A quadratic equation has a common form of ax^2 + bx + c = 0. It is polynomials with order of two. This equation is ubiquitous equation in mathematics and physics. We call the value of x that satisfies the equation as the root of the equation. In quadratic equation, there are at most two roots.
The graph of the function y = ax^2 + bx + c is a parabola. The quadratic equation impresses the intersection of the graph with the x-axis. Since there are at most two roots, the graph intersects x-axis at most at 2 points.
There are three ways to solve this equation in general. They are factorization, completing square and using quadratic formula. The factorization is a direct method to find the roots. It is the best for simple quadratic equations. We cannot apply it for all of the equation. Completing square is a little bit tedious. Quadratic formula is the easiest. We can apply this method to all of the quadratic equation.
Factorization is the quickest method for a simple quadratic equation. We can determine the roots of the equation by inspection. If a is not equal to one, divide both sides with a. We have the equation of x^2 + (b/a)x + (c/a) = 0 . We guess two numbers p and q. The product of p and q must equal to (c/a). Moreover, the sum of p and q must equal to -(b/a). If there are such two numbers p and q, the equation can be rewrite as
(x + p) (x + q) = 0.
The roots of the quadratic equation are -p and -q. This is also true for a equals to 1. The product of p and q is then equal to c. The sum of p and q is equal to – b.
We can also let the equation in its original form. We try to factorize it into the form of (mx + p).(nx + q) = 0. The product of m and n must be equal to a. Furthermore, mq + np must be equal to b. Finally, the product of p and q must equal to c. This time the roots are – p/m and – q/n.
If a equals to one, it easier to find the roots. We can directly guess two numbers p and q. The sum of p and q must equal to – b and the product must equal to c.
More often, it is hard to find two numbers of p and q. Many students get frustrated to find two numbers that satisfied the condition mentioned above. This is especially true if a is not equal to zero. The only way to overwhelm it is to practice more and more. Experience is a good teacher. The more they practice they will identify soon whether the equation can be solved by factorization or not. They will able to find the roots quickly and correctly.
The second method is completing square. This method involves a complete square. I will show how it works here. Suppose we want to find the roots of 2x^2 + 8x – 3 = 0. Here are the procedures.
1. We divide both sides with the coefficient of x^2. Here we divide both sides with 2. The equation will become
x^2 + 4x – 3/2 = 0.
2. We add both sides with the square of the half of the x coefficient. In this example, we add both sides with 4. Notice that the (4/2)^2 = 4. The equation will become
x^2 + 4x – 3/2 + 4 = 4.
3. Rearrange the terms, we have
x^2 + 4x + 4 = 3/2 + 4
x^2 + 4x + 4 = 11/2
4. We can see that the left hand side is a complete square. That is,
(x + 2)^2 = 11/4
5. We square root both sides to have
x + 2 = sqrt (11/2) and x + 2 = – sqrt (11/2)
x = – 2 + sqrt (11/2) and x = – 2 – sqrt (11/2)
When I write sqrt (x), I mean it is the square root of x.
Thus, the roots of the quadratic equation are – 2 + sqrt (11/2) and – 2 – sqrt (11/2).
We derive the quadratic formula using completing square method. We begin with our modified form x^2 +(b/a) x + c/a = 0. We add both sides with (b/2a)^2 = (b^2) / (4a^2). Rearrange the terms we have an equation of
x^2 + (b/a)x + (b^2) / (4a^2) = (b^2) / (4a^2) – c/a
The left hand side of this equation is a complete square. We can write it into {x – b/(2a)}^2. The right hand side can be written as (b^2 – 4 ac) / (4a^2). We take the square root of both sides. We will have
x + b/(2a) = {sqrt ( b^2 – 4 ac)} / (2a) and x + b/(2a) = {-sqrt (b^2 – 4ac)} / (2a)
We can isolate x in the left hand side to determine the roots. Thus we have the root of the quadratic equation is
x = – b/(2a) + {sqrt (b^2 – 4 ac)} / (2a) and x = – b/(2a) – {sqrt (b^2 – 4ac) / (2a)}
We know this as the quadratic formula.
We call the term (b^2 – 4ac) in the quadratic formula as discriminant. We symbolize it with D. The value of D determines the kind of the roots. If D is greater than zero the equation has two real roots. If D equals to zero, the equation has exactly one root. If D is less than zero, the equation has two imaginary roots. The imaginary roots are conjugate one to another. For example, the quadratic equation x^2 – x + 1 = 0 has D = – 3. The roots are then 1/2 – sqrt (3) i and 1/2 + sqrt(3) i, which are conjugate each others. The symbol i here represent an imaginary number sqrt (-1).
Further analysis reveals the following facts. We assume that p and q are the roots of the quadratic equation. Then we find that p + q = – (b/a) and p.q = c/a. Moreover, we have p – q = {sqrt (D)}/2. Suppose a quadratic equation has two positive roots. The equation must have negative value of (b/a) and positive value of (c/a). Let’s take another quadratic equation, which has two negative roots. This equation must have positive values of (b/a) and (c/a). Once more, suppose we have a quadratic equation with opposite sign roots. This quadratic equation must have a negative value of (c/a). There are no requirement for the value of (b/a) this time. A quadratic equation with b = 0 have a opposite roots. For example, x^2 – 4 = 0 has roots 2 and – 2. A quadratic equation with a = c has reciprocal roots. For example, 2x^2 – 5x – 2= 0 has roots 2 and 1/2.
That is an explanation to understand the quadratic equation. I hope this writing can help other understanding the quadratic equation better.
