Math Problem Statement

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

Solution

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

We need to find: x5+1x5x^5 + \frac{1}{x^5}

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

To find x2+1x2x^2 + \frac{1}{x^2}, we square yy: (x+1x)2=x2+1x2+2\left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 y2=x2+1x2+2y^2 = x^2 + \frac{1}{x^2} + 2 52=x2+1x2+25^2 = x^2 + \frac{1}{x^2} + 2 25=x2+1x2+225 = x^2 + \frac{1}{x^2} + 2 x2+1x2=23x^2 + \frac{1}{x^2} = 23

Next, we find x3+1x3x^3 + \frac{1}{x^3} by multiplying x+1xx + \frac{1}{x} and x2+1x2x^2 + \frac{1}{x^2}: (x+1x)(x2+1x2)=x3+1x3+x+1x\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(x2+1x2)=x3+1x3+yy \cdot (x^2 + \frac{1}{x^2}) = x^3 + \frac{1}{x^3} + y 523=x3+1x3+55 \cdot 23 = x^3 + \frac{1}{x^3} + 5 115=x3+1x3+5115 = x^3 + \frac{1}{x^3} + 5 x3+1x3=110x^3 + \frac{1}{x^3} = 110

Next, we find x4+1x4x^4 + \frac{1}{x^4} by squaring x2+1x2x^2 + \frac{1}{x^2}: (x2+1x2)2=x4+1x4+2\left( x^2 + \frac{1}{x^2} \right)^2 = x^4 + \frac{1}{x^4} + 2 232=x4+1x4+223^2 = x^4 + \frac{1}{x^4} + 2 529=x4+1x4+2529 = x^4 + \frac{1}{x^4} + 2 x4+1x4=527x^4 + \frac{1}{x^4} = 527

Finally, we find x5+1x5x^5 + \frac{1}{x^5} by multiplying x+1xx + \frac{1}{x} and x4+1x4x^4 + \frac{1}{x^4}: (x+1x)(x4+1x4)=x5+1x5+x3+1x3\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(x4+1x4)=x5+1x5+x3+1x3y \cdot (x^4 + \frac{1}{x^4}) = x^5 + \frac{1}{x^5} + x^3 + \frac{1}{x^3} 5527=x5+1x5+1105 \cdot 527 = x^5 + \frac{1}{x^5} + 110 2635=x5+1x5+1102635 = x^5 + \frac{1}{x^5} + 110 x5+1x5=2525x^5 + \frac{1}{x^5} = 2525

So, x5+1x5=2525x^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 x3+1x3x^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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Equations
Symmetric Functions

Formulas

Sum of powers of x and its reciprocal

Theorems

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Suitable Grade Level

Advanced High School