Solving Exponential, Rational, And Radical Equations A Step-by-Step Guide

In this comprehensive guide, we will delve into the intricacies of solving exponential and algebraic equations. We will explore various techniques and strategies to tackle different types of equations, providing you with the knowledge and skills to confidently solve them. This article aims to provide an in-depth understanding of the methods involved in solving these equations, ensuring clarity and accuracy in your solutions. This guide is designed to help students, educators, and anyone interested in enhancing their mathematical abilities.

1. Solving the Exponential Equation: ${7^{1+x} + 7^{1-x} = 50}$

Understanding Exponential Equations

When solving exponential equations, it’s essential to recognize the properties of exponents and how they can be manipulated to simplify the equation. In this case, the equation ${7^{1+x} + 7^{1-x} = 50}$ involves exponential terms with the same base but different exponents. The key here is to break down these terms using the laws of exponents. Specifically, we'll use the rules that state ${a^{m+n} = a^m imes a^n}$ and ${a^{m-n} = a^m / a^n}$. These rules allow us to rewrite the equation in a more manageable form, paving the way for substitution and simplification. This initial step is crucial as it transforms the equation from a complex exponential form into a more familiar algebraic structure. Understanding these transformations is paramount in solving not just this particular equation but a wide range of exponential problems.

Transforming the Equation

To effectively solve the given exponential equation, the first strategic move involves transforming the terms using exponent rules. We begin by rewriting ${7^{1+x}}$ as ${7^1 imes 7^x}$, which simplifies to ${7 imes 7^x}$. Similarly, ${7^{1-x}}$ can be rewritten as ${7^1 imes 7^{-x}}$, or ${7 / 7^x}$. This transformation is a direct application of the exponent rules mentioned earlier. By expressing the exponential terms in this manner, we introduce a common factor of ${7^x}$, which is pivotal for the subsequent steps. This transformation not only simplifies the equation but also reveals the underlying structure that allows us to apply algebraic techniques effectively. Recognizing and applying these transformations is a fundamental skill in solving exponential equations.

Substitution and Quadratic Form

Following the transformation, the equation now appears as ${7 imes 7^x + 7 / 7^x = 50}$. To simplify this further, we introduce a substitution. Let ${y = 7^x}$. This substitution is a crucial step as it converts the exponential equation into a more familiar algebraic form. Substituting ${y}$ into the equation, we get ${7y + 7/y = 50}$. This equation can be further simplified by multiplying through by ${y}$, which yields ${7y^2 + 7 = 50y}$. Rearranging the terms, we obtain a quadratic equation: ${7y^2 - 50y + 7 = 0}$. This quadratic equation is much easier to solve using standard algebraic methods, such as factoring or using the quadratic formula. The substitution technique is a powerful tool in solving exponential equations, as it allows us to leverage our knowledge of algebraic equations to find solutions.

Solving the Quadratic Equation

Now that we have the quadratic equation ${7y^2 - 50y + 7 = 0}$, we can solve it using various methods. Factoring is often the quickest method if the quadratic equation is easily factorable. In this case, the equation can be factored as ${(7y - 1)(y - 7) = 0}$. This factorization leads to two possible solutions for ${y}$: ${y = 1/7}$ and ${y = 7}$. Alternatively, if factoring is not straightforward, the quadratic formula can be used. The quadratic formula, given by ${y = [-b ± sqrt(b^2 - 4ac)] / (2a)}$, provides a reliable method for finding the roots of any quadratic equation. In this instance, applying the quadratic formula would yield the same solutions, ${y = 1/7}$ and ${y = 7}$. The ability to solve quadratic equations is a fundamental skill in algebra, and its application here demonstrates the interconnectedness of different mathematical concepts.

Back-Substitution and Final Solutions

With the values of ${y}$ found, we must now back-substitute to find the values of ${x}$. Recall that we set ${y = 7^x}$. So, we have two equations to solve: ${7^x = 1/7}$ and ${7^x = 7}$. For the first equation, ${7^x = 1/7}$, we can rewrite ${1/7}$ as ${7^{-1}}$. Thus, ${7^x = 7^{-1}}$, which implies ${x = -1}$. For the second equation, ${7^x = 7}$, we can rewrite ${7}$ as ${7^1}$. Therefore, ${7^x = 7^1}$, which implies ${x = 1}$. These values of ${x}$ are the solutions to the original exponential equation. Back-substitution is a critical step in solving equations that involve substitution, as it ensures that we find the values of the original variables. The solutions, ${x = -1}$ and ${x = 1}$, complete the process of solving the exponential equation.

