Quadratics are ubiquitous in mathematics. For the purposes of this piece a quadratic is a real-valued function $q:\mathbb{R}\rightarrow \mathbb{R}$ of the form

$q(x)=ax^2+bx+c$,

where $a,\,b,\,c\in \mathbb{R}$ such that $a\neq 0$. There is a little bit more to be said — particularly about the differences between a quadratic and a quadratic function but for those this piece is addressed to (third level: non-maths; all second level), the distinction is unimportant.

## Geometry

The basic object we study is the square function, $s:\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto x^2$:

All quadratics look similar to $x^2$. If $a>0$ then the quadratic has this $\bigcup$ geometry. Otherwise it looks like $y=-x^2$ and has $\bigcap$ geometry

The geometry dictates that quadratics can have either zero, one or two real roots. A root of a function is an input $x$ such that $f(x)=0$. As the graph of a function is of the form $y=f(x)$, roots are such that $y=f(x)=0\Rightarrow y=0$, that is where the graph cuts the $x$-axis. With the geometry of quadratics they can cut the $x$-axis no times, once (like $s(x)=x^2$), or twice.

Finding the roots of quadratic functions is a problem that occurs very often in mathematics. It is equivalent to solving a quadratic equation:

$ax^2+bx+c=0$.

Trying to solve quadratic equations in your head is fraught with problems, and trying to do so leaves one with the impression that:

Finding when a Sum is Zero is HARD

The key to solving quadratics lies in this realisation along with the principle that:

Finding when a Product is Zero is EASY

Products are numbers multiplied together: $a+b$ is a sum, while $a\cdot b$ is a product. This principle is a direct result of the following:

### No Zero Divisors

If $a$ and $b$ are two numbers such that $a\cdot b=0$, then $a=0$ or $b=0$.

Therefore if a quadratic can be writen as a product:

$q(x)=a(x)\cdot b(x)$,

then the roots of $q(x)$ are given by the (hopefully) easier $a(x)=0$ and $b(x)=0$. So how do we do this?

## Factorisation

Find a re-writing of $q(x) =ax^2 + bx + c$ as

$q(x)=ax^2+kx+mx+c$

such that $b = k + m$ (so as not to change the quadratic) and $km = ac$. From here, factorisation is inevitable using the distributive law:

$ab+ac=a(b+c)$.

### Example

Factor $6x^2-19x-7$. Hence find the roots of $q(x)=6x^2-19x-7$.

Solution: We calculate $ac=6(-7)=-42$.  Now, from one, let us list the factors of 42:

$1\times 42, 2\times 21, 3\times 14,6\times 7$.

We don’t have to go to seven as we already have seven. Now we need a pair of numbers with product minus $42$ so one of these has to negative. We also need the pair to add up to minus $19$… $latex 2\times(-21)$ does the trick:

$6x^2-19x-7=6x^2+2x-21x-7$.

Now we take something common out of the first two terms and do the same for the second two terms in such a way that we are left with the same factor twice:

$2x(3x+1)-7(3x+1)$.

Now we can take out the common factor of $(3x+1)$:

$6x^2-19x-7=(3x+1)(2x-7)$.

Now once we have the factorisation, finding the roots is easy:

$q(x)=0$

$\Rightarrow (3x+1)(2x-7)=0$

$\Rightarrow 3x+1=0$ or $2x-7=0$

$\Rightarrow 3x=-1$ or $2x=7$

$\displaystyle \Rightarrow x=-\frac13$ or $\displaystyle\frac72$.

### Example

Solve $5x^2+7x-3=0$.

Solution: We calculate $ac=5(-3)=-15$.  Now, from one, let us list the factors of 15:

$1\times 15, 3\times 5$.

Now we need a pair of numbers with product minus $15$ so one of these has to negative. We also need the pair to add up to $7$… this is impossible — we can only make the positive numbers 14 and two… factorisation has failed.

It is possible to solve the equation. It involves some gymnastics and writing it in the form:

$(10x+7)^2-109=0$.

These gymnastics can be undertaken for any quadratic and this leads us to a formula that can solve all quadratics:

## Formula

Suppose that $q(x)=ax^2+bx+c$ is a quadratic. Then the roots of $q$ are given by:

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

### Example

Find the roots of $5x^2+7x-3$.

Solution: Using the formula with $a=5,\,b=7,\,c=-3$, we have roots

$\displaystyle x_{\pm}=\frac{-7\pm\sqrt{7^2-4(5)(-3)}}{2(5)}=\frac{-7\pm\sqrt{109}}{10}$.

## Appendices: Proofs

### The Geometry of the $s(x)=x^2$ Graph

Of course, we have $s(0)=0^2=0$ and $s(1)=1^2=1$. We will now compare the geometry of the graph of $y=s(x)$ with the line $y=x$. If the scales on the $x$– and $y$– axes are the same, $y=x$ is a line through the origin at an angle of $45^\circ$.

Take now $0, then

$s(x)-x=x^2-x=\underbrace{x}_{+}\underbrace{(x-1)}_{-}<0$

$\Rightarrow s(x)-x<0\Rightarrow s(x) for $0.

Consider $x>1$. Then

$s(x)-x=x^2-x=\underbrace{x}_{+}\underbrace{(x-1)}_{+}>0$.

This gives $s(x)>x$ for $x>1$.

We can also show to other facts. The slope near $x=0$ is small in the sense that if we take the two points on the graph of $y=s(x)$ given by $(0,0)$ and $(\delta,\delta^2)$ (for $\delta>0$ but $\approx 0$), then the slope between them is given by:

$m=\frac{\delta^2}{\delta}=\delta\approx 0$.

We can also show that the slope between points is increasing from zero… this also shows that the function is increasing. Take $0 so that the slope between $(x,x^2)$ and $(x+\delta,(x+\delta)^2)$ is given by:

$\displaystyle \frac{(x+\delta)^2-x^2}{x+\delta-x}=\frac{(x+\delta+x)(x+\delta-x)}{x+\delta-x}=2x+\delta,$

which increases with $x$.

Putting these facts together, $y=s(x)$ goes through the origin and $(1,1)$, and with an increasing slope, goes from ‘below’ $y=x$ for $0 to ‘above’ for $x>1$. Finally, the graph is symmetric about the $y$-axis as

$s(-x)=(-x)^2=x^2=s(x)$.

### All Quadratics are Translations of $s(x)=x^2$

Three facts that are presented without proof:

• The graph of $y=f(x+ a)$ is a horizontal translation of the graph of $y=f(x)$.
• The graph of $y=a\cdot f(x)$ is related to the graph of $y=f(x)$:
• if $a>0$ the the graph of $y=a\cdot f(x)$ is a horizontal scaling of $y=f(x)$,
• if $a<0$ the graph of $y=a\cdot f(x)$ is an upside-down horizontal scaling of $y=f(x)$.
• The graph of $y=f(x)+a$ is vertical translation of the graph of $y=f(x)$.

By multiplying out, any quadratic $q(x)=ax^2+bx+c$ is of the form:

$\displaystyle q(x)=\underbrace{a\cdot}_{\text{scaling}}\left(x\underbrace{+\frac{b}{2a}}_{\text{horizontal translation}}\right)^2\underbrace{+\left(c-\frac{b^2}{4a}\right)}_{\text{vertical translation}}.$

All quadratic equations of the form $ax^2+bx+c=0$ can be written in the form:
$(2a+b)^2-(b^2-4ac)=0$.
(To see this is equivalent to the original, multiply out and divide by $4a$). Solve this to get the ‘$-b$‘ formula.