If x > 0 and \(\rm x^4 + \frac{1}{x^4} = 142\), what is the value of \(\rm x^7 + \frac{1}{x^7} \)?
1561\(\sqrt{14} \)
Solving Algebraic Expressions: Finding \( \rm x^7 + \frac{1}{x^7} \)
We are given an equation involving a variable \( \rm x \) and asked to find the value of a different expression involving \( \rm x \). Specifically, we are given that for \( \rm x > 0 \), \( \rm x^4 + \frac{1}{x^4} = 142 \), and we need to find the value of \( \rm x^7 + \frac{1}{x^7} \).
To solve this problem, we can use algebraic identities to work our way from the given expression (\( \rm x^4 + \frac{1}{x^4} \)) to the desired expression (\( \rm x^7 + \frac{1}{x^7} \)). This involves finding intermediate values like \( \rm x^2 + \frac{1}{x^2} \) and \( \rm x + \frac{1}{x} \), and then building up to higher powers like \( \rm x^3 + \frac{1}{x^3} \) and \( \rm x^4 + \frac{1}{x^4} \) (which is given) to combine them to get \( \rm x^7 + \frac{1}{x^7} \).
Step 1: Find the value of \( \rm x^2 + \frac{1}{x^2} \)
We know the algebraic identity \( \rm (a+b)^2 = a^2 + b^2 + 2ab \). Let \( \rm a = x^2 \) and \( \rm b = \frac{1}{x^2} \).
Then, \( \left( \rm x^2 + \frac{1}{x^2} \right)^2 = (x^2)^2 + \left(\frac{1}{x^2}\right)^2 + 2 \cdot x^2 \cdot \frac{1}{x^2} \)
\( \left( \rm x^2 + \frac{1}{x^2} \right)^2 = x^4 + \frac{1}{x^4} + 2 \)
We are given \( \rm x^4 + \frac{1}{x^4} = 142 \). Substituting this value:
\( \left( \rm x^2 + \frac{1}{x^2} \right)^2 = 142 + 2 \)
\( \left( \rm x^2 + \frac{1}{x^2} \right)^2 = 144 \)
Taking the square root of both sides:
\( \rm x^2 + \frac{1}{x^2} = \pm \sqrt{144} \)
\( \rm x^2 + \frac{1}{x^2} = \pm 12 \)
Since \( \rm x > 0 \), \( \rm x^2 \) is positive, and \( \rm \frac{1}{x^2} \) is also positive. The sum of two positive numbers must be positive.
Therefore, \( \rm x^2 + \frac{1}{x^2} = 12 \).
Step 2: Find the value of \( \rm x + \frac{1}{x} \)
Again, using the identity \( \rm (a+b)^2 = a^2 + b^2 + 2ab \). Let \( \rm a = x \) and \( \rm b = \frac{1}{x} \).
Then, \( \left( \rm x + \frac{1}{x} \right)^2 = x^2 + \left(\frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x} \)
\( \left( \rm x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 \)
We found in Step 1 that \( \rm x^2 + \frac{1}{x^2} = 12 \). Substituting this value:
\( \left( \rm x + \frac{1}{x} \right)^2 = 12 + 2 \)
\( \left( \rm x + \frac{1}{x} \right)^2 = 14 \)
Taking the square root of both sides:
\( \rm x + \frac{1}{x} = \pm \sqrt{14} \)
Since \( \rm x > 0 \), \( \rm x \) is positive, and \( \rm \frac{1}{x} \) is also positive. The sum of two positive numbers must be positive.
Therefore, \( \rm x + \frac{1}{x} = \sqrt{14} \).
Step 3: Find the value of \( \rm x^3 + \frac{1}{x^3} \)
We use the identity \( \rm (a+b)^3 = a^3 + b^3 + 3ab(a+b) \). Let \( \rm a = x \) and \( \rm b = \frac{1}{x} \).
