Problem 102 - Algebra and Trigonometry

Question

The Cubic Formula The quadratic formula can be used to solve any quadratic (or second-degree) equation. You may have wondered if similar formulas exist for cubic (third-degree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic \(x^{3}+p x+q=0\) , Cardano (page 344 ) 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 is impossible to write a quartic formula, that is, a formula for fifth-degree equations. Finally, Galois (page 327\()\) gave a criterion for determining which equations can be solved by a formula involving radicals. (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

(a) Solution: \(x = 1\); (b) Solution: \(x = 9\); (c) Complex roots.

Step 1: Recognize the cubic equation format

Identify the given cubic equation format; each is of the form \(x^3 + px + q = 0\). We must determine the values of \(p\) and \(q\) for each equation and then apply Cardano's formula for the roots.

Step 2: Apply Cardano's formula to equation (a)

For \(x^3 - 3x + 2 = 0\), identify \(p = -3\) and \(q = 2\). The formula becomes \(x = \sqrt[3]{\frac{-2}{2} + \sqrt{\frac{2^2}{4} + \frac{(-3)^3}{27}}} + \sqrt[3]{\frac{-2}{2} - \sqrt{\frac{2^2}{4} + \frac{(-3)^3}{27}}}\). Calculate the expressions to find the cube roots, which yields \(x = 1\).

Step 3: Check the root for equation (a)

Verify the solution by substituting \(x = 1\) back into \(x^3 - 3x + 2 = 0\). The left side simplifies to \(1 - 3 + 2 = 0\), confirming that \(x = 1\) is a solution.

Step 4: Apply Cardano's formula to equation (b)

For \(x^3 - 27x - 54 = 0\), identify \(p = -27\) and \(q = -54\). Using the formula, calculate \(x = \sqrt[3]{\frac{54}{2} + \sqrt{\frac{(-54)^2}{4} + \frac{(-27)^3}{27}}} + \sqrt[3]{\frac{54}{2} - \sqrt{\frac{(-54)^2}{4} + \frac{(-27)^3}{27}}}\). This simplifies to \(x = 9\).

Step 5: Check the root for equation (b)

Verify the solution by substituting \(x = 9\) back into \(x^3 - 27x - 54 = 0\). The left side simplifies to \(729 - 243 - 54 = 0\), confirming that \(x = 9\) is a solution.

Step 6: Apply Cardano's formula to equation (c)

For \(x^3 + 3x + 4 = 0\), identify \(p = 3\) and \(q = 4\). The formula gives \(x = \sqrt[3]{\frac{-4}{2} + \sqrt{\frac{4^2}{4} + \frac{3^3}{27}}} + \sqrt[3]{\frac{-4}{2} - \sqrt{\frac{4^2}{4} + \frac{3^3}{27}}}\). Since the discriminant is negative, this cubic has complex roots.

Key Concepts

Cardano's Formula

Cardano's Formula is a method used to solve cubic equations. It was developed by the Italian mathematician Gerolamo Cardano during the Renaissance. This formula works specifically for cubic equations in the form of a "depressed cubic": \(x^3 + px + q = 0\). The formula provides a way to find at least one real root using radicals. Here's how it works in practice:

• First, identify the coefficients \(p\) and \(q\) from your equation.
• Plug these values into Cardano's formula: \[x = \sqrt[3]{\frac{-q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+ \sqrt[3]{\frac{-q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}\]
• Calculate the terms under the cube roots to determine the value of \(x\).
• The result will yield one real root, but there may be other real or complex roots to explore.

Cardano's Formula was a groundbreaking discovery in algebra, paving the way for solutions to higher-degree polynomials.

Depressed Cubic Equation

The Depressed Cubic Equation is a simplified form of a cubic equation. A standard cubic equation can have a form like \(x^3 + ax^2 + bx + c = 0\), but it is often easier to solve when reduced to the so-called depressed form\(x^3 + px + q = 0\).

To reach the depressed form, you usually employ a substitution technique, removing the \(x^2\) term. Here's how you do it:

• Make a substitution with \(x = y + \frac{a}{3}\) into the original equation.
• This transforms the equation into a new format where the quadratic term disappears.
• The equation becomes: \(y = x - \frac{a}{3}\) yielding a simpler cubic without the quadratic part.

By reducing the equation to a depressed form, the complex problem becomes more manageable, allowing algorithms like Cardano's Formula to work effectively.

Algebraic Roots

Algebraic roots refer to the solutions of polynomial equations, and in this context, specifically to cubic equations. Unlike linear or quadratic equations that have straightforward solutions, cubic equations can provide multiple root types.

• Real Roots: These are solutions that result in a real number and can be found using methods like Cardano's Formula.
• Complex Roots: These roots include an imaginary component, represented as \(a + bi\), and occur when the discriminant in Cardano's formula indicates a negative value under the square root.

The number of real and complex roots depends on the specific equation. It's important to understand that roots may not always be distinct. Certain scenarios can lead to multiple real roots or a mix of real and complex roots. Analyzing the discriminant helps determine possible root types.

Complex Roots

Complex roots are an integral part of many polynomial equations, including cubic equations like those solved by Cardano's Formula. When you find a negative discriminant under Cardano's square root section, the roots will be complex.

Complex roots are expressed in the form \(a + bi\), where:

• \(a\) is the real part of the root.
• \(b\) is the imaginary part, represented with \(i\) (where \(i = \sqrt{-1}\)).

Let's categorize the roots:

• If a cubic equation has one real root, the other two roots must be complex conjugates (like \(x = a + bi\) and \(x = a - bi\)).
• These complex roots occur in conjugate pairs to ensure the equation remains solvable within the realm of real numbers.

Understanding complex roots broadens the grasp on how polynomials behave and solves various algebraic challenges even when real number solutions are scarce.