If x + 1/x = 8, then what is x 2 + 1/x 2 equal to?
62
Solving the Algebra Problem: Finding \(x^2 + \frac{1}{x^2}\)
This question asks us to find the value of \(x^2 + \frac{1}{x^2}\) given the equation \(x + \frac{1}{x} = 8\). This is a common type of algebra problem that can be solved using a basic algebraic identity.
Understanding the Key Concept
The key to solving this problem lies in recognizing the relationship between \( (a+b)^2 \) and \( a^2 + b^2 \). The algebraic identity for squaring a binomial is:
\( (a+b)^2 = a^2 + 2ab + b^2 \)
In our problem, if we let \( a = x \) and \( b = \frac{1}{x} \), the expression \( x^2 + \frac{1}{x^2} \) looks similar to the \( a^2 + b^2 \) part of the identity. The given equation is \( x + \frac{1}{x} = 8 \), which corresponds to the \( a+b \) part.
Step-by-Step Solution Process
We are given the equation:
\( x + \frac{1}{x} = 8 \)
To get terms like \( x^2 \) and \( \frac{1}{x^2} \), we can square both sides of the given equation. Remember, whatever you do to one side of an equation, you must do to the other side to maintain equality.
Squaring both sides:
\( \left(x + \frac{1}{x}\right)^2 = 8^2 \)
Now, we expand the left side using the algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \), where \( a=x \) and \( b=\frac{1}{x} \):
\( x^2 + 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 = 64 \)
Simplify the middle term. The \( x \) and \( \frac{1}{x} \) in the product cancel each other out:
\( 2 \cdot x \cdot \frac{1}{x} = 2 \cdot 1 = 2 \)
Also, \( \left(\frac{1}{x}\right)^2 = \frac{1^2}{x^2} = \frac{1}{x^2} \).
Substitute these back into the expanded equation:
\( x^2 + 2 + \frac{1}{x^2} = 64 \)
We want to find the value of \( x^2 + \frac{1}{x^2} \). To isolate this term, subtract 2 from both sides of the equation:
\( x^2 + \frac{1}{x^2} = 64 - 2 \)
Perform the subtraction:
\( x^2 + \frac{1}{x^2} = 62 \)
So, the value of \( x^2 + \frac{1}{x^2} \) is 62.
Summary of Calculation
Here is a brief summary of the steps:
- Start with the given equation: \( x + \frac{1}{x} = 8 \)
- Square both sides: \( \left(x + \frac{1}{x}\right)^2 = 8^2 \)
- Expand the left side: \( x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 64 \)
- Simplify: \( x^2 + 2 + \frac{1}{x^2} = 64 \)
- Isolate \( x^2 + \frac{1}{x^2} \): \( x^2 + \frac{1}{x^2} = 64 - 2 \)
- Calculate the result: \( x^2 + \frac{1}{x^2} = 62 \)
| Given | Operation | Result |
|---|---|---|
| \( x + \frac{1}{x} = 8 \) | Square both sides | \( \left(x + \frac{1}{x}\right)^2 = 8^2 \) |
| \( \left(x + \frac{1}{x}\right)^2 = 64 \) | Expand left side | \( x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 64 \) |
| \( x^2 + 2 + \frac{1}{x^2} = 64 \) | Subtract 2 from both sides | \( x^2 + \frac{1}{x^2} = 64 - 2 \) |
| \( x^2 + \frac{1}{x^2} = 64 - 2 \) | Calculate | \( x^2 + \frac{1}{x^2} = 62 \) |
Revision Table: Algebraic Identities
Understanding algebraic identities is crucial for solving problems like this one. Here are a few important identities:
| Identity Name | Formula |
|---|---|
| Square of a sum | \( (a+b)^2 = a^2 + 2ab + b^2 \) |
| Square of a difference | \( (a-b)^2 = a^2 - 2ab + b^2 \) |
| Difference of squares | \( a^2 - b^2 = (a+b)(a-b) \) |
| Cube of a sum | \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a+b) \) |
| Cube of a difference | \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a-b) \) |
Additional Information: Related Problems
This problem is a specific case of a more general type. If you are given \( x + \frac{1}{x} = k \), you can always find \( x^2 + \frac{1}{x^2} \) by squaring: \( (x + \frac{1}{x})^2 = k^2 \), which gives \( x^2 + 2 + \frac{1}{x^2} = k^2 \), so \( x^2 + \frac{1}{x^2} = k^2 - 2 \).
Similarly, if you are given \( x - \frac{1}{x} = k \), you can find \( x^2 + \frac{1}{x^2} \) by squaring: \( (x - \frac{1}{x})^2 = k^2 \), which gives \( x^2 - 2 + \frac{1}{x^2} = k^2 \), so \( x^2 + \frac{1}{x^2} = k^2 + 2 \).
You can also find higher powers, such as \( x^3 + \frac{1}{x^3} \) or \( x^4 + \frac{1}{x^4} \), by building upon these results and using other identities like \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \).
For example, to find \( x^3 + \frac{1}{x^3} \) when \( x + \frac{1}{x} = 8 \):
\( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 - x\cdot\frac{1}{x} + \left(\frac{1}{x}\right)^2\right) \)
\( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right) \)
We know \( x + \frac{1}{x} = 8 \) and we found \( x^2 + \frac{1}{x^2} = 62 \). Substitute these values:
\( x^3 + \frac{1}{x^3} = (8)(62 - 1) = 8 \times 61 = 488 \)
These types of problems frequently appear in competitive exams and require a good understanding of algebraic manipulation and identities.
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