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 ****ax**^{2}+ bx + c = 0

^{2}+ bx + c = 0

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

^{2}

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

**Alternatively:**

The general form of a quadratic function is: f(x) = ax^{2} +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) = ax^{2} +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) = ax^{2} + bx + c

= a(-b/2a)^{2} + b(-b/2a) + c

= ab^{2}/4a^{2} – b^{2}/2a + c

= (ab^{2}– 2ab^{2} + 4a^{2} c ) / 4a^{2}

= (4a^{2}c – ab^{2}) / 4a^{2}

= (4ac – b^{2}) / 4a

Giving the vertex as (-b/2a , (4ac – b^{2}) / 4a)

or

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) = ax^{2} + bx + c

a(x-h)^{2} + k = a(x^{2} + h^{2} -2hx) + k

= ax^{2} + ah^{2} -2ahx + k = ax^{2} + bx + c

**a = a**

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

c = ah^{2} + k => k = c – ah^{2 } = c – a (-b/2a)^{2} = c – a (b^{2}/4a^{2}) => ** k = (4ac – b ^{2}) / 4a**

**Similarly for cubic equations:**