(x+2)^2 = x^2 + 4x + 4$ which is 3 greater than $x^2 + 4x + 1$, so $-3$ is inserted: $2\left((x+2)^2 - 3\right)$; or
2. $\frac{4^2}{4} = 4$ so the error is 4: $2\left((x+2)^2 -4 +1\right)$$= 2\left((x+2)^2 -3\right)$
$a\left(\left(x + \frac{b}{2a}\right)^2 -\frac{b^2}{4a^2} + \frac{c}{a}\right)$
5. If step 2 was necessary then simplify the result a bit by expanding the outer bracket. $2(x+2)^2 -6$ $a\left(x + \frac{b}{2a}\right)^2 -\frac{b^2}{4a} + c$
6. Check that what you have obtained expands back to what you started with.

Note: You may feel confident enough to skip this step.

$2\left(x^2 + 4x + 4\right) -6$

$=2x^2 + 8x + 8 - 6$ $=2x^2 + 8x + 2$

$a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) -\frac{b^2}{4a} + c$

$=ax^2 + bx + \frac{b^2}{4a} -\frac{b^2}{4a} + c$ $=ax^2 + bx + c$

So the completed square form of $y=2x^2 + 8x + 2$ is $2(x+2)^2 -6$. The $-6$ tells us that the lowest point of the curve is at $y=-6$ and the $x+2$ tells us that the line of symmetry is at $x+2=0$ or $x = -2$. Therefore the vertex is at $(-2,-6)$, and if you look at the graph, you can see that is the case.

You can see from the graph that $y=2x^2 + 8x + 2$ has one root between $-4$ and $-3$ and another between $-1$ and $0$, where it crosses the $y$ axis. But how do you find the exact values? Using the completed square form, it can be re-arranged quite easily to find $x=0$:

Step Action Example General case
1. To solve an equation in the form $ax^2 + bx + c=0$ first complete the square using the method above. $2(x + 2)^2 - 6=0$ $a\left(x + \frac{b}{2a}\right)^2 -\frac{b^2}{4a} + c=0$
2. Isolate the $(x + k)^2$ term. $2(x + 2)^2=6$

$(x + 2)^2=3$

$a\left(x + \frac{b}{2a}\right)^2 =\frac{b^2}{4a} - c$

$\left(x + \frac{b}{2a}\right)^2 =\frac{b^2}{4a^2} - \frac{c}{a}$

3. Square root each side, including a $\pm$ as the bit inside the bracket might be negative or positive. $x + 2=\pm\sqrt{3}$ $x + \frac{b}{2a} =\pm\sqrt{\frac{b^2}{4a^2} - \frac{c}{a}}$

Then some simplification:

$x + \frac{b}{2a} =\pm\sqrt{\frac{b^2}{4a^2} - \frac{4ac}{4a^2}}$

$x + \frac{b}{2a} =\pm\sqrt{\frac{b^2-4ac}{4a^2}}$

$x + \frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$

$x + \frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{2a}$

4. Isolate $x$. $x = \pm\sqrt{3} - 2$

$x = \sqrt{3}-2$ or $-\sqrt{3}-2$ $x \approx -0.268$ or $-3.73$

$x =\pm\frac{\sqrt{b^2-4ac}}{2a} - \frac{b}{2a}$

$x =\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

The values of $x$ for this specific example are within the expected range, as seen on the graph.

The quadratic formula is derived from the general case of completing the square:

$x =\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

It can be used to find the roots of a quadratic by putting numbers directly into it. For example, for $y=2x^2 + 8x + 2$:

$x =\frac{-8 \pm \sqrt{64-16}}{4} = -2 \pm \sqrt{3}$

so $x = -2 + \sqrt{3}$ and $x = -2 - \sqrt{3}$.

The discriminant

Notice that the quadratic equation contains $b^2 - 4ac$ inside a square root sign. This part is called the discriminant and can be considered on its own to determine the number of roots of the equation.

