Problem 11 - Precalculus : Building Concepts and Connections

Question

Solve the polynomial equation. In Exercises \(7-14,\) find all solutions. In Exercises \(15-18,\) find only real solutions. Check your solutions. $$6 s^{4}-s^{2}-2=0$$

Step-by-Step Solution

Answer

The solutions are \(s = \pm \sqrt{\frac{1}{3}}\) and \(s = \pm i\).

Step 1: Simplify the Equation

The original quartic equation is given as \(6s^{4} - s^{2} - 2 = 0\). A key observation that can be made here is that if we substitute \(x = s^{2}\), the equation can be rewritten into a quadratic format: \(6x^{2} - x - 2 = 0\). This makes it more manageable.

Step 2: Solving Quadratic Equation

We can now use the quadratic formula to solve for \(x\) which is given by \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]. For our quadratic equation, \(a = 6\), \(b = -1\), and \(c = -2\). Plugging these values in, we get two solutions for \(x\): \(x_{1} = \frac{1}{3}\), \(x_{2} = -1\). Remember that \(x = s^{2}\).

Step 3: Solving Polynomial Equation

Afterwards, we solve for \(s^{2}\) by equating \(s^{2}\) to each solution of \(x\). From \(s^{2} = \frac{1}{3}\), we get \(s = \pm \sqrt{\frac{1}{3}}\). From \(s^{2} = -1\), we get \(s = \pm i\), which are imaginary solutions.

Step 4: Verifying Solutions

Finally, to verify the solutions, substitute each solution back into the original equation to check if both sides match.

Key Concepts

Quartic Equations

A quartic equation is a polynomial equation of the fourth degree. This means its highest exponent is four, typically taking the general form of \(ax^4 + bx^3 + cx^2 + dx + e = 0\). Quartic equations appear complex due to their higher degree, but they can often be simplified using substitution or factoring techniques.

There are various methods to solve quartic equations. These include:

• Factoring: Breaking down the equation into the product of lower-degree polynomials.
• Graphical Solutions: Using graphing techniques to find the roots visually.
• Substitution: Converting the quartic into a simpler form, often a quadratic, to ease the solving process.

In the provided problem, substitution was utilized by introducing \(x = s^2\), converting it effectively into a quadratic equation. This approach simplifies the task, making the solution more achievable.

Quadratic Substitution

Quadratic substitution is a technique used to simplify solving higher-degree polynomial equations. For quartic equations, this often involves identifying a substitution that converts the quartic into a quadratic.

The substitution \(x = s^2\) changes the quartic equation \(6s^4 - s^2 - 2 = 0\) into the quadratic equation \(6x^2 - x - 2 = 0\). With this new form, we apply the quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• Substitute the Coefficients: In this case, \(a = 6\), \(b = -1\), \(c = -2\).
• Calculate the Solutions: Solving gives us \(x_1 = \frac{1}{3}\) and \(x_2 = -1\).

Once these values of \(x\) are found, reversing the substitution, \(s^2 = x\), helps compute the final solutions for \(s\). This technique is immensely helpful for simplifying polynomials and finding solutions efficiently.

Complex Solutions

In mathematics, complex solutions arise when solving polynomial equations that have no real roots. They typically involve the imaginary unit \(i\), which is defined as \(i = \sqrt{-1}\).

In the given problem, after performing quadratic substitution and solving, we find one solution for \(x\) which is negative: \(x_2 = -1\). This value introduces complex numbers when we solve \(s^2 = -1\), resulting in the solutions \(s = \pm i\).

• Nature of Complex Solutions: Complex roots occur in conjugate pairs, which in this context are \(i\) and \(-i\).
• Applications: Complex numbers are vital in fields such as engineering, physics, and applied math, allowing for complex problem-solving where real numbers fall short.

Understanding complex solutions helps in comprehending the full spectrum of possibilities when solving polynomial equations, ensuring all potential solutions are covered.