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 $x=-1/3$.
Quadratic equations
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 $$ x=-\frac{b}{2a}.$$ | ➕ |
| 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 $$ x_1=\frac{-b+\sqrt{\Delta}}{2a},\quad x_2=\frac{-b-\sqrt{\Delta}}{2a}.$$ | ➕ |
| 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., $i^2=-1$. | ➕ |
| 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? | ➕ |
| 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? | ➕ |
| 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? | ➕ |
| 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.