• If $b^2 - 4ac<0$ then you will be unable to find the square roots as you don't know how to square root a negative number. The type of numbers you have encountered so far are known as real numbers and so it is said that the quadratic has no real roots.
• If $b^2 - 4ac=0$ then changing the $\pm$ sign in front of the square root won't make any difference, because it is zero either way. You will therefore get the same root twice, so it is said the quadratic has one repeated root.
• If $b^2 - 4ac>0$ then the $\pm$ will mean you get two answers, and so you can say the quadratic has two distinct (i.e. different) roots.

Inequalities

An inequality is an expression which compares the relative sizes of points, lines, or curves. Unlike an equation, where both sides of the equals sign are always equal, inequalities can have one side greater than or equal to the other side.

The four signs of inequalities

There are four main basic signs:

• $<$ less than,
• $>$ greater than,
• $\le$ less than or equal to, and
• $\ge$ greater than or equal to.

For example, $x < 4$ means that $x$ is less than 4, $x > 4$ means that $x$ is greater than 4, $x\le 4$ means that $x$ is four or any number less than this, and $x\ge 4$ means that $x$ is four or any number higher than this.

Note that $x > y$ and $y < x$ are both essentially the same statement.

If you become confused with which sign means less than and greater than, it is useful to remember that the inequality signs always point to the smaller number.

Combining inequalities

There are some cases where two inequalities can be combined into one. For example, the height of the door was said to be between $x \ge 1.95$ and $x < 2.05$. The usual way of writing these is $1.95 \le x < 2.05$. Notice that the inequality signs are in the same direction. $2.05 \ge x > 1.95$ is perfectly acceptable, but it is incorrect to combine opposite facing inequalities and they must be left as two separate inequalities.

Solving linear inequalities

These signs can be used in place of equal signs, and an equation now becomes an inequality (since both sides are not always equal).

For example, instead of $2x + 4 = 6$ we could have $2x + 4 > 6$.

In this example, $x$ may be any number which makes this inequality greater than 6. In this case $x > 1$ but $x \ne 1$. If the inequality was $2x + 4 \ge 6$, then $x$ could take the value of 1.

An inequality can be manipulated and therefore solved just like an equation, although there is an extra step you must take when you multiply or divide by a negative number.

Multiplying or dividing by a negative number

When multiplying or dividing by a negative number, you must change the direction of the inequality sign.

For example, look at the inequality $10 > 5$. This is correct since 10 is obviously greater than 5. Now if we were to multiply both sides by -1, we would get:

$-10 > -5$.

This is incorrect, since -10 is actually less than -5. By reversing the inequality sign, we now have the correct inequality:

$-10 < -5$.

To solve a quadratic inequality, you could factorise it, just like a quadratic equation.

Alternatively you could draw its graph as if it was a quadratic equation, and then shade the side that's covered by the inequality. maybe an image here, to demonstrate the method?

Indicies

You are probably already familiar with indices, for example $x^2$ is just a shorter way of writing $x \times x$ and $x^4$ is similarly $x \times x \times x \times x$. In $x^5$, $x$ is called the base and $5$ is called the power or exponent. $x^4$ is pronounced "x to the four", or "x raised to the 4th power" in full. Some powers are so useful that they have special names: $x^2$ is referred to as "x squared", $x^3$ is "x cubed" and $x^{-1}$ (which you will soon learn about if you haven't already encountered it) is called "the reciprocal of x".

Note: The "law of indices" is sometimes also called the "exponent laws" or "power rules" [1]. More generally, an index in mathematics is a superscript or subscript to a symbol.

Operations with indices

Using this notation you might notice several patterns.

