Problem 73 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$16 b^{4}-1=0$$

Step-by-Step Solution

Answer

The solutions are \(\pm \frac{1}{2}\).

Step 1: Set the Equation to Zero

Start with the given equation and set it equal to zero: \[16b^4 - 1 = 0\]

Step 2: Move the Constant to the Other Side

Add 1 to both sides to isolate the polynomial: \[16b^4 = 1\]

Step 3: Solve for the Polynomial

Divide both sides by 16 to solve for \(b^4\): \[b^4 = \frac{1}{16}\]

Step 4: Take the Fourth Root

Take the fourth root of both sides to solve for \(b\): \[b = \pm \sqrt[4]{\frac{1}{16}}\]

Step 5: Simplify the Fourth Root

Simplify the term under the fourth root: \[\sqrt[4]{\frac{1}{16}} = \sqrt[4]{\frac{1}{2^4}} = \frac{1}{2}\]Therefore: \[b = \pm \frac{1}{2}\]

Step 6: Verify Solutions

Confirm that the solutions \(\frac{1}{2}\) and \(-\frac{1}{2}\) satisfy the original equation.

Key Concepts

Polynomial Equations

Polynomial equations are fundamental in algebra and consist of variables and coefficients combined using addition, subtraction, multiplication, and non-negative integer exponents. In this exercise, the polynomial equation we are dealing with is \(16b^4 - 1 = 0\). Here, we have a single variable \(b\), raised to the fourth power. To solve such equations, our goal is to isolate the variable term by rearranging and simplifying the equation.

Fourth Roots

To solve for a variable raised to a power, we often need to take the root corresponding to that power. In this case, we take the fourth root because the variable \(b\) is raised to the fourth power. Specifically, after steps of manipulation, we reach the equation \(b^4 = \frac{1}{16}\). Taking the fourth root of both sides, we get:
\(b = \pm \sqrt[4]{\frac{1}{16}}\), which simplifies to \(\frac{\frac{1}{2}}\) because \(\frac{1}{2^4} = \frac{1}{16}\). Thus, the solutions for \(b\) are \(\frac{1}{2}\) and \(-\frac{1}{2}\).

Real and Imaginary Solutions

In this exercise, the solutions we obtained are real numbers: \(\frac{1}{2}\) and \(-\frac{1}{2}\). Real solutions are numbers that can be found on the number line, unlike imaginary solutions, which involve the imaginary unit \(i\), where \(i = \sqrt{-1}\). Imaginary solutions arise in polynomial equations when solving for higher even powers, but in this case, since we are taking the fourth root of a positive number, our results are purely real.

Verification of Solutions

Verifying solutions is a crucial step to ensure the results satisfy the original polynomial equation. To verify our solutions \(\frac{1}{2}\) and \(-\frac{1}{2}\):

• Plug \(b = \frac{1}{2}\) into the original equation: 16\((\frac{1}{2})^4 - 1 = 0\)
• Simplify: \(\frac{1}{16}\) multiplied by 16 gives 1, and 1 - 1 equals 0.

Our solution holds.

• Similarly, for \(b = -\frac{1}{2}\): \((-\frac{1}{2})^4\) is also \(\frac{1}{16}\), and thus: 1 - 1 equals 0.

So, our verification shows both values satisfy the polynomial equation
This process confirms that the solutions \(\frac{1}{2}\) and \(-\frac{1}{2}\) are indeed correct.