Then, \( \left( \rm x + \frac{1}{x} \right)^3 = x^3 + \left(\frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x + \frac{1}{x}\right) \)
\( \left( \rm x + \frac{1}{x} \right)^3 = x^3 + \frac{1}{x^3} + 3 \left(x + \frac{1}{x}\right) \)
Rearranging to solve for \( \rm x^3 + \frac{1}{x^3} \):
\( \rm x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right) \)
We found in Step 2 that \( \rm x + \frac{1}{x} = \sqrt{14} \). Substituting this value:
\( \rm x^3 + \frac{1}{x^3} = (\sqrt{14})^3 - 3(\sqrt{14}) \)
\( \rm x^3 + \frac{1}{x^3} = 14\sqrt{14} - 3\sqrt{14} \)
\( \rm x^3 + \frac{1}{x^3} = (14 - 3)\sqrt{14} \)
\( \rm x^3 + \frac{1}{x^3} = 11\sqrt{14} \)
Step 4: Find the value of \( \rm x^7 + \frac{1}{x^7} \)
Consider the product of \( \left( \rm x^3 + \frac{1}{x^3} \right) \) and \( \left( \rm x^4 + \frac{1}{x^4} \right) \).
\( \left( \rm x^3 + \frac{1}{x^3} \right) \left( \rm x^4 + \frac{1}{x^4} \right) = x^3 \cdot x^4 + x^3 \cdot \frac{1}{x^4} + \frac{1}{x^3} \cdot x^4 + \frac{1}{x^3} \cdot \frac{1}{x^4} \)
\( = x^{3+4} + x^{3-4} + x^{4-3} + x^{-3-4} \)
\( = x^7 + x^{-1} + x^1 + x^{-7} \)
\( = x^7 + \frac{1}{x} + x + \frac{1}{x^7} \)
\( = \left( \rm x^7 + \frac{1}{x^7} \right) + \left( \rm x + \frac{1}{x} \right) \)
Rearranging to solve for \( \rm x^7 + \frac{1}{x^7} \):
\( \rm x^7 + \frac{1}{x^7} = \left( x^3 + \frac{1}{x^3} \right) \left( x^4 + \frac{1}{x^4} \right) - \left( x + \frac{1}{x} \right) \)
We have the values from previous steps:
- \( \rm x^3 + \frac{1}{x^3} = 11\sqrt{14} \) (from Step 3)
- \( \rm x^4 + \frac{1}{x^4} = 142 \) (given in the question)
- \( \rm x + \frac{1}{x} = \sqrt{14} \) (from Step 2)
Substitute these values into the expression for \( \rm x^7 + \frac{1}{x^7} \):
\( \rm x^7 + \frac{1}{x^7} = (11\sqrt{14})(142) - (\sqrt{14}) \)
First, calculate \( 11 \times 142 \):
\( 11 \times 142 = 11 \times (100 + 40 + 2) = 1100 + 440 + 22 = 1562 \)
So, \( \rm x^7 + \frac{1}{x^7} = 1562\sqrt{14} - \sqrt{14} \)
Combine the terms with \( \sqrt{14} \):
\( \rm x^7 + \frac{1}{x^7} = (1562 - 1)\sqrt{14} \)
\( \rm x^7 + \frac{1}{x^7} = 1561\sqrt{14} \)
Thus, the value of \( \rm x^7 + \frac{1}{x^7} \) is \( 1561\sqrt{14} \).