Multiplication

Firstly, $x^3 \times x^2$ is $\left(x \times x \times x\right) \times \left(x \times x\right)=x \times x \times x \times x \times x$ which is $x^5$. Of course $3+2=5$ so you have added the powers together. To clarify, here is an example with numbers: $2^3 \times 2^5 = 8 \times 32 = 256 = 2^8$ (like before, $3+5=8$)

Division

Secondly $\frac{x^4}{x^2}$ is $\frac{x \times x \times x \times x}{x \times x} = x \times x = x^2$ (when $x \neq 0$). This time $4-2=2$ and so you have subtracted the powers.

Here is an example with numbers: $\frac{10^5}{10^2}=\frac{100000}{100}=1000=10^3$ and again $5-2=3$.

Base raised to two powers

Thirdly $\left(x^2\right)^3$ is $\left(x \times x\right) \times \left(x \times x\right) \times \left(x \times x\right)=x \times x \times x \times x \times x \times x$ which is $x^6$. You can see that $2 \times 3 = 6$ and so the powers have been multiplied. Here is another example with numbers: $\left(3^2\right)^4=9^4=6561=3^8$ and $2 \times 4 = 8$.

Multiple bases

Finally $\left(xy \right)^3 = xy \times xy \times xy = x \times y \times x \times y \times x \times y = x \times x \times x \times y \times y \times y$ which is the same as $x^3y^3$. Here is an example with numbers: $\left(2 \times 5\right)^2 = \left(10\right)^2 = 100 = 4 \times 25 = 2^2 \times 5^2$. There is a similar situation with division: $\left(\frac{x}{y}\right)^2 = \frac{x}{y} \times \frac{x}{y} = \frac{x \times x}{y \times y} = \frac{x^2}{y^2}$

The Laws of Indices

The rules that have been suggested above are known as the laws of indices and can be written as:

1. $x^ax^b = x^{a+b}$
2. $\frac{x^a}{x^b} = x^{a-b}$
3. $\left(x^a\right)^b = x^{ab}$
4. $\left(xy \right)^n = x^n y^n$
5. $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$

Special indices

You may well have realised that ${x^1}=x$. This can be seen by looking at the pattern of $x^3=x \times x \times x$, $x^2 = x \times x$ or by doing $\frac{x^3}{x^2}$ which is clearly $x$ but is also $x^{3-2}=x^1$ by law 2.

So far all the examples we have looked at are where the power is a positive integers, but by thinking about the laws it is possible to look at other cases.

Power 0

It is less obvious that $x^0=1$ for any $x$ (strictly speaking, any $x \neq 0$, see the note below.), but this can be shown in a similar way, for example $\frac{x^4}{x^4}$ is 1 but must be also be $x^{4-4}=x^0$ by law 2.

Negative powers

The next logical step is to ask what $x^{-1}$ is. Well by using law 2 "backwards" $x^{-1} = x^{0-1} = \frac{x^0}{x^1} = \frac{1}{x}$. A similar argument can be used for any other negative integer, for example $x^{-3} = x^{0-3} = \frac{x^0}{x^3} = \frac{1}{x^3}$.

Fractional powers

What if the power isn't even an integer? Suppose you wanted to find $x^\frac{1}{2}$, you could say that ${x^\frac{1}{2}} \times {x^\frac{1}{2}} = x^1$ (by law 3, addition of powers) which means that $x^\frac{1}{2}$ must be $\pm \sqrt x$. However it is customary to only use the positive root and so $x^\frac{1}{2}$ is defined as $\sqrt x$. You can use a similar argument for other such fractions, for example $\left(x^\frac{1}{3}\right)^3=x$ so $x^\frac{1}{3} = \sqrt[3]{x}$.

In summary:

• $x^1=x$
• $x^0=1$
• $x^{-n}=\frac{1}{x^n}$
• $x^\frac{1}{n}=\sqrt[n]{x}$

Sometimes you might have to use the laws to understand what something means, for example $x^\frac{2}{3} = \left(x^2 \right)^\frac{1}{3}$ (using law 3), and $\left(x^2\right)^\frac{1}{3} = \sqrt[3]{x^2}$ (using the definition above). It's useful to remember the general rule that $x^\frac{a}{b} = \sqrt[b]{x^a}$.

