Problem 4 - A Graphical Approach to College Algebra

Question

Find all real solutions. $$x^{4}+5=6 x^{2}$$

Step-by-Step Solution

Answer

The real solutions are \(x = 1, -1, \sqrt{5}, -\sqrt{5}\).

Step 1: Write the Original Equation

Start with the given equation: \(x^4 + 5 = 6x^2\).

Step 2: Rearrange the Equation

Rearrange the equation to set it equal to zero by moving all terms to one side, resulting in: \(x^4 - 6x^2 + 5 = 0\).

Step 3: Use Substitution

To simplify the equation, use substitution. Let \(y = x^2\), so the equation becomes \(y^2 - 6y + 5 = 0\).

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \(y^2 - 6y + 5 = 0\). You can factor it as \((y - 1)(y - 5) = 0\).

Step 5: Find Values of y

From the factored equation, find the values of \(y\): \(y - 1 = 0\) yields \(y = 1\) and \(y - 5 = 0\) yields \(y = 5\).

Step 6: Back-Substitute to x

Since \(y = x^2\), substitute back to solve for \(x\). - If \(y = 1\), then \(x^2 = 1\), giving solutions \(x = 1\) or \(x = -1\). - If \(y = 5\), then \(x^2 = 5\), giving solutions \(x = \sqrt{5}\) or \(x = -\sqrt{5}\).

Step 7: State the Real Solutions

The real solutions are \(x = 1\), \(x = -1\), \(x = \sqrt{5}\), and \(x = -\sqrt{5}\).

Key Concepts

Substitution Method

The substitution method is a clever technique often used to simplify complex equations, especially those involving higher powers of a variable. In problems involving polynomial equations, this method can transform a daunting task into a manageable one.

Let's break it down further:

• **Identify a Suitable Substitution**: Begin by identifying a part of the equation that can be replaced by a new variable, simplifying its structure. For example, when tackling an equation like \( x^4 + 5 = 6x^2 \), you can notice the expression \( x^2 \) appearing as a recurring element.
• **Change of Variables**: You then substitute this recurring part, let's say, with a new variable, \( y \), where \( y = x^2 \). This step transforms the original equation into a purely quadratic one, like \( y^2 - 6y + 5 = 0 \).
• **Simplify and Solve**: The new structure is often more straightforward and allows you to use other methods, such as factoring, to find solutions.

The substitution method is not just about equations but is a powerful tool that can be applied in various algebraic contexts to simplify and solve equations effectively.

Factoring

Factoring is an essential tool in solving quadratic equations, and it's especially effective when used in combination with the substitution method.

Here's how you approach it:

• **Convert and Read the Equation**: First, after substitution, ensure that your quadratic expression is in standard form like \( y^2 - 6y + 5 = 0 \).
• **Look for Two Numbers**: The goal is to find two numbers that multiply to give you the constant term (which is 5 in this case) and add up to the coefficient of the middle term (-6).
  • In this instance, the numbers are -1 and -5. Note how (-1) \(-5 = 5\) and \((-1) + (-5) = -6\).
• **Write in Factored Form**: Using these numbers, the expression can be rewritten in factored form as \((y - 1)(y - 5) = 0 \).
• **Solve Each Factor**: Set each factor equal to zero, resulting in solutions \( y = 1 \) and \( y = 5 \).

Factoring is sometimes seen as a puzzle: find the right pieces, and everything falls into place. It provides a straightforward path to solve quadratic equations, especially when the coefficients are simple integers.

Real Solutions

Real solutions are the outputs of an equation that are actual numbers, as opposed to imaginary numbers that you might also encounter when solving equations.

Why emphasize real solutions?

• **Accessible and Tangible**: Real solutions are the solutions you can plot on a graph, see on a number line, and often relate to in real-world applications.
• **Checking the Solutions**: Once you have successfully solved an equation (like \( y - 1 = 0 \) and \( y - 5 = 0 \) in our example, leading to real solutions), it's important to back-substitute to assure they are indeed valid.
  • Case in point, when \( y = 1 \), \( x^2 = 1 \) gives \( x = 1 \) or \( x = -1 \).
  • Similarly, \( y = 5 \), gives \( x = \sqrt{5} \) or \( x = -\sqrt{5} \).

This guarantees that the solutions are not only correct arithmetically but express the nature of the real number system. Remember, always look for these solutions unless specified otherwise in a problem statement.