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Chapter 3: Problem 39

Find all real and imaginary solutions to each equation. Check your answers. $$x^{4}-81=0$$

Short Answer

Expert verified
The solutions are \(x = 3\), \(x = -3\), \(x = 3i\), and \(x = -3i\).

Step by step solution

01

- Factor the Equation

Start by factoring the equation: The given equation is: \[x^4 - 81 = 0\] Recognize it as a difference of squares: \[x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9)\]
02

- Solve Each Factor Separately

Solve each factor separately: First, solve \(x^2 - 9 = 0\): \[x^2 - 9 = 0\] Add 9 to both sides: \[x^2 = 9\] Take the square root of both sides: \[x = \pm 3\] giving the solutions \(x = 3\) and \(x = -3\). Next, solve \(x^2 + 9 = 0\): \[x^2 + 9 = 0\] Subtract 9 from both sides: \[x^2 = -9\] Take the square root of both sides: \[x = \pm 3i\] giving the solutions \(x = 3i\) and \(x = -3i\).
03

- Verify the Solutions

Verify all solutions by substituting them back into the original equation: Check \(x = 3\): \[3^4 - 81 = 81 - 81 = 0\] Check \(x = -3\): \[(-3)^4 - 81 = 81 - 81 = 0\] Check \(x = 3i\): \[(3i)^4 - 81 = 81 - 81 = 0\] Check \(x = -3i\): \[(-3i)^4 - 81 = 81 - 81 = 0\] All values satisfy the original equation.

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Key Concepts

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

Factoring Polynomials
Factoring a polynomial involves breaking it down into simpler polynomials whose product is the original one. This makes solving the equation easier. In our exercise, we have the polynomial \(x^4 - 81\). Recognizing this as a difference of squares helps us factor it further. A difference of squares follows the form \(a^2 - b^2 = (a + b)(a - b)\).
For our problem:
  • Set \( a = x^2 \) and \( b = 9 \).
  • This allows us to rewrite \( x^4 - 81 \) as \( (x^2 + 9)(x^2 - 9) \).
We now have two simpler polynomial equations that we can solve individually.
Imaginary Numbers
Imaginary numbers come into play when we deal with the square roots of negative numbers. The imaginary unit \(i\) is defined as \(i=\sqrt{-1}\). In our second factor \(x^2 + 9 = 0\), we can isolate \(x^2\):
\[ x^2 = -9 \]
Taking the square root of both sides, we get:
  • \( x = \pm \sqrt{-9} \)
  • \( x = \pm 3i \)
This shows \( x = 3i \) and \( x = -3i \) are the solutions involving imaginary numbers. Imaginary numbers are crucial in solving polynomial equations that have no real solutions.
Difference of Squares
The difference of squares is a useful technique in factoring polynomials. It leverages the relationship \(a^2 - b^2 = (a + b)(a - b)\). In our example, we used this method to transform \(x^4 - 81\) into \( (x^2 + 9)(x^2 - 9) \).
Further factoring the second term, \(x^2 - 9 \,\) we get:
  • \( x^2 - 9 = (x + 3)(x - 3) \)
    There's no real number solution for \( x^2 + 9 \,\) which makes this further factorable over real numbers.
      These factored forms simplify the polynomial to basic linear factors and quadratics that we know how to solve.
Verifying Solutions
Verifying solutions is crucial to ensure satisfaction of the original equation. Plugging each found value back into \( x^4 - 81 = 0 \), we check:
  • For \(x = 3\): \(3^4 - 81 = 81 - 81 = 0\).
  • For \(x = -3\): \((-3)^4 - 81 = 81 - 81 = 0\).
  • For \(x = 3i\): \((3i)^4 - 81 = 81 - 81 = 0\).
  • For \(x = -3i\): \((-3i)^4 - 81 = 81 - 81 = 0\).
Each substitution returns 0, confirming that all \( \pm 3 \) and \( \pm 3i \) are indeed solutions to the polynomial equation.

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