Transforming Rational Algebraic Equations To Quadratic Form And Identifying Quadratic Equations
In the realm of mathematics, rational algebraic equations and quadratic equations hold significant importance. Understanding their properties and transformations is crucial for solving various mathematical problems. This article delves into identifying rational algebraic equations that can be transformed into quadratic equations and distinguishing quadratic equations from other types of equations. We will dissect each option provided, offering detailed explanations and insights to enhance your comprehension of these concepts.
Identifying Rational Algebraic Equations Transformable to Quadratic Equations
Rational algebraic equations involve fractions where the numerator and denominator are polynomials. A quadratic equation, on the other hand, is a polynomial equation of the second degree. The general form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants, and a ≠0. The key to identifying rational algebraic equations transformable to quadratic equations lies in manipulating the equation to fit this standard form. This often involves clearing fractions and simplifying the resulting expression. Let’s examine each option to determine if it can be transformed into a quadratic equation.
Option A: (2/4) + 1/(t+2) = 7
To ascertain if the equation
(2/4) + 1/(t+2) = 7
can be transformed into a quadratic equation, we must meticulously analyze its structure and manipulate it algebraically. The process begins by simplifying the fraction 2/4, which reduces to 1/2. The equation then becomes:
1/2 + 1/(t+2) = 7
The next crucial step is to eliminate the fractions to simplify the equation further. This is achieved by multiplying both sides of the equation by the least common denominator (LCD), which in this case is 2(t+2). Multiplying each term by the LCD gives us:
2(t+2) * (1/2) + 2(t+2) * (1/(t+2)) = 7 * 2(t+2)
Simplifying each term, we get:
(t+2) + 2 = 14(t+2)
Expanding the terms, the equation transforms into:
t + 2 + 2 = 14t + 28
Combining like terms, we have:
t + 4 = 14t + 28
Now, we rearrange the equation to bring all terms to one side, aiming to set the equation to zero:
14t - t + 28 - 4 = 0
This simplifies to:
13t + 24 = 0
Upon careful examination, the resulting equation, 13t + 24 = 0, is a linear equation, not a quadratic equation. A quadratic equation is characterized by having a term with the variable raised to the power of two (e.g., at²). In this case, the highest power of t is 1, indicating its linear nature. Therefore, despite our efforts to transform it, the original equation does not reduce to a quadratic form. This implies that option A, (2/4) + 1/(t+2) = 7, is not transformable to a quadratic equation.
Option B: 1/(p-1) + 5 = 6p
Let's examine the equation:
1/(p-1) + 5 = 6p
To determine if this rational algebraic equation can be transformed into a quadratic equation, we must follow a similar process of algebraic manipulation. The first step involves eliminating the fraction by multiplying both sides of the equation by the denominator, which is (p-1). This yields:
(p-1) * [1/(p-1) + 5] = 6p * (p-1)
Distributing (p-1) on the left side, we get:
1 + 5(p-1) = 6p(p-1)
Expanding the terms, we have:
1 + 5p - 5 = 6p² - 6p
Combining like terms on the left side, the equation simplifies to:
5p - 4 = 6p² - 6p
Now, rearrange the equation to bring all terms to one side and set the equation to zero. This involves subtracting 5p and adding 4 to both sides:
0 = 6p² - 6p - 5p + 4
Combining like terms on the right side, we obtain:
6p² - 11p + 4 = 0
Upon inspection, the resulting equation, 6p² - 11p + 4 = 0, is a polynomial equation where the highest power of the variable p is 2. This signifies that the equation is indeed a quadratic equation. The general form of a quadratic equation is ax² + bx + c = 0, and in this case, a is 6, b is -11, and c is 4. Since the equation fits this form, we can confidently conclude that the original rational algebraic equation, 1/(p-1) + 5 = 6p, can be transformed into a quadratic equation. This transformation confirms that option B is a valid candidate for a rational algebraic equation transformable to a quadratic equation.
Option C: (s+5)/2 + t/(t+3) = 3
The equation presented in option C is:
(s+5)/2 + t/(t+3) = 3
This equation is a bit different from the previous ones because it involves two different variables, s and t. To determine if this equation can be transformed into a quadratic equation, we need to analyze its structure and consider how the presence of two variables might affect the outcome. A quadratic equation, in its standard form, typically involves a single variable raised to the power of two, along with linear and constant terms. The presence of two distinct variables complicates the transformation process, especially if we aim to express the equation in a standard quadratic form.
To proceed, let’s attempt to eliminate the fractions. The least common denominator (LCD) of the fractions in the equation is 2(t+3). Multiplying both sides of the equation by this LCD, we get:
2(t+3) * [(s+5)/2 + t/(t+3)] = 3 * 2(t+3)
Distributing the LCD to each term on the left side, we have:
(t+3)(s+5) + 2t = 6(t+3)
Expanding the terms, we obtain:
ts + 5t + 3s + 15 + 2t = 6t + 18
Combining like terms, the equation becomes:
ts + 7t + 3s + 15 = 6t + 18
Now, let’s rearrange the terms to bring everything to one side:
ts + 7t - 6t + 3s + 15 - 18 = 0
Simplifying further, we get:
ts + t + 3s - 3 = 0
Upon analyzing the resulting equation, ts + t + 3s - 3 = 0, we observe a term ts, which represents the product of the two variables. This term indicates that the equation is not a standard quadratic equation in one variable. A quadratic equation in a single variable has the form ax² + bx + c = 0, where x is the variable. The presence of the ts term means that the equation is more complex and does not fit the standard quadratic form. Furthermore, there is no single variable raised to the power of two, which is a key characteristic of quadratic equations.
