Problem 36 - 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. $$\begin{aligned}&8 x^{6}+7 x^{3}-1=0\\\&[-4,4] \text { by }[-5,100]\end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{2}\) and \(x = -1\).

Step 1: Identify a Substitution

Notice that the equation is of the form \(8x^6 + 7x^3 - 1 = 0\). To simplify the equation, perform a substitution. Let \(y = x^3\). This transforms the equation into \(8y^2 + 7y - 1 = 0\).

Step 2: Solve the Quadratic Equation

With \(8y^2 + 7y - 1 = 0\), use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \(a = 8\), \(b = 7\), and \(c = -1\). Substitute these values to find \(y\):\[y = \frac{-7 \pm \sqrt{49 + 32}}{16} = \frac{-7 \pm \sqrt{81}}{16}\].This simplifies to \(y = \frac{-7 \pm 9}{16}\).

Step 3: Calculate the Values of y

Calculate for both plus and minus cases:1. \(y = \frac{-7 + 9}{16} = \frac{2}{16} = \frac{1}{8}\).2. \(y = \frac{-7 - 9}{16} = \frac{-16}{16} = -1\).

Step 4: Substitute Back to Find x

Recall the substitution \(y = x^3\). Now, solve for \(x\):1. \(x^3 = \frac{1}{8}\) implies \(x = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}\).2. \(x^3 = -1\) implies \(x = \sqrt[3]{-1} = -1\).Thus, the solutions for \(x\) are \(x = \frac{1}{2}\) and \(x = -1\).

Step 5: Confirm the Real Solutions Graphically

Graph the left side of the equation \(8x^6 + 7x^3 - 1\) as \(Y_1\) in the suggested viewing window \([-4, 4]\) by \([-5, 100]\) using a graphing calculator. The graph will intersect the x-axis at the real solutions, \(x = \frac{1}{2}\) and \(x = -1\), confirming our analytical results.

Key Concepts

Quadratic Equation

A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratic equations are fundamental in algebra because they appear in a wide range of problems—from geometry to physics to economics. The general solution method for quadratics is the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows us to solve for the variable \( y \) by plugging in the values of \( a \), \( b \), and \( c \). The part under the square root, \( b^2 - 4ac \), is known as the discriminant. It tells us about the nature of the roots:

• If the discriminant is positive, there are two different real roots.
• If it is zero, there is one real root.
• If it is negative, the roots are complex.

Understanding this helps in finding comprehensive solutions to quadratic equations.

Substitution Method

The substitution method is a powerful technique used to simplify complex equations. In cases where a straightforward solution is not possible, substitution can transform the equation into a more manageable form. For instance, in the problem \( 8x^6 + 7x^3 - 1 = 0 \), a simple direct approach is not feasible. By letting \( y = x^3 \), we substituted the cumbersome term "\( x^6 \)" into a more familiar quadratic form \( 8y^2 + 7y - 1 = 0 \).This transformation allows the use of the quadratic formula to find solutions for \( y \). After finding these solutions, we substitute \( y \) back with \( x^3 \) to uncover the values for \( x \). This method breaks down complex equations into simpler, solvable parts. Hence, the substitution method bridges the gap between recognizing complicated patterns and solving them effectively.

Graphical Analysis

Graphical analysis involves plotting the equation on a graph to visually identify the solutions. For equations like \( 8x^6 + 7x^3 - 1 = 0 \), graphing helps confirm analytical results. Here, you would plot \( 8x^6 + 7x^3 - 1 \) as \( Y_1 \) in a graphing calculator or software. In the graph, where the curve crosses the x-axis indicates the real solutions. For the interval \([-4, 4]\) along the x-axis and \([-5, 100]\) on the y-axis, we expect to see the intersections at our solution points. In our example, the intersections show \( x = \frac{1}{2} \) and \( x = -1 \). Graphing provides a visual representation that can confirm the algebraic process, adding a layer of verification to ensure the solutions' accuracy.

Cubic Roots

The concept of cubic roots is central to dealing with polynomial equations that involve terms like \( x^3 \). A cubic root of a number \( a \), symbolized as \( \sqrt[3]{a} \), is a number that, when multiplied by itself twice, equates back to \( a \). In solving equations like \( x^3 = \frac{1}{8} \) and \( x^3 = -1 \), finding cubic roots becomes essential.For \( x^3 = \frac{1}{8} \), the solution is \( x = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \), meaning the number \( \frac{1}{2} \) multiplied by itself three times equals \( \frac{1}{8} \). Similarly, for \( x^3 = -1 \), the solution is \( x = \sqrt[3]{-1} = -1 \), showing that \( -1 \) cubed is \( -1 \). Mastery of solving cubic equations helps in understanding more complex behavior of polynomial equations. This knowledge is key when transitioning from finding solutions in abstract algebra to applying these concepts in practical situations.