Problem 21 - A Graphical Approach to College Algebra

Question

Solve each equation analytically for all complex solutions, giving exact forms in your solution set. Then, graph the left side of the equation as \(y_{1}\) in the suggested viewing window and, using the capabilities of your calculator, support the real solutions. $$\begin{aligned} &x^{4}-15 x^{2}-16=0\\\ &[-5,5] \text { by }[-100,100] \end{aligned}$$

Step-by-Step Solution

Answer

Real solutions: \(x = \pm\sqrt{\frac{15 + \sqrt{305}}{2}}\). Graph verifies solutions.

Step 1: Substitute and Simplify

To solve the equation \(x^4 - 15x^2 - 16 = 0\), we use a substitution method. Let \(u = x^2\), turning the equation into a quadratic form: \(u^2 - 15u - 16 = 0\).

Step 2: Solve Quadratic Equation

Next, solve the quadratic equation \(u^2 - 15u - 16 = 0\) using the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -15\), and \(c = -16\). This gives solutions \(u = \frac{15 \pm \sqrt{305}}{2}\).

Step 3: Re-substitute to Find x

Since \(u = x^2\), this becomes \(x^2 = \frac{15 \pm \sqrt{305}}{2}\). Solving for \(x\), we get \(x = \pm\sqrt{\frac{15 + \sqrt{305}}{2}}\) and \(x = \pm\sqrt{\frac{15 - \sqrt{305}}{2}}\). These are the complex solutions for \(x\).

Step 4: Identify Real Solutions

For real solutions, the value inside the square root must be non-negative. Calculate \(\frac{15 + \sqrt{305}}{2}\), which is positive, so \(x = \pm\sqrt{\frac{15 + \sqrt{305}}{2}}\) are real solutions. Calculate \(\frac{15 - \sqrt{305}}{2}\), which is negative, so \(x = \pm\sqrt{\frac{15 - \sqrt{305}}{2}}\) are not real solutions.

Step 5: Graph and Verify Real Solutions

Graph \(y_1 = x^4 - 15x^2 - 16\) in the given window \([-5,5]\) by \([-100,100]\). Use a calculator to verify where this graph intersects the x-axis. The intersection points correspond to the real solutions found in Step 4.

Key Concepts

Quadratic Formula

The quadratic formula is a vital tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides a straightforward way to find solutions for \( x \), known as roots, by substituting the values of \( a \), \( b \), and \( c \) into the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• The symbol \( \pm \) indicates that there are generally two solutions: one by adding the square root term and one by subtracting it.
• The term inside the square root, \( b^2 - 4ac \), is called the discriminant. It reveals the nature of the roots whether they are real or complex.

For instance, in our exercise, by recognizing the equation \( x^4 - 15x^2 - 16 = 0\) as quadratically structured through substitution, we redefine it as \( u^2 - 15u - 16 = 0 \) with \( u = x^2 \). Here, the quadratic formula helps us solve for \( u \) directly, thus paving the way to find \( x \) by substitution.

Graphing Equations

Graphing equations serves as an excellent means of visualizing the solutions to algebraic expressions, particularly in understanding where these solutions occur along the x-axis. When graphing \( y_1 = x^4 - 15x^2 - 16 \), the graph's key characteristics can aid in identifying solution points:

• A graph represents the equation over a specified window, such as \([-5, 5]\) along the x-axis and \([-100, 100]\) along the y-axis in our case.
• The points where the curve crosses the x-axis denote the real solutions of the equation.
• Using a graphing calculator allows for a visual confirmation of these intersections.

Graphing provides both a qualitative and quantitative understanding of the equation’s behavior, specifically its critical points and zero crossings, which correspond to its real roots.

Real Solutions

Real solutions are those particular solutions to an equation that are not imaginary or complex. When dealing with polynomials, real solutions are the x-values where the graph of the equation intersects the x-axis. In the context of our given equation, real solutions can be specifically confirmed using the graphical method:

• The re-substituted equation \( x^2 = \frac{15 \pm \sqrt{305}}{2} \) gives insight into whether solutions are real or complex.
• The expression \( \frac{15 + \sqrt{305}}{2} \) is positive, resulting in real solutions \( x = \pm \sqrt{\frac{15 + \sqrt{305}}{2}} \).
• Conversely, \( \frac{15 - \sqrt{305}}{2} \) is negative, leading to complex solutions without real intersections.

This determination is critical, as only the positive values within a square root ensure real outcomes. Hence, verifying through both analytical and graphical methods solidifies our understanding of where real solutions exist on the graph.