Introduction
The equation v³ = 12 is a cubic equation, and our goal is to find the values of v that satisfy this equation. This involves finding both real and complex solutions. Cubic equations are a fundamental topic in algebra, and understanding how to solve them is crucial for various applications in mathematics, physics, and engineering. In this article, we will explore the methods to solve this equation, including finding the real root and the complex roots.
Finding the Real Root
To solve the equation v³ = 12 for real values of v, we first recognize that this is a simple cubic equation. The most straightforward approach to finding the real root is to take the cube root of both sides of the equation. This operation isolates v and provides us with the real solution. By taking the cube root, we eliminate the cubic power on v and directly obtain its value.
The equation v³ = 12 can be rewritten as:
v = ∛12
This indicates that v is the cube root of 12. The cube root of 12 is a real number, but it is not a perfect cube, so we need to either use a calculator or express it in simplest radical form. Approximating the cube root of 12 gives us a numerical value, while expressing it in simplest radical form provides a precise algebraic representation.
Using a calculator, we find that:
v ≈ 2.289428
Thus, the real root of the equation v³ = 12 is approximately 2.289428. This is a direct solution and is often sufficient for practical applications where a numerical approximation is acceptable. However, to fully understand the solutions to cubic equations, we must also consider the complex roots.
Finding the Complex Roots
While the real root provides one solution to the equation v³ = 12, cubic equations generally have three roots, including complex roots. To find these, we need to use complex numbers and De Moivre's Theorem. Complex roots arise from the nature of cubic equations and the properties of complex numbers, which extend the number system beyond real numbers to include imaginary numbers.
Complex Numbers and Polar Form
A complex number is expressed in the form a + bi, where a is the real part and b is the imaginary part, and i is the imaginary unit (i² = -1). Complex numbers can also be represented in polar form, which is particularly useful for finding roots of complex numbers. The polar form of a complex number z is given by:
z = r(cos θ + i sin θ)
where r is the magnitude (or modulus) of z, and θ is the argument (or angle) of z. The magnitude r is calculated as r = √(a² + b²), and the argument θ can be found using trigonometric functions.
De Moivre's Theorem
De Moivre's Theorem is a powerful tool for finding powers and roots of complex numbers. The theorem states that for any complex number in polar form z = r(cos θ + i sin θ) and any integer n:
zⁿ = rⁿ(cos(nθ) + i sin(nθ))
This theorem is particularly useful for finding roots of complex numbers. To find the n-th roots of a complex number, we use the formula:
z^(1/n) = r^(1/n) [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]
where k is an integer ranging from 0 to n-1. This formula provides n distinct roots of the complex number.
Applying De Moivre's Theorem to v³ = 12
To find the complex roots of v³ = 12, we first express 12 as a complex number. In the complex plane, 12 can be written as 12 + 0i. In polar form, this is 12(cos 0 + i sin 0), since the magnitude is 12 and the argument is 0.
Now, we apply De Moivre's Theorem to find the cube roots:
v = (12(cos 0 + i sin 0))^(1/3)
Using the formula for the n-th roots, we get:
v = 12^(1/3) [cos((0 + 2πk)/3) + i sin((0 + 2πk)/3)]
where k = 0, 1, 2. We will calculate the roots for each value of k.
For k = 0:
v₀ = ∛12 [cos(0) + i sin(0)] = ∛12 (1 + 0i) = ∛12
This is the real root we found earlier, approximately 2.289428.
For k = 1:
v₁ = ∛12 [cos(2π/3) + i sin(2π/3)]
We know that cos(2π/3) = -1/2 and sin(2π/3) = √3/2, so
v₁ = ∛12 (-1/2 + i(√3/2))
This is a complex root with a real part of -∛12/2 and an imaginary part of ∛12(√3/2).
For k = 2:
v₂ = ∛12 [cos(4π/3) + i sin(4π/3)]
We know that cos(4π/3) = -1/2 and sin(4π/3) = -√3/2, so
v₂ = ∛12 (-1/2 - i(√3/2))
This is another complex root with a real part of -∛12/2 and an imaginary part of -∛12(√3/2).
Summary of Roots
The equation v³ = 12 has three roots:
- Real root: v₀ = ∛12 ≈ 2.289428
- Complex root: v₁ = ∛12 (-1/2 + i(√3/2))
- Complex root: v₂ = ∛12 (-1/2 - i(√3/2))
These roots provide a complete solution to the equation v³ = 12. The real root is a straightforward solution, while the complex roots illustrate the broader nature of solutions in the complex plane. Understanding both real and complex roots is essential for a comprehensive understanding of cubic equations.
Conclusion
In conclusion, solving the equation v³ = 12 involves finding both its real and complex roots. The real root is ∛12, which is approximately 2.289428. The complex roots can be found using De Moivre's Theorem, resulting in ∛12 (-1/2 + i(√3/2)) and ∛12 (-1/2 - i(√3/2)). This comprehensive approach demonstrates the importance of understanding complex numbers and their role in solving polynomial equations. By exploring both real and complex solutions, we gain a deeper understanding of the mathematical concepts underlying cubic equations and their applications in various fields.
