Problem 15 - Algebra Form and Function

Question

In Exercises \(1-21,\) solve the equation for the variable. $$ 2 p^{5}+64=0 $$

Step-by-Step Solution

Answer

Question: Given the equation $$2 p^{5} + 64 = 0$$, solve for the value of p. Answer: The value of p for the given equation is $$p = -2$$.

Step 1: Subtract 64 from both sides

Subtract 64 on both sides to get: $$ 2p^5 - 64 = 0 - 64 $$ Which simplifies to: $$ 2p^5 = -64 $$

Step 2: Divide both sides by 2

Now, we will divide both sides of the equation by 2 to isolate \(p^5\). $$ \frac{2p^5}{2} = \frac{-64}{2} $$ This simplifies to: $$ p^5 = -32 $$

Step 3: Take the 5th root on both sides

To find the value of p, we will take the 5th root of both sides of the equation. $$ \sqrt[5]{p^5} = \sqrt[5]{-32} $$ This gives us the value of p: $$ p = -2 $$ The solution for the given equation is \(p = -2\).

Key Concepts

Polynomial Equations

Polynomial equations form the backbone of algebra, involving expressions with variables raised to whole number powers. You usually see them represented as terms in a specific order, such as quadratic (second degree), cubic (third degree), and so on. The equation in our exercise, \(2p^5 + 64 = 0\), is a fifth-degree polynomial because the variable \(p\) is raised to the power of five. This means it can have up to five different solutions, although in this specific case, there is only one real solution due to the nature of polynomials with a single monomial term.

• The highest exponent of the variable determines the degree.
• Polynomial equations can be simplified by combining like terms.
• The solutions of a polynomial can be real or complex numbers.

Working with polynomial equations typically involves simplifying the equation, like moving terms to one side, potentially dividing by a constant, or factoring. By understanding these basic principles, you can tackle more complex polynomial problems with greater ease.

Fifth Roots

The concept of a root in mathematics is the inverse operation of exponentiation. When we speak of a fifth root, we're specifically referring to the number that, when raised to the fifth power, yields the original number. In our equation, once we isolated \(p^5 = -32\), we found \(p\) by taking the fifth root of \(-32\). The equation \(\sqrt[5]{-32} = p\) equals \(-2\) because

• \((-2)^5 = -32\)

This showcases the nature of roots: they shrink higher powers back down to their base values. Calculating roots, especially for higher powers like a fifth root, can seem tricky. However, for small integers, it can often be done intuitively or with the aid of a calculator.

• In mathematics, only odd roots of negative numbers exist in the set of real numbers.
• Even roots of negative numbers result in complex numbers.

Isolation of Variables

To solve any equation, isolating the variable is a key technique. It means manipulating the equation to get the unknown variable on one side by itself. This often involves inverse operations, such as:

• Adding or subtracting terms to both sides.
• Multiplying or dividing terms across the equation.
• Taking roots or powers as appropriate.

In our given problem, we first subtracted 64 from both sides, then divided both sides by 2 to reach the form \(p^5 = -32\). This logical step-by-step process simplifies solving even complicated equations by focusing each step on isolating the variable until you find its value. Consistent practice with these steps instills confidence to solve increasingly complex equations and understand the underlying algebraic principles.