When you apply the right technique, solving quadratic equations is easy. In this write-up, I provide an easy to follow guide on finding solutions to these solutions.
A genre of algebraic problems, which I have personally enjoyed solving in my school years were the quadratic equations. In applied mathematics and particularly in the physical sciences, quadratic equations naturally arise when solving actual problems. Ergo, solving quadratic equations is an essential part of a budding scientist’s or mathematician’s training. Through this Buzzle article, I provide a guide on solving quadratic equations using two different techniques.
What are Quadratic Equations?
Let me define what is meant by a quadratic equation as a math term, before elucidating simple techniques of finding a solution. A polynomial, single variable equation, with variables that have the highest power to be 2 or highest degree to be second, are known as quadratic equations. A typical quadratic equation can be written in the following form.
ax^{2} + bx + c = 0
Here a, b and c are constants or pure numbers, while ‘x’ is the variable. Notice that there is a single term of second degree and there are no expressions like x^{3} or x^{4} with powers greater than 2. Being a second degree equation, such a quadratic equation has two solutions. These two solutions may be real or imaginary numbers. It’s good practice to convert any quadratic equation into the above presented standard form, before solving them. Now that you are acquainted with the nature of a quadratic equation, let me outline strategies to solve them, in the following section.
Techniques of Solving Quadratic Equations
There are more than one ways in which you can find the solution of a quadratic equation. In what follows, I briefly explain each solution method and illustrate its usage through the solving of an actual example.
Solve By Factoring
Factoring is the simplest method of solving quadratic equations. The technique works as follows. Firstly, bring the equation in standard form presented above. Then inspect the coefficient of ‘x’ term in particular, along with the coefficient of x^{2} and the constant term.
Factoring method works by splitting the ‘x’ term into two parts, such that a common factor can be found by grouping together each of its parts with the other two terms (which includes the x^{2} and the constant term). If the common factors can be found in such a fashion, that the quadratic equation can be converted into a product of two first degree or linear equations, you directly have your solutions. By equating the two first degree equation separately to zero, two solutions are found. This method is best illustrated through examples.
Example: Find the two roots of the equation – ‘x^{2} – 6x + 8 = 0′. |
Solution: x^{2} – 6x + 8 = 0 ∴ x^{2} – (4 + 2)x + 8 = 0 (Splitting central term into two parts to derive common factors) ∴ x^{2} – 4x – 2x + 8 = 0 ∴ x(x – 4) – 2(x – 4) = 0 (Factoring out the common terms from first and last two terms) ∴ (x – 2)(x – 4) = 0 ∴ x = 2 or x = 4 |
This technique of factoring will not work when common factors cannot be found. In that case, there is a second line of attack you could follow, which is explained in the next section.
Use the Standard Solution Formula
If the above technique doesn’t work, there is a surefire solution that is bound to work, which directly provides you with a solution. Using ‘Complete the Square‘ method, the solution for any type of quadratic equation is directly worked out for you. For any quadratic equation written in the form:
ax^{2} + bx + c = 0
Two solutions are: [-b + √(b^{2} – 4ac)] / 2a & -b – √(b^{2} – 4ac)] / 2a
To find the two values, all you have to do is substitute the values of the constants – a, b and c in the above two expressions to get the solution. Depending on whether the term (b^{2} – 4ac), referred to as ‘Δ’, is negative, positive or zero, you can predict the nature of the solutions or ‘roots’. Here are the three conditions you must remember:
- if Δ is negative, the solutions of the quadratic equation are imaginary.
- if Δ is positive, the solutions of the quadratic equation are real and distinct.
- if Δ is zero, the solutions of the quadratic equation are real and equal.
Let me now demonstrate the usage of this formula with an example in the following lines.
Example: Find the roots of the equation: x^{2} + 6x + 1 = 0 |
Solution: Comparing ‘x^{2} + 6x + 1 = 0′ with ‘ax^{2} + bx + c = 0‘, here a=1, b=6 and c=1
∴ Δ = (b^{2} – 4ac) = 6^{2} – 4 (1 x 1) = 32 Since Δ is positive, the roots are real and distinct. Root 1: x_{1} = [-b + √(Δ)] / [2a] = (-6 + √32)/2 Root 2: x_{2} = [-b – √(Δ)] / [2a] = (-6 – √32)/2 |
Thus, you can simply write out the roots of the equation using this method. That concludes this short tutorial on solving quadratic equations. The key to getting better at solving them or to get better at anything is PRACTICE. There is no other way out. Take up a host of quadratic equation examples and go after them one after the other. By the time you are a pro at solving these kinds of equations, you will be able to read off the solutions directly after a glance at the quadratic equation, when expressions can be factored. Happy solving!