C1: Algebra & Equations

Written: 12/09/2008 Last edited: 29/07/2009 Filed under: WJEC Mathematics A-level C1

The following guide could probably always be improved... if it can, shout.

Laws of Indices

You are probably familiar with the notation $x\times x\times x\times \dots \times x=x^n$ where there are $n$ occurences of the $x$ terms, and also with the notation $\sqrt{x}\times\sqrt{x}=x$ and $\sqrt{x}\times\sqrt{x}\times\sqrt{x}\times\sqrt{x}=x^2$ etc,, as well as $\sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x}=x$ .

However, we can readily combine these two differenct notations to simplify some of our calculations.

The principle is readily seen from expressions like $\sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x}=x^2$ . We have six occurences of third roots multiplying to give a power of 2. Our logic is then to write $\sqrt[3]{x}=x^{1/3}$ so that $(x^{1/3})^6=x^{1/3\times6}=x^2$ .

Also, we use the rule $x^a\times x^b=x^{a+b}$ to see that $x^3\times\frac{1}{x^2}=x^1$ should be written $x^3\times x^{-2}=x^1$ .

The laws of indices are then as follows, where the exponents (powers) can be any rational numbers (fractions) which are positive or negative.

$x^{a/b}=\sqrt[b]{x^a}=(\sqrt[b]{x})^a$

$x^{-a}=\frac{1}{x^a}=(\frac{1}{x})^a$

$x^{a+b}=x^ax^b$

$x^{a-b}=\frac{x^a}{x^b}$

$(x^{a})^b=x^{ab}$

Surds

Sometimes, we come across numbers which incorporate square roots of whole numbers which are not perfect squares, such as $\sqrt{5}+1$ . Since the decimal expansions of these numbers neither stop nor repeat (since they are not rational - fractions of whole numbers), we often find it useful to be able to manipulate these numbers in this same, concise form.

We begin with the basic rules $\sqrt{x}\times\sqrt{y}=\sqrt{xy}$ and $\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$

Now we can easily figure out how to add, subtract and multiply surds:

$\begin{align*} (3 + \sqrt{7}) + (1 + 4\sqrt{7}) &= (3 + 1) + (1 + 4)\sqrt{7} \\ &= 4 + 5\sqrt{7} \\ (2 + 3\sqrt{3}) - (2 + 4\sqrt{3}) &= (2 - 2) + (3 - 4)\sqrt{3} \\ &= -\sqrt{3} \\ (3 + \sqrt{7}) \times (1 + 4\sqrt{7}) &= (3 \times 1) + (3 \times 4\sqrt{7}) \\ & + (1 \times 1\sqrt{7}) + (1\sqrt{7} \times 4\sqrt{7}) \\ &= 3 + 12\sqrt{7} + \sqrt{7} + 4\times (\sqrt{7})^2 \\ &= 31 + 13\sqrt{7} \end{align*}$

However, this leaves us with the problem of simplifying expressions such as

$\frac{a + b\sqrt{n}}{c + d\sqrt{n}}$

The trick needed is called:

Rationalizing the Denominator

The problem in the division is that the square root on the bottom cannot be easily be divided into the numerator on top.

To resolve this, we would like to transform the denominator into a single number. We do this by multiplying the top and bottom by $c - d\sqrt{n}$ , because the denominator becomes $(c + d\sqrt{n})(c - d\sqrt{n})$ which expands to $c^2 + cd\sqrt{n} - cd\sqrt{n} - d^2\sqrt{n}\sqrt{n} = c^2 - nd^2$ which involves no square roots.

For example: $\begin{align*} \frac{2 + \sqrt{5}}{3 - \sqrt{5}} &= \frac{(2 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} \\ &= \frac{11 + 5\sqrt{5}}{9 - 5} \\ &= \frac{11 + 5\sqrt{5}}{4} \\ &= \frac{11}{4} + \frac{5}{4}\sqrt{5} \end{align*}$ which is much easier to work with in any following manipulations.

Quadratics

A quadratic expression is one like this: $2x^2 - x + 3$ which has no powers of $x$ higher than 2.

Now the easiest way of understanding how an expression like this relates to the value of $x$ is almost always to plot a graph. So have a look around online for some graphs.

Graphs

What you find is that the graphs all either point up or down in a curve which is symmetrical about the point it stops at (in fact, this particular graph is called a parabola). The ones that point up are those with a negative coefficient of (number multiplying with) $x^2$ , and those that point down have a positive coefficient.

Also, because at $x=0$ the rest of the formula is zero, the last term (without any x) gives the 'height' at which the curve crosses the (vertical) y axis.

Now, since the graph has one turning point (at its minimum or maximum, depending), it can either:

stay completely on one side of the (horizontal) x axis and not touch it
just touch the x-axis at one point
cross the x-axis at two points

Now, finding where - if anywhere - the curve intersects (crosses) the x-axis is called finding the solution of the equation $y = ax^2 + bx + c = 0$

Factorization by Inspection

One of the first methods taught for solving quadratics is by inspection. This means, basically, guesswork or trial-and-error.

This is easiest to explain using an example. Take $y=x^2+11x+24$ What we want to do is write this as $y=(x + a)(x+b)$ for some new numbers. Let's have a go first.

