Problem 31 - Algebra 2

Question

Solve each equation. $$ x^{4}-15 x^{2}-16=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 4 \) and \( x = -4 \).

Step 1: Substitute Variables

Start by letting \( y = x^2 \). This substitution simplifies the equation by changing it to \( y^2 - 15y - 16 = 0 \). This new equation is a quadratic in terms of \( y \).

Step 2: Use the Quadratic Formula

To solve the quadratic equation \( y^2 - 15y - 16 = 0 \), apply the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -15 \), \( c = -16 \).

Step 3: Calculate the Discriminant

Calculate the discriminant, \( \Delta = b^2 - 4ac = (-15)^2 - 4 \cdot 1 \cdot (-16) = 225 + 64 = 289 \). The discriminant is positive, so there are two real solutions.

Step 4: Solve for y

Using the positive discriminant, apply the quadratic formula: \( y = \frac{15 \pm \sqrt{289}}{2} \). Evaluate the square root to obtain \( y = \frac{15 \pm 17}{2} \).

Step 5: Find the Roots for y

Calculate the roots: \( y_1 = \frac{15 + 17}{2} = 16 \) and \( y_2 = \frac{15 - 17}{2} = -1 \). These are the potential values for \( y = x^2 \).

Step 6: Solve for x

Since \( y = x^2 \), solve for \( x \) by considering both roots:- For \( y = 16 \), solve \( x^2 = 16 \), which gives \( x = \pm 4 \).- For \( y = -1 \), solve \( x^2 = -1 \). No real solutions exist since \( x^2 \) cannot be negative for real numbers.

Key Concepts

Substitution Method

The substitution method is an effective strategy to simplify complex equations. To tackle the original problem, which is not a typical quadratic equation, we first transform it into one by substituting a part of the expression. In our case, we've let \( y = x^2 \). This substitution converts the quartic equation \( x^4 - 15x^2 - 16 = 0 \) into a simpler quadratic: \( y^2 - 15y - 16 = 0 \).
This transformed equation is far easier to handle.

• Start by identifying a substitution that simplifies the equation.
• Replace the complex term with a single variable.
• Solve the resulting simpler equation.

This method not only simplifies but also streamlines the solving process, making it approachable and less intimidating.

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation, \( ax^2 + bx + c = 0 \). It's expressed as: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\)Applying this formula to \( y^2 - 15y - 16 = 0 \) with \( a = 1 \), \( b = -15 \), and \( c = -16 \) allows us to determine the values of \( y \). This approach is especially useful as it provides a straightforward path to solutions, circumventing potentially complex factorizations.
The formula involves:

• Computing the discriminant \(b^2 - 4ac\).
• Calculating both the addition and subtraction scenarios.
• Dividing by \(2a\) for the final solutions.

Using the quadratic formula is universal in that it can solve any quadratic equation, real or complex.

Discriminant

The discriminant, \( \Delta = b^2 - 4ac, \)plays a crucial role in determining the nature of the roots of a quadratic equation. Calculating the discriminant for our equation, we have \( (-15)^2 - 4 \times 1 \times (-16) = 289 \). A positive discriminant implies two distinct real roots, zero indicates a repeated real root, and a negative discriminant suggests imaginary roots.
Determining the discriminant allows us to predict the types of solutions before even solving the equation:

• Positive discriminant: two real and distinct solutions.
• Zero discriminant: one real double root.
• Negative discriminant: no real solutions, although two complex ones.

Understanding this can guide which methods or approaches might best solve the problem.

Polynomial Equations

Polynomial equations, by definition, contain terms in the form of \( ax^n \), where \(n\) is a non-negative integer, and these equations can vary in degree. In our example, the polynomial \( x^4 - 15x^2 - 16 = 0 \) is initially of degree 4. By employing substitution, we've reduced it to a quadratic equation. Each polynomial's degree determines the maximum number of solutions it can have.
When handling polynomial equations:

• Identify the degree for understanding the maximum possible roots.
• Consider possible simplifications, like substitution or factoring.
• Distinguish between real and complex solutions based on discriminant and term structure.

Recognizing a polynomial's nature and structure can simplify the solving process and reveal solution strategies.