2. Solving the Rational Equation: rac{x}{1+x} + rac{1+x}{x} = rac{13}{6}

Introduction to Rational Equations

Rational equations involve fractions where the numerator and/or the denominator contain variables. To solve the equation {rac{x}{1+x} + rac{1+x}{x} = rac{13}{6}}, the primary strategy involves eliminating the fractions. This is typically achieved by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions present. In this case, the denominators are ${1+x}$, ${x}$, and ${6}$, making the LCD ${6x(1+x)}$. Multiplying by the LCD clears the fractions, transforming the equation into a more manageable algebraic form, usually a quadratic equation. The process of eliminating fractions is a fundamental technique in solving rational equations and simplifies the subsequent steps significantly. Understanding this initial step is crucial for tackling a wide variety of rational equations.

Clearing the Fractions

To begin solving the rational equation, the most crucial step is to eliminate the fractions. This is achieved by multiplying both sides of the equation by the least common denominator (LCD). In this case, the LCD of the denominators ${1+x}$, ${x}$, and ${6}$ is ${6x(1+x)}$. Multiplying both sides of the equation {rac{x}{1+x} + rac{1+x}{x} = rac{13}{6}} by ${6x(1+x)}$ is a strategic move to simplify the equation. This results in:

{6x(1+x) imes rac{x}{1+x} + 6x(1+x) imes rac{1+x}{x} = 6x(1+x) imes rac{13}{6}}

Simplifying each term, we get:

${6x^2 + 6(1+x)^2 = 13x(1+x)}$

This step is pivotal as it transforms the equation from a complex rational form into a more manageable algebraic form, specifically a quadratic equation. Eliminating fractions makes the equation easier to work with and sets the stage for subsequent algebraic manipulations.

Expanding and Simplifying

Following the elimination of fractions, the equation we have is:

${6x^2 + 6(1+x)^2 = 13x(1+x)}$

The next step involves expanding the terms to remove the parentheses and prepare the equation for further simplification. Expanding ${(1+x)^2}$ gives us ${1 + 2x + x^2}$, and expanding ${13x(1+x)}$ gives us ${13x + 13x^2}$. Substituting these back into the equation, we get:

${6x^2 + 6(1 + 2x + x^2) = 13x + 13x^2}$

Further expanding, we have:

${6x^2 + 6 + 12x + 6x^2 = 13x + 13x^2}$

Now, we combine like terms to simplify the equation. On the left side, we combine the ${x^2}$ terms and the constant terms:

${12x^2 + 12x + 6 = 13x + 13x^2}$

This simplification is crucial for rearranging the equation into a standard form, making it easier to solve.

Rearranging into Quadratic Form

After expanding and simplifying, the equation is ${12x^2 + 12x + 6 = 13x + 13x^2}$. To solve for ${x}$, we need to rearrange the equation into a standard quadratic form, which is ${ax^2 + bx + c = 0}$. This involves moving all terms to one side of the equation. Subtracting ${12x^2}$, ${12x}$, and ${6}$ from both sides, we get:

${0 = 13x^2 - 12x^2 + 13x - 12x - 6}$

Simplifying, we obtain the quadratic equation:

${x^2 + x - 6 = 0}$

This quadratic form allows us to apply standard methods for solving quadratic equations, such as factoring or using the quadratic formula. Recognizing and rearranging the equation into this form is a key step in finding the solutions.

Solving the Quadratic Equation

With the equation now in quadratic form, ${x^2 + x - 6 = 0}$, we can proceed to solve for ${x}$. There are several methods to solve quadratic equations, but factoring is often the quickest if the equation is easily factorable. In this case, we look for two numbers that multiply to -6 and add to 1. These numbers are 3 and -2. Therefore, the quadratic equation can be factored as:

${(x + 3)(x - 2) = 0}$

Setting each factor equal to zero gives us two possible solutions for ${x}$:

${x + 3 = 0 ext{ or } x - 2 = 0}$

Solving these equations, we find:

${x = -3 ext{ or } x = 2}$

Alternatively, if factoring is not straightforward, the quadratic formula, ${x = [-b ± sqrt(b^2 - 4ac)] / (2a)}$, can be used. This formula provides a reliable method for finding the roots of any quadratic equation. In this instance, applying the quadratic formula would yield the same solutions, ${x = -3}$ and ${x = 2}$. The ability to solve quadratic equations is a fundamental skill in algebra, and its application here demonstrates the interconnectedness of different mathematical concepts.

Verification of Solutions

After finding potential solutions, it's crucial to verify that they are valid by substituting them back into the original equation. This step is particularly important for rational equations because extraneous solutions can arise due to the initial multiplication by the LCD. Extraneous solutions are values that satisfy the transformed equation but not the original equation.

Substituting ${x = -3}$ into the original equation {rac{x}{1+x} + rac{1+x}{x} = rac{13}{6}}, we get:

{rac{-3}{1+(-3)} + rac{1+(-3)}{-3} = rac{-3}{-2} + rac{-2}{-3} = rac{3}{2} + rac{2}{3} = rac{9 + 4}{6} = rac{13}{6}}

Thus, ${x = -3}$ is a valid solution.

Next, substituting ${x = 2}$ into the original equation, we get:

{rac{2}{1+2} + rac{1+2}{2} = rac{2}{3} + rac{3}{2} = rac{4 + 9}{6} = rac{13}{6}}

Thus, ${x = 2}$ is also a valid solution. Therefore, the solutions to the equation are ${x = -3}$ and ${x = 2}$. Verification is an essential step in solving rational equations to ensure the accuracy of the solutions.

3. Solving the Radical Equation:

${ \sqrt{3x^2 + 1} + \frac{4}{\sqrt{3x^2 + 1}} = 5 }$

Introduction to Radical Equations

Radical equations involve variables inside a radical, typically a square root. To solve the equation ( \sqrt{3x^2 + 1} + \frac{4}{\sqrt{3x^2 + 1}} = 5 ), the key strategy is to eliminate the radical by using substitution and algebraic manipulation. The equation includes a square root term and its reciprocal, suggesting a substitution to simplify the structure. Recognizing the presence of the radical and planning to eliminate it is the first crucial step in solving these types of equations.

Substitution to Simplify the Equation

To solve the given radical equation, the initial step involves a strategic substitution to simplify the equation’s structure. We observe that the term ( \sqrt{3x^2 + 1} ) appears both as a direct term and in the denominator of another term. This pattern suggests that substituting a single variable for this radical term will help reduce the complexity of the equation. Let’s set:

${ y = \sqrt{3x^2 + 1} }$

This substitution transforms the original equation into:

${ y + \frac{4}{y} = 5 }$

This new equation is a rational equation, which is generally easier to handle than the original radical equation. The substitution technique is a powerful tool in solving radical equations, as it allows us to leverage our knowledge of other types of equations to find solutions.

Clearing the Fraction

Following the substitution, our equation is now ${y + \frac{4}{y} = 5}$. To proceed, we need to eliminate the fraction. This is achieved by multiplying every term in the equation by ${y}$, the denominator of the fractional term. This multiplication yields:

${ y imes y + y imes \frac{4}{y} = 5 imes y}$

Simplifying, we get:

${ y^2 + 4 = 5y }$

This step is pivotal as it transforms the rational equation into a simpler polynomial form, specifically a quadratic equation. Eliminating fractions is a common strategy in solving equations involving rational terms, making the subsequent steps more straightforward.

Rearranging into Quadratic Form

After eliminating the fraction, the equation we have is ${y^2 + 4 = 5y}$. To solve for ${y}$, we need to rearrange this equation into the standard quadratic form, which is ${ay^2 + by + c = 0}$. This involves moving all terms to one side of the equation, leaving zero on the other side. Subtracting ${5y}$ from both sides, we get:

${ y^2 - 5y + 4 = 0 }$

This rearranged form allows us to apply standard methods for solving quadratic equations, such as factoring or using the quadratic formula. Recognizing and transforming the equation into this form is a critical step in finding the solutions.

Solving the Quadratic Equation

Now that we have the quadratic equation ${y^2 - 5y + 4 = 0}$, we can proceed to solve for ${y}$. Factoring is often the quickest method if the quadratic equation is easily factorable. In this case, we look for two numbers that multiply to 4 and add to -5. These numbers are -1 and -4. Therefore, the quadratic equation can be factored as:

${ (y - 1)(y - 4) = 0 }$

Setting each factor equal to zero gives us two possible solutions for ${y}$:

${ y - 1 = 0 \text{ or } y - 4 = 0 }$

Solving these equations, we find:

${ y = 1 \text{ or } y = 4 }$

Alternatively, if factoring is not straightforward, the quadratic formula, ${y = [-b ± sqrt(b^2 - 4ac)] / (2a)}$, can be used. Applying the quadratic formula would yield the same solutions, ${y = 1}$ and ${y = 4}$. The ability to solve quadratic equations is a fundamental skill in algebra, and its application here demonstrates the interconnectedness of different mathematical concepts.

Back-Substitution and Solving for x

With the values of ${y}$ found, we must now back-substitute to find the values of ${x}$. Recall that we set ${y = \sqrt{3x^2 + 1}}$. So, we have two equations to solve:

1. ${ \sqrt{3x^2 + 1} = 1 }$
2. ${ \sqrt{3x^2 + 1} = 4 }$

For the first equation, ( \sqrt{3x^2 + 1} = 1 ), we square both sides to eliminate the square root:

${ 3x^2 + 1 = 1 }$

Subtracting 1 from both sides gives:

${ 3x^2 = 0 }$

Dividing by 3, we get:

${ x^2 = 0 }$

Thus, ${x = 0}$ is a solution.

For the second equation, ( \sqrt{3x^2 + 1} = 4 ), we again square both sides:

${ 3x^2 + 1 = 16 }$

Subtracting 1 from both sides gives:

${ 3x^2 = 15 }$

Dividing by 3, we get:

${ x^2 = 5 }$

Taking the square root of both sides, we find:

${ x = \pm\sqrt{5} }$

So, ${x = \sqrt{5}}$ and ${x = -\sqrt{5}}$ are also solutions. Back-substitution is a critical step in solving equations that involve substitution, as it ensures that we find the values of the original variables.

Verification of Solutions

After finding potential solutions, it is crucial to verify that they are valid by substituting them back into the original equation. This step is particularly important for radical equations, as squaring both sides can sometimes introduce extraneous solutions.

First, let's check ${x = 0}$. Substituting into the original equation:

${ \sqrt{3(0)^2 + 1} + \frac{4}{\sqrt{3(0)^2 + 1}} = \sqrt{1} + \frac{4}{\sqrt{1}} = 1 + 4 = 5 }$

So, ${x = 0}$ is a valid solution.

Next, let's check ${x = \sqrt{5}}$:

${ \sqrt{3(\sqrt{5})^2 + 1} + \frac{4}{\sqrt{3(\sqrt{5})^2 + 1}} = \sqrt{3(5) + 1} + \frac{4}{\sqrt{3(5) + 1}} = \sqrt{16} + \frac{4}{\sqrt{16}} = 4 + \frac{4}{4} = 4 + 1 = 5 }$

So, ${x = \sqrt{5}}$ is also a valid solution.

Finally, let's check ${x = -\sqrt{5}}$:

${ \sqrt{3(-\sqrt{5})^2 + 1} + \frac{4}{\sqrt{3(-\sqrt{5})^2 + 1}} = \sqrt{3(5) + 1} + \frac{4}{\sqrt{3(5) + 1}} = \sqrt{16} + \frac{4}{\sqrt{16}} = 4 + \frac{4}{4} = 4 + 1 = 5 }$

So, ${x = -\sqrt{5}}$ is also a valid solution. Therefore, the solutions to the original radical equation are ${x = 0}$, ${x = \sqrt{5}}$, and ${x = -\sqrt{5}}$. Verification ensures the accuracy of the solutions and confirms that no extraneous solutions are included.

Conclusion

In this comprehensive guide, we've explored techniques for solving exponential, rational, and radical equations. Each type requires specific strategies, such as using substitution, eliminating fractions, or dealing with radicals. Mastering these methods enhances your mathematical toolkit and problem-solving skills. Remember to always verify solutions to avoid extraneous results, ensuring the accuracy of your answers. The ability to tackle these equations confidently opens doors to more advanced mathematical concepts and applications. By understanding the underlying principles and practicing diligently, you can excel in solving a wide range of mathematical problems.