Problem 59 - Algebra and Trigonometry

Question

\(47-70\) The given equation involves a power of the variable. Find all real solutions of the equation. $$ x^{4}-16=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 2\) and \(x = -2\).

Step 1: Understanding the Equation

The given equation is \( x^4 - 16 = 0 \). Our task is to find all real values of \(x\) which satisfy this equation.

Step 2: Isolating the Power Expression

Add 16 to both sides of the equation to isolate the power expression. This gives us \( x^4 = 16 \).

Step 3: Taking the Fourth Root

Determine \(x\) by taking the fourth root of both sides of the equation \( x^4 = 16 \). This implies \( x = \pm \sqrt[4]{16} \).

Step 4: Simplifying the Fourth Root

Since \( \sqrt[4]{16} = \sqrt{\sqrt{16}} = \sqrt{4} = 2 \), our solutions become \( x = 2 \) and \( x = -2 \).

Step 5: Verifying the Solutions

Substitute both solutions back into the original equation to verify. For \(x = 2\), \(2^4 = 16\). For \(x = -2\), \((-2)^4 = 16\). Therefore, both solutions satisfy the equation.

Key Concepts

Fourth Roots

When you encounter an equation like \(x^4 - 16 = 0\), you are dealing with fourth roots. The term \(x^4\) means you raise \(x\) to the power of four. Solving for \(x\) involves reversing this operation, which is known as finding the fourth root.

The equation simplifies to \(x^4 = 16\). We find the fourth root to determine what values of \(x\) make this equation true. Taking the fourth root of a number means finding a number that, when multiplied by itself four times, equals the original number.

In this case, the fourth roots of 16 are both positive and negative 2. This is because both \(2^4\) and \((-2)^4\) equal 16. The positive and negative roots are important because they give us all potential solutions in a real number system.

Real Solutions

Real solutions refer to values of \(x\) that solve the equation and belong to the set of real numbers, which includes all the numbers on the number line (not including complex numbers). For the equation \(x^4 = 16\), the real solutions are determined by finding the fourth roots.

In the context of this exercise, we find that \(x = 2\) and \(x = -2\) are the real solutions, since these are the roots we obtain from \(\sqrt[4]{16}\).

It's crucial when solving polynomial equations to remember that some equations might have non-real (complex) solutions. However, the given exercise specifically asks for real solutions, so we focus on those.

Verification of Solutions

Verifying solutions is an essential step to ensure the values we've found are correct. After calculating that \(x = 2\) and \(x = -2\) are solutions to the equation \(x^4 = 16\), we need to check them by substituting back into the original equation.

Let's verify:

• For \(x = 2\), substituting gives \(2^4 - 16 = 0\). Calculating \(2^4\) gives 16, and hence \(16 - 16 = 0\), confirming that this solution is correct.
• For \(x = -2\), substituting gives \((-2)^4 - 16 = 0\). Similarly, \((-2)^4 = 16\), and \(16 - 16 = 0\), which confirms this solution too.

Verification helps to make sure our calculations align with the original equation, providing confidence that both solutions \(x = 2\) and \(x = -2\) are indeed real and correct.