Comparing with Options
Let's check the calculated value against the given options:
- Option 1: \( 1562\sqrt{14} \)
- Option 2: \( 1563\sqrt{14} \)
- Option 3: \( 1561\sqrt{14} \)
- Option 4: \( 1560\sqrt{14} \)
Our calculated value \( 1561\sqrt{14} \) matches Option 3.
| Step | Calculation | Result |
|---|---|---|
| Given | \( \rm x^4 + \frac{1}{x^4} = 142 \) | \( \rm x^4 + \frac{1}{x^4} = 142 \) |
| Find \( \rm x^2 + \frac{1}{x^2} \) | \( \left( \rm x^2 + \frac{1}{x^2} \right)^2 = x^4 + \frac{1}{x^4} + 2 \) | \( \rm x^2 + \frac{1}{x^2} = 12 \) |
| Find \( \rm x + \frac{1}{x} \) | \( \left( \rm x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 \) | \( \rm x + \frac{1}{x} = \sqrt{14} \) |
| Find \( \rm x^3 + \frac{1}{x^3} \) | \( \rm x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right) \) | \( \rm x^3 + \frac{1}{x^3} = 11\sqrt{14} \) |
| Find \( \rm x^7 + \frac{1}{x^7} \) | \( \rm x^7 + \frac{1}{x^7} = \left( x^3 + \frac{1}{x^3} \right) \left( x^4 + \frac{1}{x^4} \right) - \left( x + \frac{1}{x} \right) \) | \( \rm x^7 + \frac{1}{x^7} = 1561\sqrt{14} \) |
Revision Table: Key Identities
Understanding key algebraic identities is crucial for solving problems like this involving powers of a variable and its reciprocal.
| Identity | Formula |
|---|---|
| Square of a sum | \( \rm (a+b)^2 = a^2 + b^2 + 2ab \) |
| Cube of a sum | \( \rm (a+b)^3 = a^3 + b^3 + 3ab(a+b) \) |
| Sum of cubes (derived) | \( \rm a^3 + b^3 = (a+b)^3 - 3ab(a+b) \) |
| Sum of powers (derived) | \( \rm a^m + b^m = (a^k + b^k)(a^{m-k} + b^{m-k}) - ab^k a^{m-k} - ba^k b^{m-k} \) (General form) |
| Sum of 7th powers (specific) | \( \rm x^7 + \frac{1}{x^7} = \left( x^3 + \frac{1}{x^3} \right) \left( x^4 + \frac{1}{x^4} \right) - \left( x + \frac{1}{x} \right) \) |
Additional Information: Generalizing Power Sums
Problems asking for \( \rm x^n + \frac{1}{x^n} \) given \( \rm x + \frac{1}{x} \) or a similar expression are common in algebra. We can find \( \rm x^n + \frac{1}{x^n} \) for various integer values of \( \rm n \) if we know \( \rm x + \frac{1}{x} \).
- If \( \rm x + \frac{1}{x} = k \), then \( \rm x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = k^2 - 2 \).
- If \( \rm x + \frac{1}{x} = k \), then \( \rm x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = k^3 - 3k \).
- If \( \rm x^2 + \frac{1}{x^2} = m \), then \( \rm x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = m^2 - 2 \).
- If we have values for \( \rm x^a + \frac{1}{x^a} \) and \( \rm x^b + \frac{1}{x^b} \), we can find \( \rm x^{a+b} + \frac{1}{x^{a+b}} \) or \( \rm x^{a-b} + \frac{1}{x^{a-b}} \) using the product identity: \( \left( \rm x^a + \frac{1}{x^a} \right) \left( \rm x^b + \frac{1}{x^b} \right) = \left( \rm x^{a+b} + \frac{1}{x^{a+b}} \right) + \left( \rm x^{a-b} + \frac{1}{x^{a-b}} \right) \), assuming \( \rm a > b \).
In this problem, we used the case where \( \rm a=4 \) and \( \rm b=3 \) to find \( \rm x^{4+3} + \frac{1}{x^{4+3}} \) using \( \left( \rm x^4 + \frac{1}{x^4} \right) \left( \rm x^3 + \frac{1}{x^3} \right) = \left( \rm x^7 + \frac{1}{x^7} \right) + \left( \rm x^{4-3} + \frac{1}{x^{4-3}} \right) \), which is \( \left( \rm x^7 + \frac{1}{x^7} \right) + \left( \rm x + \frac{1}{x} \right) \).
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