If \(\rm (x+\frac{1}{x})=2\), then \(\rm x^7+\frac{1}{x^{117}}=\) ___________.
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Understanding the Algebra Problem: Finding the Value of an Expression
The problem asks us to find the value of the expression \( \rm x^7+\frac{1}{x^{117}} \) given the condition \( \rm (x+\frac{1}{x})=2 \). This type of problem often involves first solving the given equation for the variable \( \rm x \) and then substituting that value into the expression.
Step-by-Step Solution to Find x
We are given the equation:
\( \rm x+\frac{1}{x}=2 \)
To solve for \( \rm x \), we can multiply the entire equation by \( \rm x \) to eliminate the fraction. Note that if \( \rm x=0 \), the original expression \( \rm x+\frac{1}{x} \) would be undefined, so \( \rm x \) cannot be zero.
Multiplying by \( \rm x \), we get:
\( \rm x(x) + x(\frac{1}{x}) = 2(x) \)
\( \rm x^2 + 1 = 2x \)
Now, rearrange the terms to form a standard quadratic equation:
\( \rm x^2 - 2x + 1 = 0 \)
This quadratic equation is a perfect square trinomial. It can be factored as:
\( \rm (x-1)^2 = 0 \)
Taking the square root of both sides:
\( \rm x-1 = 0 \)
Solving for \( \rm x \):
\( \rm x = 1 \)
So, the only value of \( \rm x \) that satisfies the given condition \( \rm x+\frac{1}{x}=2 \) is \( \rm x=1 \).
Evaluating the Expression \( \rm x^7+\frac{1}{x^{117}} \)
Now that we have found \( \rm x=1 \), we need to substitute this value into the expression \( \rm x^7+\frac{1}{x^{117}} \).
Substitute \( \rm x=1 \):
\( \rm (1)^7+\frac{1}{(1)^{117}} \)
We know that any positive integer power of 1 is equal to 1. That is, \( \rm 1^n = 1 \) for any positive integer \( \rm n \).
So, \( \rm 1^7 = 1 \) and \( \rm 1^{117} = 1 \).
Substituting these values back into the expression:
\( \rm 1 + \frac{1}{1} \)
\( \rm 1 + 1 = 2 \)
Therefore, the value of \( \rm x^7+\frac{1}{x^{117}} \) when \( \rm (x+\frac{1}{x})=2 \) is 2.
Summary of the Solution
Given \( \rm x+\frac{1}{x}=2 \), we found that \( \rm x=1 \). Substituting \( \rm x=1 \) into the expression \( \rm x^7+\frac{1}{x^{117}} \) gives \( \rm 1^7 + \frac{1}{1^{117}} = 1 + \frac{1}{1} = 1 + 1 = 2 \).
| Given Equation | \( \rm x+\frac{1}{x}=2 \) |
|---|---|
| Solution for \( \rm x \) | \( \rm x=1 \) |
| Expression to Evaluate | \( \rm x^7+\frac{1}{x^{117}} \) |
| Value at \( \rm x=1 \) | \( \rm 1^7+\frac{1}{1^{117}} = 1+1=2 \) |
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Solving Algebraic Equations | Finding the value(s) of the variable that satisfy the equation. | We solved \( \rm x+\frac{1}{x}=2 \) for \( \rm x \). |
| Quadratic Equations | Equations of the form \( \rm ax^2+bx+c=0 \). Can be solved by factoring, completing the square, or quadratic formula. | \( \rm x^2-2x+1=0 \) is a quadratic equation. |
| Perfect Square Trinomial | A trinomial that is the square of a binomial, e.g., \( \rm a^2 \pm 2ab + b^2 = (a \pm b)^2 \). | \( \rm x^2-2x+1 \) is \( \rm (x-1)^2 \). |
| Properties of Exponents | Rules governing operations with exponents, e.g., \( \rm a^1=a \), \( \rm 1^n=1 \). | Used to evaluate \( \rm 1^7 \) and \( \rm 1^{117} \). |
| Substitution | Replacing a variable with its known value in an expression or equation. | We substituted \( \rm x=1 \) into the expression. |
Additional Information: Special Cases of \( \rm x+\frac{1}{x} \)
The expression \( \rm x+\frac{1}{x} \) is interesting in algebra. Here are a few points:
- For real numbers \( \rm x \), the minimum value of \( \rm x+\frac{1}{x} \) (for \( \rm x>0 \)) is 2, which occurs when \( \rm x=1 \).
- For real numbers \( \rm x \), the maximum value of \( \rm x+\frac{1}{x} \) (for \( \rm x<0 \)) is -2, which occurs when \( \rm x=-1 \).
- The given condition \( \rm x+\frac{1}{x}=2 \) specifically leads to \( \rm x=1 \) for real \( \rm x \). If the condition were \( \rm x+\frac{1}{x}=-2 \), the solution for real \( \rm x \) would be \( \rm x=-1 \).
- If the condition were \( \rm x+\frac{1}{x}=k \) where \( \rm |k|<2 \), there would be no real solutions for \( \rm x \), only complex solutions.
- Problems involving \( \rm x^n + \frac{1}{x^n} \) given \( \rm x+\frac{1}{x} \) are common. For integer \( \rm n \), these can often be solved using formulas or by finding \( \rm x \) if \( \rm x+\frac{1}{x} \) is a specific value like 2 or -2.
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