Solving $y^4 - 20y^2 + 64 = 0$ A Step-by-Step Guide

This article provides a step-by-step guide to solving the quartic equation $y^4 - 20y^2 + 64 = 0$. Quartic equations, which are polynomial equations of the fourth degree, might seem daunting at first glance. However, many quartic equations can be solved using clever algebraic techniques. In this particular case, we'll demonstrate how to reduce the quartic equation to a quadratic equation, making it much simpler to solve. We will explore the method of substitution to transform the equation into a more manageable form and then apply standard techniques for solving quadratic equations. This approach highlights the power of algebraic manipulation in simplifying complex problems. Understanding how to solve such equations is fundamental in various areas of mathematics, physics, and engineering. Let's delve into the solution process and uncover the values of y that satisfy this equation.

1. Recognizing the Quadratic Form

To effectively tackle the quartic equation $y^4 - 20y^2 + 64 = 0$, the first crucial step is to recognize its underlying quadratic form. By observing the structure of the equation, we can see that it closely resembles a quadratic equation. Specifically, we have terms with $y^4$ (which is $(y^2)^2$), $y^2$, and a constant term. This pattern suggests that a suitable substitution can transform the quartic equation into a quadratic equation, which we can then solve using familiar methods. Recognizing this form is a key insight that simplifies the entire solution process. The ability to identify such patterns is a valuable skill in algebra, allowing us to apply known techniques to seemingly complex problems. By understanding the relationship between quartic and quadratic equations in this context, we can leverage our knowledge of quadratic equations to find the solutions for the given quartic equation. The next step will involve making a formal substitution to make this quadratic form explicit.

2. Substitution: Transforming to a Quadratic

Now that we've identified the quadratic form, we can use substitution to explicitly transform the quartic equation into a quadratic equation. Let's introduce a new variable, say $z$, such that $z = y^2$. This substitution is the cornerstone of our approach. By replacing every instance of $y^2$ with $z$, we also have $y^4 = (y^2)^2 = z^2$. Substituting these into our original equation, $y^4 - 20y^2 + 64 = 0$, we obtain: $z^2 - 20z + 64 = 0$. Notice that this is a standard quadratic equation in terms of z. The power of this substitution is that it allows us to apply well-known methods for solving quadratic equations to what initially appears as a more complex problem. This technique is a common strategy in algebra, where substitutions are used to simplify equations and make them more amenable to solution. By performing this substitution, we've effectively reduced the problem's complexity, setting the stage for solving the quadratic equation and subsequently finding the values of y.

3. Solving the Quadratic Equation

Having transformed the quartic equation into a quadratic equation $z^2 - 20z + 64 = 0$, we now focus on finding the values of z. There are several methods available for solving quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is a particularly efficient approach. We seek two numbers that multiply to 64 and add up to -20. These numbers are -4 and -16. Thus, we can factor the quadratic equation as follows: $(z - 4)(z - 16) = 0$. This factorization directly leads to two possible solutions for z: $z - 4 = 0$ or $z - 16 = 0$. Solving these linear equations gives us $z = 4$ and $z = 16$. These values of z are crucial intermediate results. They represent the solutions for the substituted variable, but we're ultimately interested in the values of y. Therefore, the next step involves reversing the substitution to find the corresponding values of y. This process highlights the importance of carefully tracking substitutions and their implications for the final solution.

4. Reversing the Substitution to Find y

With the solutions for z in hand, $z = 4$ and $z = 16$, we now need to reverse our earlier substitution to find the corresponding values of y. Recall that we defined $z = y^2$. This relationship is the key to unlocking the solutions for y. For $z = 4$, we have $y^2 = 4$. Taking the square root of both sides, we get $y = ext{±}2$. This gives us two solutions: $y = 2$ and $y = -2$. Similarly, for $z = 16$, we have $y^2 = 16$. Taking the square root of both sides, we get $y = ext{±}4$. This yields another two solutions: $y = 4$ and $y = -4$. Thus, by reversing the substitution, we've successfully found four possible values for y. It's important to note that taking the square root introduces both positive and negative solutions, which is a crucial consideration when solving equations involving even powers. These four solutions represent the complete set of roots for the original quartic equation.

5. Verifying the Solutions

Having obtained the potential solutions for y, it's always a good practice to verify them by substituting them back into the original equation. This step helps ensure that we haven't made any errors during the solution process and that our solutions are indeed correct. Let's verify each solution:

• For $y = 2$: $(2)^4 - 20(2)^2 + 64 = 16 - 80 + 64 = 0$ (Correct)
• For $y = -2$: $(-2)^4 - 20(-2)^2 + 64 = 16 - 80 + 64 = 0$ (Correct)
• For $y = 4$: $(4)^4 - 20(4)^2 + 64 = 256 - 320 + 64 = 0$ (Correct)
• For $y = -4$: $(-4)^4 - 20(-4)^2 + 64 = 256 - 320 + 64 = 0$ (Correct)

Since all four values satisfy the original equation, we can confidently conclude that they are the correct solutions. This verification step underscores the importance of rigor in mathematical problem-solving. By checking our answers, we increase our confidence in the correctness of our solution and catch any potential errors. In this case, all solutions are verified, giving us a complete and accurate solution set.

6. Conclusion: The Solutions

In conclusion, by employing the method of substitution, we successfully solved the quartic equation $y^4 - 20y^2 + 64 = 0$. We transformed the equation into a quadratic form, solved for the intermediate variable z, and then reversed the substitution to find the solutions for y. The solutions we obtained are $y = -4$, $y = -2$, $y = 2$, and $y = 4$. We also verified these solutions by substituting them back into the original equation. This process demonstrates a powerful technique for solving quartic equations that possess a quadratic-like structure. The key takeaway is the ability to recognize patterns and use substitutions to simplify complex problems. This approach is not only applicable to this specific equation but also provides a general strategy for tackling other similar algebraic challenges. Understanding these techniques enhances our problem-solving skills in mathematics and related fields. The solutions we have found represent the complete set of roots for the given quartic equation, providing a thorough and comprehensive answer.

The solutions to the equation $y^4 - 20y^2 + 64 = 0$ are:

$y = -4, -2, 2, 4$