If we expand the brackets in what we want, we get $\begin{array}{rcl} y&=&(x+a)(x+b) \\ &=& x^2 + ax + bx + ab \\ &=& x^2 + (a + b)x + ab \\ &=& x^2 + 11x + 24 \end{array}$ Now by 'comaprison of coefficients' (looking at the bits we multiply by different powers of $x$ ) we get $\begin{array}{rcl} a + b &=& 11 \\ ab &=& 24 \end{array}$ So we want two numbers that add to give 11 and multiply to give 24. Since these are nice whole numbers, we could try listing all the factors of 24 and seeing if they add to give 11. It turns out that $3\times 8=24$ and $3+8=11$ , so $y=x^2+11x+24=(x+3)(x+8)$

What use is this? Well, what we have here is two numbers, $x+3$ and $x+8$ , multiplying together to give $y$ . But if $y=0$ , then at least one of $x+3$ and $x+8$ must be zero. This means that either $\begin{array}{rcl} x+3 &=& 0 \\ x &=& -3 \end{array}$ or $\begin{array}{rcl} x+8 &=& 0 \\ x &=& -8 \end{array}$ so we have our solutions!

Completing the Square

However, this doesn't always work. Imagine getting decimals or fractions, or something like $x^2-x-1$ where you can't find whole numbers that multiply and add right. Our next strategy is to complete the square.

What this means is writing the original expression $ax^2+bx+c$ in a different form again. This time, we use $a(x+j)^2+k$ for two new numbers. Let's do what we did for inspection, and expand this: $\begin{array}{rcl} y&=& a(x+j)^2+k \\ &=& a(x+j)(x+j) + k \\ &=& ax^2 + ajx + ajx + aj\times j + k \\ &=& ax^2 + 2ajx + (aj^2 + k) \\ &=& 3x^2 + 7x + 1 \end{array}$ where we've picked a new example. Now, using our old 'comparison of coefficients', $\begin{array}{rcl} a &=& 3 \\ 2aj &=& 7 \\ aj^2 + k &=& 1 \end{array}$

Well, the first line tells us $a=3$ . So in the second line, $2\times 3j = 7$ and $j=\frac{7}{6}$ . Then in the third line, $3\times\left(7\over 6\right)^2+k=1$ so $k=1-3\times\left(7\over 6\right)^2$ or $k=1-\frac{3\times 49}{36}=1-\frac{49}{12}=\frac{-37}{12}$ .

Lovely - we now know that $\begin{array}{rcl} y &=& 3x^2 + 7x + 1 \\ &=& a(x+j)^2+k \\ &=& 3\left(x+\frac{7}{6}\right)^2 + \frac{-37}{12} \end{array}$ The clever bit here is separating $x$ from the rest of the equation. This allows us to do several tricks.

The Quadratic Formula

We can also work $x$ out backwards from $y$ when we've completed the square - so if we set $y=0$ ... $\begin{array}{rcl} y &=& 3\left(x+\frac{7}{6}\right)^2 - \frac{37}{12} \\ &=& 0 \\ 3\left(x+\frac{7}{6}\right)^2 &=& \frac{37}{12} \\ \left(x+\frac{7}{6}\right)^2 &=& \frac{37}{3\times 12} \end{array}$ Now we can take square roots (which can be positive or negative!) to get $\begin{array}{rcl} x+\frac{7}{6} &=& \pm\sqrt{\frac{37}{36}} \\ x &=& \pm\sqrt{\frac{37}{36}} - \frac{7}{6} \\ &=& \pm\frac{\sqrt{37}}{6} - \frac{7}{6} \\ &=& \frac{\pm\sqrt{37} - 7}{6} \\ &=& \frac{\sqrt{37} - 7}{6} \text{ or } \frac{-\sqrt{37} - 7}{6} \end{array}$

... and those are the solutions!

Now, we can actually use this to derive the next really clever formula from scratch (don't worry - you don't ever have to do this yourself!). Let's say we have $y=ax^2+bx+c$ Now to complete the square, we know from above that we end up with $\begin{array}{rcl} a &=& a \\ 2aj &=& b \\ aj^2 + k &=& c \end{array}$ Can you can see why we chose $a$ as the number outside the brackets now? Anyway, solving these to work out $j$ and $k$ we get $\begin{array}{rcl} 2aj &=& b \\ j &=& \frac{b}{2a} \end{array}$ and $\begin{array}{rcl} aj^2 + k &=& c \\ k &=& c - aj^2 \\ &=& c - a\left(\frac{b}{2a}\right)^2 \\ &=& c - a\frac{b^2}{4a^2} \\ &=& c - \frac{b^2}{4a} \\ &=& \frac{4ac-b^2}{4a} \end{array}$

So we know $\begin{array}{rcl} y &=& ax^2 + bx + c \\ &=& a\left(x+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a} \\ &=& 0 \end{array}$ from which we can work out that $\begin{array}{rcl} a\left(x+\frac{b}{2a}\right)^2 &=& \frac{b^2-4ac}{4a} \\ \left(x+\frac{b}{2a}\right)^2 &=& \frac{b^2-4ac}{4a^2} \\ x+\frac{b}{2a} &=& \pm\sqrt{\frac{b^2-4ac}{4a^2}} \\ x &=& \pm\sqrt{\frac{b^2-4ac}{4a^2}} - \frac{b}{2a} \\ &=& \pm\frac{\sqrt{b^2-4ac}}{2a} - \frac{b}{2a} \\ &=& \frac{\pm\sqrt{b^2-4ac} - b}{2a} \end{array}$ or $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ which is known as the quadratic formula, and allows you to work out the solutions of any quadratic equation.