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


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.


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:


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?


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


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



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:


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:


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


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


\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.


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:


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


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}.


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


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

Consider x>1. Then


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


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}}.

Quadratic Formula

All quadratic equations of the form ax^2+bx+c=0 can be written in the form:


(To see this is equivalent to the original, multiply out and divide by 4a). Solve this to get the ‘-b‘ formula.