Problem 37 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation. $$ x^{4}-5 x^{2}+6=0 $$

Step-by-Step Solution

Answer

The equation \(x^4 - 5x^2 + 6 = 0\) has the following four solutions: \(x_1 = -\sqrt{2}\), \(x_2 = \sqrt{2}\), \(x_3 = -\sqrt{3}\), and \(x_4 = \sqrt{3}\).

Step 1: Substitute a new variable

Let's substitute a new variable, \(u\), such that \(u = x^2\). The equation \(x^4 - 5x^2 + 6 = 0\) becomes: \(u^2 - 5u + 6 = 0\)

Step 2: Solve the quadratic equation

We can solve the equation \(u^2 - 5u + 6 = 0\) using factoring, completing the square, or the quadratic formula. Here, we'll use factoring to find the possible values of u: \((u - 2)(u - 3) = 0\) Which gives us two possible values for u: \(u_1 = 2\) and \(u_2 = 3\)

Step 3: Find the possible values of x

We can now substitute back the original variable \(x^2\), and find the possible values of x for each value of u. For \(u_1 = 2\): \(x^2 = 2\) \(x_{1,2} = \pm\sqrt{2}\) For \(u_2 = 3\): \(x^2 = 3\) \(x_{3,4} = \pm\sqrt{3}\)

Step 4: Write the solution

The possible solutions for the given equation \(x^4 - 5x^2 + 6 = 0\) are: \(x_1 = -\sqrt{2}\) \(x_2 = \sqrt{2}\) \(x_3 = -\sqrt{3}\) \(x_4 = \sqrt{3}\)

Key Concepts

Substitution Method

The substitution method is a handy tool for solving complex equations by transforming them into a simpler form. In the provided exercise, the substitution method makes it possible to break down the quartic equation into a quadratic one. The idea is simple.

• First, identify the expression that can be substituted with a single variable. Here, you transform the original equation by letting \( u = x^2 \).
• Next, plug this substitution back into the equation. This turns the quartic term \( x^4 \) into \( u^2 \), resulting in the quadratic equation \( u^2 - 5u + 6 = 0 \).

This approach transforms the original problem into solving a quadratic equation, which is generally easier. After solving for \( u \), you revert the substitution to find the original variable values. Substitution thus acts like a stepping stone in solving more intricate mathematical equations, making them approachable by reducing complexity.

Factoring Quadratics

Factoring quadratics is a key method for solving quadratic equations. It involves expressing the quadratic equation in product form so that it can be set to zero. In the equation from the original problem, \( u^2 - 5u + 6 = 0 \) was factored as:
\[ (u - 2)(u - 3) = 0 \]
Here's how the factoring works:

• You identify two numbers that multiply to give the constant term (6) and add up to give the middle coefficient (-5).
• In this case, 2 and 3 satisfy both conditions because \( 2 \times 3 = 6 \) and \( 2 + 3 = 5 \).
• Write the factored form as \((u - 2)(u - 3)\).

Once factored, it becomes simple to solve the quadratic equation by setting each factor equal to zero:

• \( u - 2 = 0 \)
• \( u - 3 = 0 \)

Solving these gives the roots \( u_1 = 2 \) and \( u_2 = 3 \), which are essential in finding the solutions for \( x \) after reversing the substitution.

Algebraic Equations

Algebraic equations form the backbone of many mathematical problem-solving scenarios. They include variables and constants, and their solution often requires operations such as substitution and factoring.
For the equation in the exercise, understanding algebraic equations involves recognizing patterns and structures. The original equation \( x^4 - 5x^2 + 6 = 0 \) is a higher-dimensional polynomial that initially seems complex.

• By recognizing it can be transformed into a standard quadratic form through substitution, mathematical manipulation becomes possible.
• Solving such equations then involves using techniques like substitution and factoring to simplify the problem.
• Ultimately, it's about breaking down complex expressions into easier ones where basic algebraic techniques can be applied effectively.

Solving algebraic equations is all about systematic approaches and applying correct mathematical operations to unravel the problem.