Therefore, based on our analysis, the equation (s+5)/2 + t/(t+3) = 3 cannot be transformed into a standard quadratic equation. The presence of two variables and the resulting ts term prevent it from being expressed in the form ax² + bx + c = 0. Thus, option C is not a rational algebraic equation transformable to a quadratic equation.
Option D: 2/(r+1) + 4 = 7r
The given equation is:
2/(r+1) + 4 = 7r
Our objective is to determine if this rational algebraic equation can be manipulated and transformed into a quadratic equation. To achieve this, we will follow the standard procedure of eliminating the fraction and rearranging the terms to see if the resulting equation fits the quadratic form, which is ax² + bx + c = 0.
The first step is to eliminate the fraction by multiplying both sides of the equation by the denominator (r+1). This yields:
(r+1) * [2/(r+1) + 4] = 7r * (r+1)
Distributing (r+1) on the left side, we get:
2 + 4(r+1) = 7r(r+1)
Expanding the terms, we have:
2 + 4r + 4 = 7r² + 7r
Combining like terms on the left side, the equation simplifies to:
4r + 6 = 7r² + 7r
Now, rearrange the equation to bring all terms to one side and set the equation to zero. This involves subtracting 4r and 6 from both sides:
0 = 7r² + 7r - 4r - 6
Combining like terms on the right side, we obtain:
7r² + 3r - 6 = 0
Upon inspection, the resulting equation, 7r² + 3r - 6 = 0, is a polynomial equation where the highest power of the variable r is 2. This confirms that the equation is indeed a quadratic equation. The general form of a quadratic equation is ax² + bx + c = 0, and in this case, a is 7, b is 3, and c is -6. Since the equation fits this form, we can conclude that the original rational algebraic equation, 2/(r+1) + 4 = 7r, can be transformed into a quadratic equation. This transformation validates that option D is a suitable candidate for a rational algebraic equation transformable to a quadratic equation.
Identifying Quadratic Equations
A quadratic equation is a polynomial equation of the second degree. This means the highest power of the variable in the equation is 2. The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants, and a ≠0. Let’s analyze the given options to identify which one is a quadratic equation.
Option A: x^3 - 27 = 0
The given equation is:
x³ - 27 = 0
To determine whether this equation is a quadratic equation, we need to examine the highest power of the variable x. In this case, the variable x is raised to the power of 3, which is denoted as x³. This indicates that the equation is a cubic equation, not a quadratic equation. A quadratic equation is characterized by having the highest power of the variable as 2, fitting the general form ax² + bx + c = 0, where a, b, and c are constants, and a is not equal to zero.
In contrast, a cubic equation is a polynomial equation of the third degree, meaning the highest power of the variable is 3. The general form of a cubic equation is ax³ + bx² + cx + d = 0, where a, b, c, and d are constants, and a is not equal to zero. The given equation, x³ - 27 = 0, clearly matches this cubic form, with a being 1, b, c being 0, and d being -27. The presence of the x³ term is the defining characteristic that classifies this equation as cubic rather than quadratic.
Therefore, based on the degree of the variable, we can confidently conclude that the equation x³ - 27 = 0 is not a quadratic equation. It is a cubic equation due to the x³ term, which sets it apart from the quadratic form. Thus, option A does not represent a quadratic equation.
Option B: 3q - 6 = 0
The equation in question is:
3q - 6 = 0
To identify the type of equation, we examine the highest power of the variable q. In this equation, the variable q appears only to the first power, which means the equation is linear, not quadratic. A linear equation is characterized by having a variable raised to the power of 1, and its general form is ax + b = 0, where a and b are constants, and a is not equal to zero. In our case, the equation 3q - 6 = 0 fits this form perfectly, with a being 3 and b being -6.
A quadratic equation, on the other hand, is distinguished by having the highest power of the variable as 2. Its general form is ax² + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. The absence of a term with q² in the given equation immediately rules out the possibility of it being a quadratic equation.
Since the highest power of the variable q in the equation 3q - 6 = 0 is 1, it clearly fits the definition of a linear equation. This equation represents a straight line when graphed, which is a characteristic feature of linear equations. Therefore, we can definitively conclude that option B, 3q - 6 = 0, is not a quadratic equation; it is a linear equation.
Conclusion
In summary, after a thorough analysis of the given rational algebraic equations, we identified that options B and D can be transformed into quadratic equations. Option B, 1/(p-1) + 5 = 6p, transforms to 6p² - 11p + 4 = 0, and option D, 2/(r+1) + 4 = 7r, transforms to 7r² + 3r - 6 = 0. Options A and C, however, do not transform into quadratic equations. Furthermore, among the given equations, option B, 3q - 6 = 0, is not a quadratic equation, while option A, x³ - 27 = 0, is a cubic equation.
Understanding how to transform rational algebraic equations into quadratic equations and identifying different types of equations is a fundamental skill in algebra. This knowledge enables us to solve complex problems and gain a deeper appreciation for mathematical structures.
