Problem 41 - College Algebra

Question

Solve each equation by making an appropriate substitution. $$x^{4}-5 x^{2}+4-0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 2, -2, 1, -1\).

Step 1: Apply Substitution

Let's substitute \(u = x^2\). Doing so transforms the equation into \(u^2 - 5u + 4 = 0\). This is a simpler, quadratic equation in terms of \(u\) instead of \(x\).

Step 2: Solve the Quadratic Equation

To solve the quadratic equation \(u^2 - 5u + 4 = 0\), it can be factored into \((u-4)(u-1) = 0\). The solutions to this equation are \(u = 4\) and \(u = 1\), which we get by setting each factor equal to zero and solving for \(u\).

Step 3: Substitute Back and Solve for \(x\)

Bring back the \(x\)'s by replacing \(u\) with \(x^2\). This gives the two equations \(x^2 = 4\) and \(x^2 = 1\). Solving these equations gives four solutions: \(x = \sqrt{4}, -\sqrt{4}, \sqrt{1}, -\sqrt{1}\), or \(x = 2, -2, 1, -1\).

Key Concepts

The Substitution Method for Quadratic Equations

The substitution method is a vital tool in algebra, especially when dealing with higher-order polynomial equations that might seem overwhelming at first glance. By introducing a temporary variable, complex equations can become more manageable. For example, consider the equation \(x^4 - 5x^2 + 4 = 0\). By substituting \(u = x^2\), the equation is transformed into a quadratic equation in terms of \(u\), \(u^2 - 5u + 4 = 0\), which is much easier to handle.

The beauty of the substitution method lies in its ability to simplify the equation without losing any information. After solving for \(u\), we can easily find \(x\) by reversing the substitution, effectively 'unmasking' the variable we were originally seeking. This method not only simplifies calculations but also helps in understanding the structure of complex polynomials.

Factoring Quadratics as a Solution Technique

Factoring is a critical skill when solving quadratic equations. It requires expressing the quadratic in the form \(ax^2 + bx + c = 0\) as a product of two binomials. In our example, we factored the equation \(u^2 - 5u + 4 = 0\) into \(u - 4)(u - 1) = 0\).

The key to successful factoring often hinges on finding two numbers that multiply to the constant term (in this case, 4) and add to the linear coefficient (in this case, -5). Factoring quadratics can be an efficient method to find zeros of a polynomial function since setting each factor equal to zero provides potential solutions. Additionally, mastering factoring allows students to solve equations more quickly and to understand more deeply the properties of quadratic functions.

Finding Roots of a Polynomial

Discovering the roots of a polynomial, which are the values of \(x\) that make the polynomial equal to zero, is central to understanding the behavior of polynomial functions. In our exercise, after substituting and factoring, we obtained the roots of the \(u\) equation as \(u = 4\) and \(u = 1\). By substituting \(u\) back with \(x^2\), we solved \(x^2 = 4\) and \(x^2 = 1\), yielding the roots \(x = 2, -2, 1, -1\).

Each root represents where the graph of the polynomial intersects the \(x\)-axis. Quadratic equations will always have up to two real roots, but higher-degree polynomials can have more, as demonstrated by the four real roots in this situation. Understanding the concept of roots is also essential for grasping other mathematical ideas, such as graphing functions and performing more advanced calculus operations.