Solving The Biquadratic Equation Y^4 - 17y^2 + 16 = 0

In this article, we will delve into the process of solving the biquadratic equation $y^4 - 17y^2 + 16 = 0$. Biquadratic equations, also known as quartic equations lacking odd-degree terms, can often be elegantly solved by employing a substitution method that transforms them into quadratic equations. This approach simplifies the problem, allowing us to utilize familiar techniques such as factoring, completing the square, or the quadratic formula to find the solutions. We will explore each step in detail, ensuring a clear and comprehensive understanding of the solution process. This mathematical exploration is not just an academic exercise; it highlights the power of algebraic manipulation and its applications in various fields, ranging from physics and engineering to computer science and economics. Mastering the technique to solve biquadratic equations enriches our mathematical toolkit, equipping us with the ability to approach more complex problems with confidence and precision.

At its core, a biquadratic equation is a quartic (fourth-degree) polynomial equation that lacks the terms with odd powers of the variable. Its general form can be expressed as $ay^4 + by^2 + c = 0$, where 'a', 'b', and 'c' are constants, and 'y' is the variable. The absence of the cubic ($y^3$) and linear (y) terms is what distinguishes a biquadratic equation from a general quartic equation. This specific structure allows for a strategic substitution that simplifies the equation into a quadratic form, which is far easier to handle. The recognition of this form is the first key step in solving these equations efficiently.

Why are biquadratic equations important? They appear in various contexts, including problems in geometry, physics, and engineering. For instance, they can arise when finding the dimensions of a shape with certain area and perimeter constraints or in calculations involving projectile motion. The ability to solve biquadratic equations is therefore a valuable skill in many scientific and technical disciplines. Moreover, understanding these equations deepens our appreciation of the elegance and symmetry often found in mathematical structures. The techniques used to solve them, such as substitution and factorization, are fundamental in algebra and provide a foundation for tackling more advanced mathematical problems.

The biquadratic equation we aim to solve is $y^4 - 17y^2 + 16 = 0$. The standard approach to solving such equations involves a clever substitution that transforms it into a more manageable quadratic equation. Let's outline the steps:

1. Substitution: Introduce a new variable, say $z$, such that $z = y^2$. This substitution is the cornerstone of the method, effectively reducing the degree of the equation we need to solve. By replacing every instance of $y^2$ with $z$, we will convert the quartic equation into a quadratic equation in terms of $z$.
2. Form the Quadratic Equation: Replace $y^2$ with $z$ in the original equation. The term $y^4$ can be written as $(y^2)^2$, which then becomes $z^2$. Our biquadratic equation now transforms into a quadratic equation in $z$, which we can solve using standard methods.
3. Solve the Quadratic Equation: Use techniques such as factoring, completing the square, or the quadratic formula to find the values of $z$. These methods are well-established and provide a direct way to find the roots of the quadratic equation. The choice of method often depends on the specific coefficients of the equation; some quadratics are easily factored, while others may require the quadratic formula.
4. Back-Substitute to Find y: Once we have the values of $z$, we need to reverse the substitution to find the values of $y$. Since $z = y^2$, we take the square root of each value of $z$ to find the corresponding values of $y$. Remember that taking the square root can yield both positive and negative solutions, so we must consider both possibilities.
5. List All Solutions: Collect all the values of $y$ obtained in the previous step. These values represent all the solutions (or roots) of the original biquadratic equation. A quartic equation can have up to four roots, so we should expect to find up to four solutions for $y$.

Let's apply the method described above to solve the equation $y^4 - 17y^2 + 16 = 0$ step by step:

1. Substitution: Let $z = y^2$.
2. Form the Quadratic Equation: Substituting $z$ into the equation, we get: $z^2 - 17z + 16 = 0$.
3. Solve the Quadratic Equation: We can solve this quadratic equation by factoring. We look for two numbers that multiply to 16 and add up to -17. These numbers are -1 and -16. So, we can factor the equation as: $(z - 1)(z - 16) = 0$. This gives us two possible values for $z$: $z = 1$ or $z = 16$.
4. Back-Substitute to Find y: Now, we substitute back $y^2$ for $z$ and solve for $y$:
   • If $z = 1$, then $y^2 = 1$. Taking the square root of both sides, we get $y = ±1$.
   • If $z = 16$, then $y^2 = 16$. Taking the square root of both sides, we get $y = ±4$.
5. List All Solutions: The solutions for $y$ are -4, -1, 1, and 4.

