Problem 108 - Precalculus Mathematics for Calculus

Question

The quadratic formula can be used to solve any quadratic (or second-degree) equation. You might have wondered whether similar formulas exist for cubic (thirddegree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic \(x^{3}+p x+q=0,\) Cardano (page 274) found the following formula for one solution: $$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$ A formula for quartic equations was discovered by the Italian mathematician Ferrari in \(1540 .\) In 1824 the Norwegian mathematician Niels Henrik Abel proved that it is impossible to write a quintic formula, that is, a formula for fifth-degree equations. Finally, Galois (page 254) gave a criterion for determining which equations can be solved by a formula involving radicals. Use the cubic formula to find a solution for the following equations. Then solve the equations using the methods you learned in this section. Which method is easier? (a) \(x^{3}-3 x+2=0\) (b) \(x^{3}-27 x-54=0\) (c) \(x^{3}+3 x+4=0\)

Step-by-Step Solution

Answer

Traditional methods are generally easier for clear rational roots.

Step 1: Identify p and q

For each cubic equation given, identify the values of \(p\) and \(q\) in the form \(x^3 + px + q = 0\).

Step 2: Apply Cardano's Formula for Equation (a)

For \(x^3 - 3x + 2 = 0\), identify \(p = -3\) and \(q = 2\). Substitute these values into Cardano's formula: \[ x = \sqrt[3]{\frac{-2}{2} + \sqrt{\frac{2^2}{4} + \frac{(-3)^3}{27}}} + \sqrt[3]{\frac{-2}{2} - \sqrt{\frac{2^2}{4} + \frac{(-3)^3}{27}}} \]

Step 3: Simplify Cardano's Formula for Equation (a)

Calculate the inside of the cube roots in Cardano's formula:1. Compute \(\frac{q^2}{4} = \frac{2^2}{4} = 1\).2. Compute \(\frac{p^3}{27} = \frac{(-3)^3}{27} = -1\).3. Find the discriminant: \(\frac{q^2}{4} + \frac{p^3}{27} = 1 - 1 = 0\).4. Thus, \(x = \sqrt[3]{0} + \sqrt[3]{-2} = 0 + (-1) = -1\).The solution for this method is \(x = 1\).

Step 4: Try Traditional Methods for Equation (a)

Factor the cubic equation: \(x^3 - 3x + 2 = 0\) using trial and error or rational root theorem:Find a root: \(x = 1\).Factor: \((x - 1)(x^2 + x - 2) = 0\).Solve quadratic: \(x^2 + x - 2 = 0\).\Solutions from factoring are \((x - 1) = 0 \Rightarrow x = 1\) and \((x^2 + x - 2) = 0 \Rightarrow x = 1, x = -2\).

Step 5: Apply Cardano's Formula for Equation (b)

For \(x^3 - 27x - 54 = 0\), identify \(p = -27\) and \(q = -54\). Substitute these values into Cardano's formula.

Step 6: Simplify Cardano's Formula for Equation (b)

Evaluate the discriminant for \(x^3 - 27x - 54 = 0\): After simplifying, continue substituting and solving to find that one solution is \(x = 6\).

Step 7: Try Traditional Methods for Equation (b)

Factor out a root by the rational root theorem and synthetic division. You find that \(x = 6\) is a root. Then, use synthetic division to factor out the cubic to find: \((x - 6)(x^2 + 6x + 9) = 0\).Simplify the quadratic: \((x + 3)^2 = 0 \Rightarrow x = -3\) is a repeated root.

Step 8: Apply Cardano's Formula for Equation (c)

For \(x^3 + 3x + 4 = 0\), identify \(p = 3\) and \(q = 4\). Substitute these values into Cardano's formula.

Step 9: Simplify Cardano's Formula for Equation (c)

Evaluate the discriminant for \(x^3 + 3x + 4 = 0\).Here, calculations reveal that the roots are complex. Using Cardano’s formula won't directly help without complex calculations. Finding roots manually could confirm that real rational solutions are not possible.

Step 10: Compare Methods

Using Cardano's formula for simple roots is straightforward after calculation but can be cumbersome with complex results or multiple roots. Traditional factoring methods or quadratic solutions are often quicker for simple integer roots.

Key Concepts

Cardano's Formula

Cardano's formula is a method for solving cubic equations, specifically designed for equations in the form of \(x^3 + px + q = 0\). This formula was developed by the Italian mathematician Gerolamo Cardano and allows for finding at least one root of a cubic equation. The formula itself is rather complex: \[x = \sqrt[3]{\frac{-q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{\frac{-q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}\]. Its use involves identifying the correct \(p\) and \(q\) values from the given cubic equation and then solving for \(x\). While powerful, using Cardano's formula can require solving complex expressions, especially when the discriminant \(\frac{q^2}{4} + \frac{p^3}{27}\) leads to irrational or complex numbers. This formula represents one of the first algebraic solutions to cubic equations, marking a significant advancement in algebra.

Roots of Cubic Equations

Finding the roots—or solutions—of a cubic equation like \(x^3 + px + q = 0\) involves determining the values of \(x\) that satisfy the equation. The nature of the roots can vary:

• Real and distinct: Three different real numbers.
• Real and repeated: At least one real number occurs more than once.
• Complex: At least one pair of complex numbers.

The discriminant \(\Delta = \frac{q^2}{4} + \frac{p^3}{27}\) helps determine the nature of the roots:

• If \(\Delta > 0\), one real root and two complex conjugate roots.
• If \(\Delta = 0\), all roots are real, and there is at least one repeated root.
• If \(\Delta < 0\), all three roots are real and distinct.

Determining these roots can be done through Cardano's formula, factoring, or numerical methods when analytic solutions are cumbersome or impractical.

Polynomial Factorization

Polynomial factorization is a technique used for breaking down a polynomial equation into simpler "factor" polynomials that, when multiplied together, give the original polynomial. When you solve a cubic equation like \(x^3 + px + q = 0\), factorization can reveal the equation's roots. This is especially useful when the roots are integers or can be found using the rational root theorem.For example, consider the equation \(x^3 - 3x + 2 = 0\). By applying the rational root theorem and testing possible roots, you might find that \(x = 1\) is a root. This allows you to express the cubic equation as \((x - 1)(x^2 + x - 2) = 0\). Further factoring of the quadratic polynomial \(x^2 + x - 2\) will completely solve the equation providing additional roots such as \(x = -2\).Factorization, therefore, simplifies solving polynomials and is generally easier for students proficient in recognizing and manipulating polynomial expressions, especially when dealing with integer or rational roots.