Problem 42 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ x^{6}-2 x^{3}-3=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = \sqrt[3]{3} \) and \( x = -1 \).

Step 1: Set Substitution

In order to simplify the equation, let's use the substitution \( y = x^3 \). Then, the equation becomes \( y^2 - 2y - 3 = 0 \).

Step 2: Solve the Quadratic Equation

The equation \( y^2 - 2y - 3 = 0 \) is a quadratic equation in \( y \). Use the quadratic formula to solve for \( y \): \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -2 \), and \( c = -3 \). Substituting these into the formula gives: \[ y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2} \] This results in \( y = 3 \) and \( y = -1 \).

Step 3: Substitute Back to x

Now, substitute back to \( x \) using \( y = x^3 \). For \( y = 3 \), solve \( x^3 = 3 \) which gives \( x = \sqrt[3]{3} \). For \( y = -1 \), solve \( x^3 = -1 \) which gives \( x = -1 \).

Step 4: List the Real Solutions

The real solutions from the equation are \( x = \sqrt[3]{3} \) and \( x = -1 \). These are the values for which the original equation \( x^6 - 2x^3 - 3 = 0 \) holds true.

Key Concepts

Real Solutions

In mathematics, not all equations will have solutions that are real numbers. Real solutions are values that can be either positive, negative, or zero. These numbers are found on the number line. Unlike complex or imaginary numbers, real solutions do not contain the imaginary unit \(i\). Determining whether an equation has real solutions involves examining the nature of its roots.
Consider a quadratic equation of the form \(y^2 - 2y - 3 = 0\). One way to identify real solutions is to calculate the discriminant, \(b^2 - 4ac\).

• If the discriminant is positive, there are two distinct real solutions.
• If it equals zero, there's exactly one real solution.
• If negative, no real solutions exist; instead, the solutions are complex.

For our equation, the discriminant is \(( -2 )^2 - 4 \times 1 \times -3 = 16\), which is positive. Thus, we have two real solutions, \(y = 3\) and \(y = -1\). These solutions are real and applicable using the substitution made earlier in the equation.

Substitution Method

The substitution method is a valuable tool in simplifying complex equations by transforming them into an easier form to solve. In this context, we used substitution to handle a higher-degree polynomial.
For the given equation \(x^6 - 2x^3 - 3 = 0\), the first substitution step involves replacing \(x^3\) with \(y\). This is because dealing with cubic terms can be more straightforward when treated as a quadratic equation.
Upon substituting \(y = x^3\), the equation becomes \(y^2 - 2y - 3 = 0\), a quadratic equation much easier to solve. Using this method:

• Simplifies complex equations to forms where standard formulas, like the quadratic formula, can be applied.
• Reduces the risk of calculation errors by transforming higher-degree terms into simpler variables.
• The substitution choice often depends on the middle or repeated variable, like \(x^3\) in this case, providing insight into the equation’s overall structure.

Following the solution of the quadratic equation, it is vital to substitute back to the original variable, ensuring that we solve for the values in the context of the initial equation, leading us back to solutions for \(x\).

Cubic Roots

The concept of cubic roots involves finding a number that, when multiplied by itself three times, results in the given number. Cubic roots are essential when solving equations involving cubic terms like \(x^3\).
For example, in solving \(x^3 = 3\), finding the cubic root gives us \(x = \sqrt[3]{3}\), while for \(x^3 = -1\), the solution is \(x = -1\). Key characteristics of cubic roots include:

• Any real number has precisely one real cubic root.
• Unlike square roots, cubic roots of negative numbers are also real.
• The notation \(\sqrt[3]{a}\) denotes the principal (main) cubic root of a number \(a\).

Cubic roots help provide solutions to otherwise challenging polynomial equations, expanding the range of numbers available to us when finding solutions. In the context of polynomial equations, recognizing cubic roots allows for a more intuitive solution-finding process, especially when these terms simplify the equation considerably.