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Chapter 3: Problem 9

Find all real and imaginary solutions to each equation. Check your answers. $$a^{4}-16=0$$

Short Answer

Expert verified
The solutions are \[ \text{\textpm}2 \] and \[ \text{\textpm}2i \].

Step by step solution

01

Identify the equation

The given equation is \[ a^4 - 16 = 0 \].
02

Rewrite equation in standard form

Rewrite the equation as \[ a^4 = 16 \].
03

Find the fourth roots of 16

The equation \[ a^4 = 16 \] suggests that we need to find the fourth roots of 16. Remember, 16 can be written as \[ 16 = (2^4) \]. So, \[ a^4 = (2^4) \]. This gives the solutions \[ a = \boxed{\text{\textpm}2, \text{\textpm}2i} \].
04

Verifying the solutions

Verify each solution: \[ (2)^4 = 16, (-2)^4 = 16, (2i)^4 = (i^4 \times 2^4) = (1 \times 16) = 16, (-2i)^4 = (i^4 \times (-2)^4) = (1 \times 16) = 16 \]. All solutions are correct thus verifying the answers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

fourth roots
Fourth roots are values that, when raised to the power of four, give the original number. In our exercise, we have to solve \[ a^4 - 16 = 0 \]. To do this, we first set the equation in the form \[ a^4 = 16 \]. Now, we need to find all possible values of \[ a \] that satisfy the equation. Since \[ 16 \] can be expressed as \[ 2^4 \], we get the roots \[ a = 2, -2, 2i, -2i \]. This means that the fourth roots of 16 include both real and imaginary numbers. Each of these solutions, when raised to the power of four, equals 16.
imaginary numbers
Imaginary numbers are quantities that, when squared, give a negative result. They are usually represented with the symbol \[ i \], where \[ i = \sqrt{-1} \]. In the context of our exercise, the imaginary roots are \[ 2i \] and \[ -2i \]. When dealing with imaginary numbers, it's important to remember their properties:
  • \[ i^2 = -1 \]
  • \[ i^3 = -i \]
  • \[ i^4 = 1 \]
As shown in our solution verification step, \{(2i)^4 = (i^4 \times 2^4) = (1 \times 16) = 16 \}\. Similarly, \[ (-2i)^4 = (i^4 \times (-2)^4) = (1 \times 16) = 16 \]. These calculations show why \[ 2i \] and \[ -2i \] are solutions to the given polynomial equation.
verification of solutions
Verification of solutions ensures that our answers are correct. This involves substituting the solutions back into the original equation to see if they satisfy it. For our exercise, the given equation is \[ a^4 - 16 = 0 \]. We'll verify the solutions \[ 2, -2, 2i, \text{ and } -2i \]:
  • \[ (2)^4 = 16 \]: True
  • \[ (-2)^4 = 16 \]: True
  • \[ (2i)^4 = (i^4 \times 2^4) = (1 \times 16) = 16 \]: True
  • \[ (-2i)^4 = (i^4 \times (-2)^4) = (1 \times 16) = 16 \]: True
Since all these calculations result in 16, each solution satisfies the initial equation, confirming the correctness of our answers.

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