Problem 38 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: 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: Identify the Substitution Variable

Recognize that the equation is a quadratic in form with respect to the expression \( y^2 \). We'll let \( u = y^2 \).

Step 2: Substitute and Form a Quadratic Equation

Substitute \( u = y^2 \) into the equation, transforming it into \( 3u^2 - 5u + 1 = 0 \). This new equation is a typical quadratic equation.

Step 3: Apply the Quadratic Formula

Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = -5 \), and \( c = 1 \). First, find the discriminant: \( b^2 - 4ac = (-5)^2 - 4(3)(1) = 25 - 12 = 13 \).

Step 4: Solve for u

Plug the discriminant into the quadratic formula: \( u = \frac{-(-5) \pm \sqrt{13}}{2(3)} = \frac{5 \pm \sqrt{13}}{6} \). Thus, we have the solutions \( u = \frac{5 + \sqrt{13}}{6} \) and \( u = \frac{5 - \sqrt{13}}{6} \).

Step 5: Solve for y

Since \( u = y^2 \), solve for \( y \) by taking the square root of each solution for \( u \). Calculate \( y = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \) and \( y = \pm \sqrt{\frac{5 - \sqrt{13}}{6}} \).

Step 6: Verify and Compile the Solutions

Ensure each calculation is accurate. The complete set of solutions for \( y \) is \( y = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \), and \( y = \pm \sqrt{\frac{5 - \sqrt{13}}{6}} \).

Key Concepts

Polynomial Equations

Polynomial equations, which are crucial in algebra, involve variables raised to whole-number powers. The exercise begins by introducing a polynomial equation of the form \(3y^4-5y^2+1=0\). These equations feature terms with varying powers, often arranged in decreasing order. For example:

• \(3y^4\) is the highest power term, known as the leading term.
• \(-5y^2\) is the middle term.
• \(+1\) is the constant term.

Polynomial equations can have multiple solutions or roots. Solving them typically involves techniques like factoring, using the quadratic formula, or other algebraic methods to find variable values that satisfy the equation. Here, we transform a quartic polynomial into a familiar quadratic form to simplify our solution.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of quadratic equations, expressed as \(ax^2 + bx + c = 0\). With the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]you can solve any quadratic equation. The components are:

• \(a\): Coefficient of \(u^2\) or \(x^2\)
• \(b\): Coefficient of \(u\) or \(x\)
• \(c\): Constant term

In our exercise, after substituting \(u = y^2\) and transforming the equation into \(3u^2 - 5u + 1 = 0\), the quadratic formula reveals the solutions for \(u\). The discriminant, calculated as \(b^2 - 4ac = 13\), shows a positive value indicating two distinct real roots: \( \frac{5 + \sqrt{13}}{6} \) and \( \frac{5 - \sqrt{13}}{6} \).

Substitution Method

The substitution method simplifies complex polynomial equations by changing variables. Here, we use substitution to transform a quartic equation into a quadratic form, making it easier to solve. The strategy involves:

• Identifying a substitution variable that simplifies the equation.
• Replacing complex expressions with the substitution variable.

For example, in this exercise:

• Identify \( y^2 \) as a component to substitute.
• Let \( u = y^2 \), transforming \(3y^4-5y^2+1=0\) into \(3u^2 - 5u + 1 = 0\).

This method allows us to apply the quadratic formula, solving easily for \(u\), and then back-substituting to find \(y\). It streamlines the solving process for otherwise complicated polynomials.

Roots of Equations

Finding the roots of an equation means identifying values for variables that make the equation true. For polynomials, these roots depend on the degree and type of equation. In our specific problem with \(3y^4 - 5y^2 + 1 = 0\), once simplified, we find the solutions through:

• Determining \(u\) via the quadratic formula, yielding two values.
• Back-substituting to find \(y\) using \( y^2 = u \).

The final roots for \(y\) become:

• \( y = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \)
• \( y = \pm \sqrt{\frac{5 - \sqrt{13}}{6}} \)

These reflect the solutions to the original equation, illustrating how polynomial and quadratic techniques translate into practical solutions for complex expressions.