Quadratic Formula derived by Completing the Square

quadratic formula

One of the most important formulas encountered by students of second level maths is the quadratic formula, or what some students call the “minus B” formula.

The video below gives a very nice intuitive description of how this can be derived using the method known as “completing the Square”.


Completing the Square to write ax2+ bx + c = 0

in the form a(x-h)2 + k = 0  (Vertex Form)

complete the square image2

In the more general case where the coefficient of the squared term is not 1, we divide across by this coefficient first.


complete the square image3



The general form of a quadratic function is: f(x) = ax2 +bx +c. The formula to find the vertex of a quadratic function in general form is: (-b/2a, f(-b/2a)).

This can be derived from the formula used to find the roots of the equation.


The roots are the values of x where the graph cuts the x-axis. The x-axis is described by y=0. In other words the above formula gives the values of x when f(x) = ax2 +bx +c equals 0.

More generally the above formula gives the values of x when the graph cuts any horizontal line described by a constant value of y for which the graph exists. For all but one point on the graph their will be two solutions because a horizontal line will cut the parabola in two places except at the vertex. There only one point will be touched by a horizontal line. So here the discriminant (the bit under the root sign) will be zero as the + or – must give the same answer. The discriminant will be zero. The y ordinate for any point on the graph is a function of the x ordinate. Our coordinates can be written in the form (x, f(x)).

So at the apex: x = -b/2a and y = f(x)

f(x) = ax2 + bx + c

       = a(-b/2a)2 + b(-b/2a) + c

       = ab2/4a2 – b2/2a + c

       = (ab2– 2ab2 + 4a2 c ) / 4a2

      = (4a2c – ab2) / 4a2

      = (4ac – b2) / 4a

Giving the vertex as (-b/2a , (4ac – b2) / 4a)




You could start with the general vertex form f(x) = a(x-h)2 + k and expand it out. Then equate the resulting coefficients with those of f(x) = ax2 + bx + c

a(x-h)2 + k = a(x2 + h2 -2hx) + k

= ax2 + ah2 -2ahx + k = ax2 + bx + c

a = a

b = -2ah => h = -b/2a

c = ah2 + k => k = c – ah2 = c – a (-b/2a)2 = c – a (b2/4a2) =>   k =  (4ac – b2) / 4a


Similarly for cubic equations: