Solving The Quartic Equation: X^4 - 13x^2 + 36 = 0

Let's dive into solving this interesting quartic equation. Quartic equations, those with a highest power of four, might seem daunting at first, but often they can be tackled using clever substitutions and factoring techniques. In this article, we'll break down the steps to find the solutions for x in the equation x^4 - 13x^2 + 36 = 0. We'll use a method that transforms this equation into a more manageable quadratic form, making it easier to solve. This approach not only simplifies the problem but also gives us a clear pathway to the roots of the equation. So, grab your pencils and let's get started!

Understanding Quartic Equations

Before we jump into the solution, let's briefly discuss quartic equations. A quartic equation is a polynomial equation of the fourth degree. The general form of a quartic equation is ax^4 + bx^3 + cx^2 + dx + e = 0, where a, b, c, d, and e are constants and a ≠ 0. Solving quartic equations can be complex, but some, like the one we're addressing, can be simplified using specific techniques. The key here is recognizing patterns and making strategic substitutions that reduce the equation's complexity. For instance, if we notice that the equation only contains even powers of x, it suggests a possible substitution that can transform it into a quadratic equation. This is precisely the strategy we'll employ to solve x^4 - 13x^2 + 36 = 0. By understanding the structure of quartic equations and recognizing these patterns, we can make the process of finding solutions significantly more straightforward. This foundation will help us appreciate the elegance and efficiency of the substitution method we're about to use.

Step 1: Substitution to Simplify

The equation we're tackling is x^4 - 13x^2 + 36 = 0. Notice that it only contains even powers of x, which suggests a substitution. Let's substitute y = x^2. This clever move will transform our quartic equation into a quadratic equation, which we know how to solve. Replacing x^2 with y, we get y^2 - 13y + 36 = 0. This new equation is much friendlier, isn't it? It's a quadratic equation in terms of y, and we can use various methods to solve it, such as factoring, completing the square, or using the quadratic formula. This substitution is a powerful technique because it simplifies the structure of the equation, making the roots easier to find. By reducing the degree of the polynomial, we've essentially made the problem more accessible and less intimidating. This step is crucial in our solution process, as it lays the groundwork for the subsequent steps where we'll find the values of y and then ultimately solve for x.

Step 2: Solving the Quadratic Equation

Now that we've made the substitution y = x^2, our equation has transformed into a quadratic equation: y^2 - 13y + 36 = 0. To solve this quadratic equation, we can use factoring. We're looking for two numbers that multiply to 36 and add up to -13. Those numbers are -4 and -9. Therefore, we can factor the quadratic equation as (y - 4)(y - 9) = 0. This factored form makes it easy to identify the solutions for y. Setting each factor equal to zero gives us y - 4 = 0 and y - 9 = 0. Solving these simple equations, we find that y = 4 and y = 9. These are the solutions for y, but remember, we're trying to find the values of x. We'll need to reverse our substitution to get there. This step highlights the power of factoring in simplifying quadratic equations, and it's a crucial bridge to finding the ultimate solutions for x in our original quartic equation.

Step 3: Back-Substitution to Find x

We've found the values of y, which are y = 4 and y = 9. Now, we need to back-substitute x^2 for y to find the values of x. Remember, we made the substitution y = x^2 in the first step. So, we now have two equations to solve: x^2 = 4 and x^2 = 9. Let's solve the first equation, x^2 = 4. Taking the square root of both sides, we get x = ±2. This means we have two solutions: x = 2 and x = -2. Next, let's solve the second equation, x^2 = 9. Taking the square root of both sides, we get x = ±3. This gives us two more solutions: x = 3 and x = -3. This back-substitution is a critical step, as it connects our intermediate solutions in terms of y back to the original variable x, allowing us to fully answer the question at hand. This process demonstrates how a strategic substitution, followed by back-substitution, can simplify complex equations and make them solvable.

Step 4: The Solutions

After performing the substitution, solving the quadratic equation, and back-substituting, we've arrived at the solutions for the original quartic equation x^4 - 13x^2 + 36 = 0. We found four solutions for x: x = -3, x = -2, x = 2, and x = 3. These are the roots of the equation, the values of x that make the equation true. We can verify these solutions by plugging them back into the original equation and confirming that the result is zero. This set of solutions completes our journey, showcasing the effectiveness of our approach. By transforming the quartic equation into a more manageable quadratic form, we were able to systematically find all the roots. This method underscores the importance of strategic problem-solving in mathematics, where a clever substitution can significantly simplify a seemingly complex problem. Now we have a complete understanding of the solutions to the equation.

Conclusion

In this article, we successfully solved the quartic equation x^4 - 13x^2 + 36 = 0 by using a strategic substitution and factoring techniques. We transformed the quartic equation into a quadratic equation, making it easier to solve. Then, we back-substituted to find the values of x. The solutions we found are x = -3, x = -2, x = 2, and x = 3. This process demonstrates a powerful approach to tackling higher-degree polynomial equations. By recognizing patterns and making appropriate substitutions, we can often simplify complex problems into more manageable forms. This method not only helps in solving equations but also enhances our problem-solving skills in mathematics. Remember, the key to solving complex equations often lies in finding the right approach and breaking the problem down into smaller, more solvable steps. We encourage you to explore more mathematical problem-solving techniques and apply them to different types of equations. For further learning and resources on algebraic equations, you might find helpful information on websites like Khan Academy Algebra.