While the substitution method is the most common and straightforward approach for solving biquadratic equations, it's worth noting that other methods exist, which can provide alternative perspectives and solutions. Understanding these methods enhances our problem-solving toolkit and offers different angles to tackle similar mathematical challenges.

Completing the Square

Completing the square is a powerful technique for solving quadratic equations, and it can be adapted to solve biquadratic equations as well. This method involves manipulating the equation to create a perfect square trinomial on one side. For the equation $y^4 - 17y^2 + 16 = 0$, we can rewrite it as:

$y^4 - 17y^2 = -16$

To complete the square, we need to add and subtract $(17/2)^2$ on the left side:

$y^4 - 17y^2 + (17/2)^2 = -16 + (17/2)^2$

$(y^2 - 17/2)^2 = -16 + 289/4$

$(y^2 - 17/2)^2 = 225/4$

Now, take the square root of both sides:

$y^2 - 17/2 = ±15/2$

This gives us two quadratic equations:

1. $y^2 = 17/2 + 15/2 = 16$, which leads to $y = ±4$
2. $y^2 = 17/2 - 15/2 = 1$, which leads to $y = ±1$

Factoring Directly (if Possible)

In some cases, biquadratic equations can be factored directly without the need for substitution. This is often the quickest method when the equation has integer roots and the coefficients allow for easy factorization. In our example, $y^4 - 17y^2 + 16 = 0$, we already saw how the substituted quadratic equation factored nicely. However, we could also think of factoring the original quartic expression directly, by recognizing that it might factor into a product of two quadratics:

$(y^2 - a)(y^2 - b) = y^4 - (a + b)y^2 + ab$

We need to find $a$ and $b$ such that $a + b = 17$ and $ab = 16$. By inspection, we can see that $a = 1$ and $b = 16$ (or vice versa) satisfy these conditions. Thus, the equation factors as:

$(y^2 - 1)(y^2 - 16) = 0$

This gives us the same solutions as before: $y = ±1$ and $y = ±4$.

Graphical Methods

While not a precise algebraic method, graphing the function $f(y) = y^4 - 17y^2 + 16$ can provide a visual representation of the solutions. The real roots of the equation correspond to the points where the graph intersects the y-axis. Graphing can be particularly useful for estimating solutions or for understanding the number and nature of the roots.

To ensure the accuracy of our solutions, it's crucial to verify them by substituting each value back into the original equation. This process confirms that the solutions satisfy the equation and that no errors were introduced during the solution process. Let's verify each of our solutions: y = -4, -1, 1, and 4.

1. For $y = -4$:
   • $(-4)^4 - 17(-4)^2 + 16 = 256 - 17(16) + 16 = 256 - 272 + 16 = 0$
2. For $y = -1$:
   • $(-1)^4 - 17(-1)^2 + 16 = 1 - 17(1) + 16 = 1 - 17 + 16 = 0$
3. For $y = 1$:
   • $(1)^4 - 17(1)^2 + 16 = 1 - 17(1) + 16 = 1 - 17 + 16 = 0$
4. For $y = 4$:
   • $(4)^4 - 17(4)^2 + 16 = 256 - 17(16) + 16 = 256 - 272 + 16 = 0$

As we can see, all four solutions satisfy the original equation, confirming their validity. This verification step is a critical part of the problem-solving process, particularly in mathematics, as it provides a high degree of confidence in the correctness of the solutions.

In summary, the biquadratic equation $y^4 - 17y^2 + 16 = 0$ can be solved effectively using the substitution method, which transforms the equation into a quadratic form. By letting $z = y^2$, we obtained the quadratic equation $z^2 - 17z + 16 = 0$, which we then solved to find $z = 1$ and $z = 16$. Substituting back to find $y$, we obtained the solutions $y = -4, -1, 1,$ and $4$. We also explored alternative methods such as completing the square and direct factoring, which provided additional perspectives on solving the equation. Finally, we verified our solutions by substituting them back into the original equation, ensuring their accuracy.

This process not only demonstrates the solution to a specific problem but also highlights the broader techniques and strategies applicable to solving a range of algebraic equations. The ability to recognize and apply these methods is a fundamental skill in mathematics and has practical applications in various fields. Mastering these techniques enhances our mathematical proficiency and equips us with the tools to tackle more complex problems with confidence and precision. The journey through solving biquadratic equations reinforces the importance of algebraic manipulation, substitution, and verification in the world of mathematics and beyond.