Problem 19 - 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. $$\begin{aligned}&4 x^{4}-25 x^{2}+36=0\\\&[-5,5] \text { by }\lfloor- 5,100\rfloor\end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \(x = -2, -\frac{3}{2}, \frac{3}{2}, 2\).

Step 1: Substitute Variable

Let us substitute a new variable to make the equation easier to solve. Let \( u = x^2 \). Then our equation becomes \( 4u^2 - 25u + 36 = 0 \).

Step 2: Solve the Quadratic Equation

We now solve the quadratic equation \( 4u^2 - 25u + 36 = 0 \) using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -25 \), and \( c = 36 \).

Step 3: Compute Discriminant

Calculate the discriminant: \( b^2 - 4ac = (-25)^2 - 4 \times 4 \times 36 = 625 - 576 = 49 \).

Step 4: Find Possible Solutions for u

Using the quadratic formula, substitute the values: \( u = \frac{25 \pm 7}{8} \). Thus, \( u = \frac{32}{8} = 4 \) or \( u = \frac{18}{8} = \frac{9}{4} \).

Step 5: Back Substitute for x

Since \( u = x^2 \), for \( u = 4 \), \( x = \pm \sqrt{4} = \pm 2 \). For \( u = \frac{9}{4} \), \( x = \pm \sqrt{\frac{9}{4}} = \pm \frac{3}{2} \).

Step 6: Verify Real Solutions with Graphing

Graph the equation \( Y_1 = 4x^4 - 25x^2 + 36 \) using the interval \([-5, 5]\) for \( x \). The intersections with the x-axis at \( x = -2, -\frac{3}{2}, \frac{3}{2}, \text{ and } 2 \) confirm the real solutions.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool in algebra for solving quadratic equations. A quadratic equation is typically in the form of \( ax^2 + bx + c = 0 \). The quadratic formula allows you to find the values of \( x \) (also known as the roots or solutions of the equation) by substituting the coefficients \( a \), \( b \), and \( c \) into the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]It is derived from completing the square on the standard form of a quadratic equation. This formula offers a straightforward method to solve not just real solutions but also complex ones if the roots are not real. Complex roots appear when the value inside the square root (the discriminant) is negative, and in such cases, the solutions will include imaginary numbers.
The quadratic formula efficiently solves any quadratic equation and ensures all potential solutions are found, both real and imaginary.

Discriminant

The discriminant is part of the quadratic formula expressed as \( b^2 - 4ac \). It determines the nature of the roots of a quadratic equation:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, it has one real double solution (or repeated root).
• If the discriminant is negative, the quadratic has two complex solutions.

In our exercise, with the equation \( 4u^2 - 25u + 36 = 0 \), calculating the discriminant using \( b = -25 \), \( a = 4 \), and \( c = 36 \), we found it to be 49. Since 49 is positive, this reveals that the equation has two real and distinct solutions.
The discriminant provides insight into what type of number system (real or complex) the solutions will belong to, which is essential for understanding the behavior of quadratic functions.

Graphing Equations

Graphing quadratic equations can visually confirm the solutions found algebraically. The graph of a quadratic equation \( y = ax^2 + bx + c \) is a parabola. Depending on the sign of \( a \):

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), it opens downwards.

In the original exercise, the equation \( Y_1 = 4x^4 - 25x^2 + 36 \) is graphed. Although it is not a simple quadratic equation, it is transformed into a quadratic form \( 4u^2 - 25u + 36 \) by substituting \( u = x^2 \), which allows us to handle it as a standard quadratic equation. This graph showcases points where it intersects the x-axis, corresponding to the real solutions found: \( x = -2, -\frac{3}{2}, \frac{3}{2}, \text{ and } 2 \).
Using graphing calculators or software can also provide visual support to ensure those solutions are accurate, enhancing the understanding of the quadratic's behavior over specified ranges.

Substituting Variables

Substituting variables can simplify complex algebraic problems, making them more manageable. In this particular exercise, the original quartic equation \( 4x^4 - 25x^2 + 36 = 0 \) was simplified by substituting a new variable \( u = x^2 \). This transformation turns the problem into a quadratic equation \( 4u^2 - 25u + 36 = 0 \), which is much easier to solve.
The process of substituting variables reduces complicated equations into familiar forms you know how to handle, like quadratics. After solving for the new variable \( u \), you can back substitute to find the solution for the original variable \( x \):

• If \( u = 4 \), then \( x = \pm 2 \).
• If \( u = \frac{9}{4} \), then \( x = \pm \frac{3}{2} \).

This approach exemplifies how variable substitution can streamline the path to finding solutions, whether they are real or complex.