Surds

In mathematics, a Surd is an expression containing a root with an irrational solution that can not be expressed exactly – for example, √3 = 1.732050808... .

Sometimes it is useful to work in square roots, rather than using an approximate decimal value. Square roots can be manipulated just like algebraic expressions and sometimes it may be possible to eliminate the square root (called rationalising the expression), which may have not been possible if you tried to work with the approximate value. When asked to give the exact value, approximate decimal answers will not do and you will have to manipulate surds in order to give a final answer in simplified surd form.

Simplification of surds

Because surds can be manipulated like algebraic expressions, you can easily multiply out the terms and add the like terms. However, there are also a few rules that will be useful when simplifying surds.

Basic rule of surds

Because $\sqrt {x} \times \sqrt {x} = x$, it is useful to know that it can be rearranged to give $\sqrt {x} = \frac {x} {\sqrt {x}}$ and $\frac {1} {\sqrt {x}} = \frac {\sqrt {x}} {x}$.

Surds as indices

Because $\sqrt[n]{x} = x^\frac{1}{n}$ the laws of indices also apply to any n-th root. The most frequently used instances of this are laws 4 and 5 with square roots:

• $\left(xy \right)^n = x^n y^n$ becomes $\sqrt{xy} = \sqrt{x} \times \sqrt{y}$
• $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$ becomes $\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}$

The first of these points is often used to simplify a square root, for example $\sqrt{200}=\sqrt{100 \times 2}=\sqrt{100} \times \sqrt{2}=10\sqrt{2}$. In an exam, you will be expected to write all square roots with the smallest possible number inside the square root (i.e. the number inside the root shouldn't have any square factors).

Rationalising the denominator

Another technique to simplify expressions involving square roots is to rationalise the denominator. This means getting rid of square roots from the bottom of a fraction. In the case of a fraction such as $\frac{5}{\sqrt{3}}$, both numerator and denominator can be multiplied by $\sqrt{3}$ to give $\frac{5\sqrt{3}}{3}$.

If the fraction is of the form$\frac{a}{b+\sqrt{c}}$ the strategy used in the previous paragraph will only work if it is modified slightly. This time you should multiply the numerator and denominator by ${b-\sqrt{c}}$. If you are familiar with the standard difference of two squares expansion you should already know what happens next:

$\frac{a}{b+\sqrt{c}} \times \frac{b-\sqrt{c}}{b-\sqrt{c}} = \frac{ab-a\sqrt{c}}{b^2 + b\sqrt{c} - b\sqrt{c} - \sqrt{c}^2} = \frac{ab - a\sqrt{c}}{b^2-c}$. As you can see the denominator now does not contain any square roots. For example: $\frac{2}{\sqrt{3} - 1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{2\sqrt{3}+2}{3-1} = \sqrt{3}+1$

Common questions and mistakes

Splitting up roots

A common mistake is to split $\sqrt{x + y}$ into $\sqrt{x} + \sqrt{y}$ or $\left(x+y\right)^2$ into $x^2 + y^2$, usually whilst moving it to the other side of the equals. Trying a few examples will quickly convince you that this is not possible:

• $\sqrt{25}$$\sqrt{9} + \sqrt{16}$
• $\sqrt{64}$$\sqrt{32} + \sqrt{32}$

And so on

What is the value of $0^0$?

The short answer is that for this course you don't need to know, and you can safely skip this section. If you're still interested then read on:

The question arises because $x^0=1$ for any $x$ and yet you would expect that $0^y=0$ for any $y$ as $0 \times 0 \times 0 \dots = 0$. It turns out using a value of 1 is quite useful (perhaps even necessary) in various parts of algebra, whereas making it zero doesn't help at all. Almost all mathematicians would therefore either say that $0^0$ is 1 or that it is undefined (that is, it can't be given a value). A more technical discussion can be found at http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/