x to the power of 5 added to reciprocal of x to the power of 5, when x added to the reciprocal of x is 5

Given:
\[ x + \frac{1}{x} = 5 \]

We need to find:
\[ x^5 + \frac{1}{x^5} \]

Let's denote \( y = x + \frac{1}{x} \). Then, \( y = 5 \).

To find \( x^2 + \frac{1}{x^2} \), we square \( y \):
\[ \left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 \]
\[ y^2 = x^2 + \frac{1}{x^2} + 2 \]
\[ 5^2 = x^2 + \frac{1}{x^2} + 2 \]
\[ 25 = x^2 + \frac{1}{x^2} + 2 \]
\[ x^2 + \frac{1}{x^2} = 23 \]

Next, we find \( x^3 + \frac{1}{x^3} \) by multiplying \( x + \frac{1}{x} \) and \( x^2 + \frac{1}{x^2} \):
\[ \left( x + \frac{1}{x} \right) \left( x^2 + \frac{1}{x^2} \right) = x^3 + \frac{1}{x^3} + x + \frac{1}{x} \]
\[ y \cdot (x^2 + \frac{1}{x^2}) = x^3 + \frac{1}{x^3} + y \]
\[ 5 \cdot 23 = x^3 + \frac{1}{x^3} + 5 \]
\[ 115 = x^3 + \frac{1}{x^3} + 5 \]
\[ x^3 + \frac{1}{x^3} = 110 \]

Next, we find \( x^4 + \frac{1}{x^4} \) by squaring \( x^2 + \frac{1}{x^2} \):
\[ \left( x^2 + \frac{1}{x^2} \right)^2 = x^4 + \frac{1}{x^4} + 2 \]
\[ 23^2 = x^4 + \frac{1}{x^4} + 2 \]
\[ 529 = x^4 + \frac{1}{x^4} + 2 \]
\[ x^4 + \frac{1}{x^4} = 527 \]

Finally, we find \( x^5 + \frac{1}{x^5} \) by multiplying \( x + \frac{1}{x} \) and \( x^4 + \frac{1}{x^4} \):
\[ \left( x + \frac{1}{x} \right) \left( x^4 + \frac{1}{x^4} \right) = x^5 + \frac{1}{x^5} + x^3 + \frac{1}{x^3} \]
\[ y \cdot (x^4 + \frac{1}{x^4}) = x^5 + \frac{1}{x^5} + x^3 + \frac{1}{x^3} \]
\[ 5 \cdot 527 = x^5 + \frac{1}{x^5} + 110 \]
\[ 2635 = x^5 + \frac{1}{x^5} + 110 \]
\[ x^5 + \frac{1}{x^5} = 2525 \]

So, \( x^5 + \frac{1}{x^5} = 2525 \).

Would you like any details or have any questions?

Here are 5 questions you might ask next:
1. How do you solve other similar polynomial equations?
2. Can you explain the method used to solve for \( x^3 + \frac{1}{x^3} \) in more detail?
3. How can these methods be applied to other types of equations?
4. What are some common applications of these types of mathematical problems?
5. Can you show the derivation of the binomial theorem used in such expansions?

**Tip:** When dealing with expressions involving reciprocals and powers, leveraging identities and symmetry in algebra can simplify the problem significantly.

Explore how to solve polynomial equations involving powers and reciprocals with detailed step-by-step solutions. This problem covers algebraic concepts like symmetric functions and is suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation