Problem 37 - College Algebra

Question

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. $$ x^{4}-10 x^{2}+9=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 1, -1, 3, -3\).

Step 1: Identify Quadratic Form

Recognize that the equation \(x^4 - 10x^2 + 9 = 0\) can be written in a quadratic form by substituting \(u = x^2\). This substitution transforms the equation into \(u^2 - 10u + 9 = 0\).

Step 2: Factor the Quadratic Equation

Factor the quadratic equation \(u^2 - 10u + 9 = 0\). We can do this by finding two numbers that multiply to 9 and add up to -10. The numbers are -1 and -9. Thus, the equation can be factored as \((u - 1)(u - 9) = 0\).

Step 3: Solve for Substitute Variable

Set each factor equal to zero to solve for \(u\). This gives \(u - 1 = 0\) or \(u - 9 = 0\). Solving these gives \(u = 1\) and \(u = 9\).

Step 4: Substitute Back and Solve for 'x'

Replace \(u\) with \(x^2\) to find the real solutions for \(x\). For \(u = 1\), solve \(x^2 = 1\). The solutions are \(x = \pm 1\). For \(u = 9\), solve \(x^2 = 9\). The solutions are \(x = \pm 3\).

Step 5: List All Real Solutions

Compile all the solutions from the above steps. The real solutions to the original equation \(x^4 - 10x^2 + 9 = 0\) are \(x = 1, -1, 3, -3\).

Key Concepts

Quadratic Form

When dealing with equations like \( x^4 - 10x^2 + 9 = 0 \), it's crucial to first recognize the structure known as the quadratic form. A quadratic form refers to situations where an equation can be restructured to involve a squared term, even though it might initially seem more complex.

To do this, a technique called substitution can be applied. We let \( u = x^2 \), which dramatically alters and simplifies the equation to resemble a typical quadratic equation: \( u^2 - 10u + 9 = 0 \). This form resembles the standard quadratic \( ax^2 + bx + c = 0 \), making factors easier to identify.

Factoring Quadratics

After identifying the quadratic form, the next step is factoring the new equation \( u^2 - 10u + 9 = 0 \). Factoring is a key concept in solving quadratic equations, as it involves breaking down the equation into simpler expressions.

To factor, we need two numbers that add to the middle coefficient (-10) and multiply to the constant term (9). In this case, those numbers are -1 and -9, leading us to the factors \((u - 1)(u - 9) = 0\).

This factored form will allow us to find solutions by setting each factor equal to zero and solving for \( u \).

Substitution Method

The substitution method is a helpful technique when dealing with equations that seem too complex on the surface. By substituting \( u = x^2 \), we convert the higher-degree polynomial into a manageable quadratic equation \( u^2 - 10u + 9 = 0 \).

Once we've factored the quadratic, substituting back lets us solve for the original variable \( x \). Thus, for each solution of \( u \), we solve \( u = x^2 \) and obtain \( x = \pm 1 \) and \( x = \pm 3 \), accounting for all real possibilities of \( x \).

This dual stage approach not only simplifies the problem but also ensures all potential solutions are identified and tested.

Real Solutions

The term "real solutions" in mathematics refers to numbers that can be positioned on the number line, essentially encompassing all numbers except those involving the square root of negative values (which are imaginary numbers).

In the context of this exercise, real solutions are precisely what we find and verify via solving the factored equations \((u - 1)(u - 9) = 0\).

Converting back using \( u = x^2 \), we confirm the real solutions \( x = 1, -1, 3, -3 \). These are the "roots" or "zeroes" of the original polynomial equation. They are pivotal as they signify the x-intercepts where the graph of the equation touches or crosses the x-axis.