Problem 20 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation analyrically 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 vour calculater, suppert the real solutions. $$\begin{aligned}&4 x^{4}-29 x^{2}+25=0\\\&[-5,5] \text { by }\lfloor- 50,100\rfloor\end{aligned}$$

Step-by-Step Solution

Answer

Solutions: \(x = \pm \frac{5}{2}, \pm 1\).

Step 1: Recognize the Equation as a Quadratic in Disguise

Notice the equation is in the form of a quadratic equation when we set \(u = x^2\). This transforms the original equation into \(4u^2 - 29u + 25 = 0\). The goal is now to solve for \(u\), and then substitute back to find \(x\).

Step 2: Apply the Quadratic Formula

The quadratic formula is \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a=4, b=-29,\) and \(c=25\). Substitute these values to find \(u\):\[u = \frac{-(-29) \pm \sqrt{(-29)^2 - 4 \cdot 4 \cdot 25}}{2 \cdot 4}\]\[u = \frac{29 \pm \sqrt{841 - 400}}{8}\]\[u = \frac{29 \pm \sqrt{441}}{8}\]\[u = \frac{29 \pm 21}{8}\]

Step 3: Simplify to Find Values for u

There are two possible solutions for \(u\):1. \(u = \frac{29 + 21}{8} = \frac{50}{8} = \frac{25}{4}\)2. \(u = \frac{29 - 21}{8} = \frac{8}{8} = 1\)

Step 4: Substitute Back to Solve for x

Now substitute \(u = x^2\) back into each solution:- For \(u = \frac{25}{4}\): \[x^2 = \frac{25}{4} \Rightarrow x = \pm \frac{5}{2}\]- For \(u = 1\): \[x^2 = 1 \Rightarrow x = \pm 1\]

Step 5: Listing all Complex Solutions

The solutions to the equation \(4x^4 - 29x^2 + 25 = 0\) are \(x = \pm \frac{5}{2}, \pm 1\). These are all real numbers, but since we are solving for complex solutions, we note that the imaginary part is 0 for these solutions.

Key Concepts

Quadratic Equations

Often, when we hear about quadratic equations, we imagine something like the well-known form: \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. However, sometimes these equations can be cleverly disguised, as in the original exercise where the equation looks like \( 4x^4 - 29x^2 + 25 = 0 \). By making a smart substitution, such as \( u = x^2 \), we can turn this higher degree polynomial into a more familiar quadratic form: \( 4u^2 - 29u + 25 = 0 \). Important traits of quadratic equations include:

• They have a degree of 2 (unless disguised within a higher-degree equation, like \( 4x^4 - 29x^2 + 25 \)).
• They typically yield two solutions, which might be real or complex numbers.
• The solutions can be found using a variety of methods: factorization, completing the square, or applying the quadratic formula.

Quadratic equations are pivotal in algebra because they pave the way for understanding more complex polynomials.

Quadratic Formula

The quadratic formula is a universal method for finding the roots of quadratic equations where factoring isn't straightforward. This miraculous formula is: \[u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].Here, \( a, b, \) and \( c \) represent the coefficients from the equation \( ax^2 + bx + c = 0 \), and the variable \( \pm \) shows there are typically two unique solutions. Using the quadratic formula involves:

• Identifying the coefficients \( a, b, \) and \( c \).
• Plugging these values into the formula.
• Carefully following the syntax to ensure each algebraic step is correct.
• Solving for \( u \) first and if needed, substituting back to find the variable \( x \) or its equivalent.

In the exercise solution, we applied this formula with \( a = 4 \), \( b = -29 \), and \( c = 25 \). Solving it gives us two possible values for \( u \), namely \( \frac{25}{4} \) and \( 1 \). The beauty of the quadratic formula is that it exposes the nature of the roots through its discriminant \( b^2 - 4ac \). If this value is positive, the solutions are real and distinct; if zero, the solutions are real and repeated; if negative, the solutions are complex.

Graphical Solution

Visualizing solutions to equations can often provide intuitive insight that algebraic manipulation alone cannot. By graphing the equation \( 4x^4 - 29x^2 + 25 = 0 \) as \( Y_1 \), we can observe where the graph intersects the x-axis.Graphing Highlights:

• The roots or solutions of the equation correspond to the x-axis points where the graph meets the horizontal line \( y = 0 \).
• In this specific context, we found that \( x = \pm \frac{5}{2}, \pm 1 \) are the real roots by solving the equation analytically.
• By setting the viewing window (like \([-5,5]\) for x-values), we can focus on a specific portion of the graph that highlights the intersection points.

Though all solutions in this exercise are real, graphing can also display if solutions would exist as complex (non-x-axis intersections). While graphical solutions are immensely helpful for confirming algebraic results or identifying approximate solutions, they should be used alongside analytical methods for complete accuracy.