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.
Deriving the quadratic formula
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).
Applying the quadratic formula
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