Problem 39 - College Algebra with Corequisite Support

Question

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. $$ \left(x^{2}-1\right)^{2}+\left(x^{2}-1\right)-12=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 2\) and \(x = -2\).

Step 1: Identify the Quadratic Form

Notice that the expression \((x^2 - 1)^2 + (x^2 - 1) - 12 = 0\) resembles a quadratic equation in the form \(u^2 + u - 12 = 0\) with \(u = x^2 - 1\). This substitution simplifies the expression.

Step 2: Substitute the Variable

Let \(u = x^2 - 1\). Substitute into the equation to get \(u^2 + u - 12 = 0\). Now, we have a standard quadratic equation in terms of \(u\).

Step 3: Factor the Quadratic Equation

We need to factor \(u^2 + u - 12 = 0\). Find two numbers that multiply to \(-12\) and add to \(1\). These numbers are \(4\) and \(-3\). Thus, the equation factors to \((u + 4)(u - 3) = 0\).

Step 4: Solve for the Substitute Variable

Set each factor equal to zero: \(u + 4 = 0\) or \(u - 3 = 0\). Solving these gives \(u = -4\) or \(u = 3\).

Step 5: Substitute Back for \(x\) and Solve

Replace \(u\) back with \(x^2 - 1\): for \(u = -4\), we have \(x^2 - 1 = -4\), which simplifies to \(x^2 = -3\) (no real solutions); for \(u = 3\), we have \(x^2 - 1 = 3\), which simplifies to \(x^2 = 4\).

Step 6: Find the Real Solutions for \(x\)

Solve \(x^2 = 4\) by taking the square root of both sides, giving \(x = 2\) or \(x = -2\). These are the real solutions.

Key Concepts

Factoring Quadratics

Factoring quadratic equations is a method used to find the values of the variable that satisfy the equation. In general, a quadratic equation takes the form of \(ax^2 + bx + c = 0\). The goal is to express it as a product of two binomials, like \((x + p)(x + q) = 0\). To do this, you need to find two numbers that:

• Multiply to give the constant term \(c\).
• Add up to give the middle coefficient \(b\).

In our problem, the equation \(u^2 + u - 12 = 0\) needs to be factored. We are searching for two numbers that multiply to \(-12\) and sum to \(1\). These numbers are \(4\) and \(-3\). Thus, the quadratic equation factors easily to \((u + 4)(u - 3)\). This step is crucial because once factored, it's simpler to solve, as each factor can be set to zero to find possible solutions.

Substitution Method

The substitution method is a powerful tool for simplifying complex equations, particularly useful in quadratic equations when they are not in their usual form. The basic idea is to substitute a part of the expression with a single variable, making it easier to work with.

In the given exercise, the expression \((x^2 - 1)^2 + (x^2 - 1) - 12 = 0\) is identified as having a quadratic form. Here, by substituting \(u = x^2 - 1\), the equation transforms into the simpler form \(u^2 + u - 12 = 0\). This conversion makes it manageable to factor and solve.

After solving for \(u\), we reverse the substitution back to the original terms. For solving the final results in terms of \(x\), substitute back \(x^2 - 1\) for each obtained value of \(u\). Solving the remaining equations will then provide solutions related to the original problem.

Real Solutions

After performing operations like substitution and factoring, we arrive at equations that need to be solved for real solutions. In our exercise, after substituting and factoring, we find \(u = -4\) or \(u = 3\).

However, when substituting back for \(x\), only \(u = 3\) yields real solutions. Solving \(x^2 - 1 = 3\) simplifies to \(x^2 = 4\), allowing us to find real solutions: \(x = 2\) or \(x = -2\).

Real solutions indicate actual values of \(x\) that satisfy the initial equation. Unlike complex solutions, real numbers can be plotted on a number line, offering tangible results. It's important to understand that not every value of \(u\) will provide real solutions once reverted back to the original variable \(x\). In such cases, the result from \(u = -4\) led to a negative under the square root, so no real solutions exist for that scenario.