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Magazine Chapter 8: Problem 71
Find all real and imaginary solutions to each equation. $$m^{4}-16=0$$
Short Answer
Expert verified
The solutions are \( m = 2, -2, 2i, -2i \).
Step by step solution
01
- Set the equation to zero
First, note that the equation given is \( m^4 - 16 = 0 \). We start by moving the constant term to the other side: \( m^4 = 16 \).
02
- Take the fourth root of both sides
To solve for \( m \), take the fourth root of both sides. Remember, there are two possible solutions for any even root: a positive and a negative. Thus, we have \( m = \pm 2 \).
03
- Write complex solutions
Since this involves a fourth-degree polynomial, look for complex solutions as well. For complex solutions, we use \( m = 2i \) and \( m = -2i \), where \( i \) is the imaginary unit.
04
- Combine all solutions
Combine the real and imaginary solutions to list all solutions: \( m = 2, -2, 2i, -2i \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourth Root
When solving an equation like \( m^{4} - 16 = 0 \), it's important to understand the concept of a fourth root. A root of a number is a value that, when raised to a certain power, gives the original number. For example, if \( x^{2} = 25 \), then \( x \) is \( \pm 5 \), because both 5 and -5, when squared, give 25. Here, we are dealing with the fourth root, which means we must find a number that, when raised to the power of four, equals 16.
Remember:
For real numbers, \( 2^{4} = 16 \), so the fourth root of 16 is \( \pm 2 \). For complex numbers, \( (2i)^{4} = 16 \), so \( 2i \) and \( -2i \) also satisfy the equation.
Remember:
- The fourth root of 16 has two pairs of solutions: one pair of real numbers and one pair of imaginary numbers because the polynomial degree is four (an even number).
- Each pair includes both a positive and a negative solution.
For real numbers, \( 2^{4} = 16 \), so the fourth root of 16 is \( \pm 2 \). For complex numbers, \( (2i)^{4} = 16 \), so \( 2i \) and \( -2i \) also satisfy the equation.
Real Solutions
Real solutions are those that we can locate on the number line, without involving any imaginary parts. They are the actual x-values for which the polynomial equals zero.
In the equation \( m^{4} - 16 = 0 \), we find that the real solutions are \( m = 2 \) and \( m = -2 \). These are found by taking the fourth root of both sides after setting up the equation as \( m^{4} = 16 \).
In the equation \( m^{4} - 16 = 0 \), we find that the real solutions are \( m = 2 \) and \( m = -2 \). These are found by taking the fourth root of both sides after setting up the equation as \( m^{4} = 16 \).
- Here are some points to remember:
- Real solutions are the values you find when you use standard arithmetic and do not need imaginary numbers.
- For \( m^{4} = 16 \), the two real solutions are the fourth roots: \( m = 2 \) and \( m = -2 \).
Imaginary Solutions
Imaginary solutions involve \( i \), the imaginary unit, where \( i^{2} = -1 \). They are crucial for solving equations that don鈥檛 have real roots.
For the given equation \( m^{4} - 16 = 0 \), the imaginary solutions are determined by considering complex numbers. When we solve \( m^{4} = 16 \), the complex roots can be derived as follows: since \( (2i)^{4} = 16 \), \( m = 2i \) and \( m = -2i \) are the relevant solutions.
Important notes include:
For the given equation \( m^{4} - 16 = 0 \), the imaginary solutions are determined by considering complex numbers. When we solve \( m^{4} = 16 \), the complex roots can be derived as follows: since \( (2i)^{4} = 16 \), \( m = 2i \) and \( m = -2i \) are the relevant solutions.
Important notes include:
- Imaginary solutions occur when the polynomial degree exceeds the number of real solutions.
- \( i \) (the imaginary unit) helps in expressing roots of negative numbers.
- The imaginary roots \( 2i and -2i \) provide additional solutions to higher-degree polynomials like the fourth-degree polynomial here.
Complex Numbers
Complex numbers combine real and imaginary parts in the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit.
When solving polynomial equations, complex numbers play a critical role, especially if the degree of the polynomial suggests multiple roots. For the equation \( m^{4} - 16 = 0 \), all solutions (real and imaginary) must be considered.
To fully grasp complex numbers, remember:
Thus, the complete set of solutions includes both real (\( 2, -2 \)) and complex (\( 2i, -2i \)), demonstrating their utility in solving higher-degree polynomials.
When solving polynomial equations, complex numbers play a critical role, especially if the degree of the polynomial suggests multiple roots. For the equation \( m^{4} - 16 = 0 \), all solutions (real and imaginary) must be considered.
To fully grasp complex numbers, remember:
- Complex numbers are written as \( a + bi \).
- They help solve polynomials that real numbers alone can't solve, by introducing imaginary parts.
- For \( m^{4} = 16 \), the complex roots include \( 2i \) and \( -2i \), showing the power of complex numbers in extending the scope of solutions.
Thus, the complete set of solutions includes both real (\( 2, -2 \)) and complex (\( 2i, -2i \)), demonstrating their utility in solving higher-degree polynomials.
