An algebraic equation can be defined as a mathematical statement in which two expressions are set equal to each other. Algebraic equations usually consist of a variable, coefficients, and constants.

## Types of Algebraic Equations

Algebraic equations are mathematically defined as equations of the form $P(x)=0$, where $P$ is a polynomial and $x$ is the solution of the equation. The classification of these equations is dependent on the degree of the polynomial $P$. This article aims to explore three commonly utilized types of algebraic equations.

### Linear equations

Linear equations are mathematical expressions of the form $P(x)=0$, where the degree of $P$ is equal to one. Specifically, they can be represented as $ax+b=0$. The process of solving linear equations typically involves isolating the variable $x$ by performing operations on both sides of the equation. In cases where the coefficient $a$ is non-zero, the solution to the linear equation can be expressed as $x=-\frac{b}{a}$. It is noteworthy that the solution is a fraction.

The equation $3x+6=5$ can be transformed into an equivalent form, namely $3x+1=0$, through the process of subtracting $5$ from both sides of the equation. The solution to this linear equation is then determined to be $x=-1/3$.

The quadratic equation, denoted by $P(x)=0$, where $P$ is a polynomial of degree 2, is a fundamental concept in mathematics. It is expressed in the form $ax^2+bx+c=0$, where $a$, $b$, and $c$ are real numbers, and $a\neq 0$. The solution of this type of equation is closely linked to the discriminant, denoted by $\Delta$, which is defined as $\Delta=b^2-4ac$. The discriminant plays a crucial role in determining the nature of the solutions of the quadratic equation. Specifically, depending on the sign of $\Delta$, we can distinguish three types of solutions.

 If the discriminant $\Delta=0$, the the quadratic equation has one solution “root” given $$x=-\frac{b}{2a}.$$ β Given that $a$ is non-zero, the quadratic equation can be expressed as $$x^2+\frac{b}{a}x+\frac{c}{a}=0.$$ Additionally, since the discriminant $\Delta$ is equal to zero, it follows that $c=\frac{b^2}{4a}$. Substituting this value of $c$ into the quadratic equation yields $$x^2+2\frac{b}{2a}x+\left(\frac{b}{2a}\right)^2=0.$$ This is a well-known remarkable identity, which can be simplified to $$\left(x+\frac{b}{2a}\right)^2=0.$$ Consequently, we obtain $x+\frac{b}{2a}=0$, and thus $x=-\frac{b}{2a}$.
 If $\Delta>0$, the quadratic equation has two distinct solutions given by $$x_1=\frac{-b+\sqrt{\Delta}}{2a},\quad x_2=\frac{-b-\sqrt{\Delta}}{2a}.$$ β With some effort, one can see that the quadratic equation is equivalent to \begin{align*} \left(x+\frac{b}{2a}\right)^2= \frac{b^2-4ac}{4a^2}=\left( \frac{\sqrt{\Delta}}{2a}\right)^2.\end{align*} Thus we have $$x+\frac{b}{2a}= \frac{\sqrt{\Delta}}{2a}\;\text{or}\; x+\frac{b}{2a}= – \frac{\sqrt{\Delta}}{2a}.$$ This ends the proof.
 In the event that the discriminant$\Delta$ is less than zero, the quadratic equation possesses two unique complex solutions, which are expressed as follows: $$x_1=\frac{-b+i\sqrt{-\Delta}}{2a},\quad x_2=\frac{-b-i\sqrt{-\Delta}}{2a}.$$ It is important to note that $i$ represents the imaginary unit, which is defined as the square root of negative one, i.e., $i^2=-1$. β Given that $\Delta < 0$, it follows that $\Delta= -(-\Delta)=i^2(-\Delta)$. Proceeding analogously to the preceding case, the quadratic equation may be expressed as \begin{align*} \left(x+\frac{b}{2a}\right)^2= \frac{i^2(-\Delta)}{4a^2}=\left( \frac{i\sqrt{-\Delta}}{2a}\right)^2.\end{align*} This establishes the conclusion of the proof.

## The polynomial equations

A polynomial equation is a mathematical expression in the form of $P(x)=0$, where $P$ is a polynomial function of degree greater than two. To solve such equations, a range of techniques can be employed, such as factoring and synthetic division.

## Q&A on algebraic equations

 Q1: Can algebraic equations have more than one variable? β A1: Yes, algebraic equations can involve multiple variables, creating systems of equations. Solving such systems requires finding values for all variables that satisfy all equations simultaneously.
 Q2: Are algebraic equations only relevant in mathematics? β A2: No, algebraic equations have far-reaching applications in science, engineering, economics, and many other fields. They are essential tools for modeling and problem-solving.
 Q3: How can I improve my skills in solving algebraic equations? β A3: Practice is key to mastering algebraic equations. Start with simpler equations and gradually tackle more complex ones. Seek guidance from textbooks, online resources, and educational institutions.

In conclusion, algebraic equations serve as the fundamental basis for mathematical problem-solving and offer a flexible approach to comprehending and representing real-world phenomena. As an educator, one has the potential to motivate students by highlighting the elegance and applicability of algebraic equations, thereby providing them with indispensable problem-solving abilities for their academic and vocational pursuits. By exploring the intriguing realm of mathematics collaboratively, we can foster a deeper understanding of the subject and equip forthcoming generations with the tools necessary to confront the obstacles of the future.

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