Quadratics: A First Look

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, $s:\mathbb{R}\rightarrow \mathbb{R}$ , $x\mapsto x^2$ :

graph1

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 $0<x<1$ , then

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

$\Rightarrow s(x)-x<0\Rightarrow s(x)<x$ for $0<x<1$ .

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 $0<x<x+\delta$ 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 $0<x<1$ 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 is related to the graph of :
- 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:

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

Quadratic Formula

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.

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Quadratics: A First Look