# The Quadratic Formula: its origin and application The quadratic formula is one of students’ most favourite mathematical formulas. It can be quite easily memorized. Students eagerly use it to solve quadratic equations of the form ax2 + bx + c = 0. However, not everyone knows why we use it and where the quadratic formula comes from.

When we are trying to determine the x-intercepts of a parabola represented by a certain quadratic equation, we are solving that equation. Quadratic equations may have 1 solutions, 2 solutions or no solutions. This means that the corresponding parabola has 1, 2 or no x-intercepts. Sometimes, the x-intercepts are not “nice” numbers and it is difficult to use factoring methods to solve an equation to determine the x-intercepts. This is when the quadratic formula could be used.

Now, let’s see where the quadratic formula comes from, by first solving a general quadratic equation of the form ax2 + bx + c = 0, where a is the leading coefficient and c is the constant (the y-intercept of the parabola).

To start, move the constant to the other side of the equation

ax2 + bx = -c

Next, divide both parts of the equation by the leading coefficient

(ax)2/a + (bx)/a = (-c)/a

Then, complete the square

x2 + 2bx/2a + (b/2a)2 = (-c)/a + (b/2a)2

(x + (b/2a))2 = b2/4a2 – c/a

(x + (b/2a))2 = (b2 – 4ac)/4a2

The number of roots of this equation depends on the sign of the rational expression (b2 – 4ac)/4a2. The denominator of this expression is always positive. Therefore, the sign of the whole expression is determined by its numerator, i.e. b2 – 4ac.

The expression b2 – 4ac is called the Discriminant of any given quadratic equation of the form ax2 + bx + c = 0. It is labeled with the capital letter D.

Therefore, D = b2 – 4ac and it is a part of the quadratic formula.

When D is a positive number (greater than 0), then the quadratic equation has two possible solutions and we can continue deriving the quadratic formula as follows:

When D = 0, the equation has only one possible solution (two repeating x-intercepts that are also the vertex of a parabola).

Whereas, when D turns out to be a negative value, the equation has no solutions (the parabola does not have x-intercepts).

Solve an equation 9x2 + 6x + 1 = 0.

a = 9, b = 6, c = 1

D = b2 – 4ac = (6)2 – 4(9)(1) = 0

Since D = 0, we know that the equation will have two identical roots.

x1 = (-b + 0)/2a and x2 = (-b – 0)2a

x1 = -6/18 and x2 = -6/18

x1,2 = -1/3

Therefore, there is only one repeating solution x1,2 = -1/3

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