Problem 38 - Precalculus: Functions and Graphs

Question

Solve the equation. \(3 y^{4}-5 y^{2}+1=0\)

Step-by-Step Solution

Answer

The solutions are \( y = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \) and \( y = \pm \sqrt{\frac{5 - \sqrt{13}}{6}} \).

Step 1: Substitution

Notice that the equation is a quadratic in form. To simplify, let us make the substitution \( z = y^2 \). This gives us a new equation \(3z^2 - 5z + 1 = 0\).

Step 2: Solving the Quadratic Equation

Now solve the new quadratic equation \(3z^2 - 5z + 1 = 0\) using the quadratic formula. The quadratic formula is given by: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 3\), \(b = -5\), and \(c = 1\).

Step 3: Applying the Quadratic Formula

Substitute \(a\), \(b\), and \(c\) into the quadratic formula: \[ z = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]This simplifies to:\[ z = \frac{5 \pm \sqrt{25 - 12}}{6} \]\[ z = \frac{5 \pm \sqrt{13}}{6} \]

Step 4: Back-Substitution

Recall the substitution \(z = y^2\). Therefore, \(y^2 = \frac{5 + \sqrt{13}}{6}\) or \(y^2 = \frac{5 - \sqrt{13}}{6}\).

Step 5: Solving for y

Solve for \(y\) by taking the square root on both sides of each equation:\[ y = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \] and \[ y = \pm \sqrt{\frac{5 - \sqrt{13}}{6}} \].

Step 6: Conclusion

The solutions to the original equation \(3y^4 - 5y^2 + 1 = 0\) are: \[ y = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \] and \[ y = \pm \sqrt{\frac{5 - \sqrt{13}}{6}} \].

Key Concepts

Substitution Method

The substitution method is an effective technique used to simplify equations by introducing a new variable. This is especially useful when dealing with higher degree polynomials that exhibit a quadratic form, such as in the equation given: \(3 y^{4} - 5 y^{2} + 1 = 0\). By recognizing that the equation can be rewritten using a squared variable, it becomes manageable.

• Start by identifying part of the equation that resembles a quadratic structure. For example, note the term \(y^2\) could fit here.
• Introduce a new variable, often called 'z', where \(z = y^2\).
• The original equation then transforms into a more straightforward quadratic equation: \(3z^2 - 5z + 1 = 0\).

This substitution reduces the complexity of handling a fourth-degree polynomial, allowing us to focus on a solvable quadratic equation. Upon finding the solutions for \(z\), we can backtrack to solve for \(y\). This method is a prime example of reducing complexity to find solutions efficiently.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2+bx+c=0\). This universal formula provides solutions by computing specific values of \(x\) that satisfy the equation:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a\), \(b\), and \(c\) are coefficients from the equation. This formula remains a cornerstone of quadratic solutions due to its broad applicability:

• Identifying the coefficients: With our example, \(a = 3\), \(b = -5\), and \(c = 1\).
• Plugging these into the quadratic formula provisions the solutions:
  
  • Calculate the discriminant: \(\sqrt{b^2 - 4ac}\)
  • Evaluate for \(x\) using the formula to get two potential solutions.

This method systematically provides values for \(z\), which transitions us to retrieve the solution for \(y\) using our earlier substitution.

Solving Higher Degree Polynomials

A fourth-degree polynomial, like \(3 y^{4} - 5 y^{2} + 1 = 0\), might seem daunting at first. But with the techniques discussed, particularly substitution and the quadratic formula, we can break down complex polynomials into more manageable components.

• Recognize that high-degree polynomials can sometimes be reformulated to lower-degree through substitution.
• This reformulation helps leverage simpler solving techniques, such as the quadratic formula.
• Once the simpler form is solved, connect the solutions back to the original variable through strategies such as back-substitution.

Lastly, confirm all solutions to ensure they satisfy the original polynomial equation. This validation ensures that every potential solution works within the context of the given problem, confirming mathematical accuracy and consistent application of solving techniques.