V5 + v6 Find a monic integral polynomial for which x is a root and prove that x is algebraically irrational.
V5 + v6 Find a monic integral polynomial for which x is a root and prove that x is algebraically irrational.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find a monic integral polynomial for which \( x \) is a root, where \( x = \sqrt{5} + \sqrt{6} \). Prove that \( x \) is algebraically irrational.
To solve this problem:
1. **Set Up the Expression:**
Let \( x = \sqrt{5} + \sqrt{6} \).
2. **Square Both Sides:**
\[
x^2 = (\sqrt{5} + \sqrt{6})^2 = 5 + 2\sqrt{30} + 6 = 11 + 2\sqrt{30}
\]
3. **Isolate the Square Root:**
\[
x^2 - 11 = 2\sqrt{30}
\]
4. **Square Again to Eliminate the Square Root:**
\[
(x^2 - 11)^2 = (2\sqrt{30})^2 = 120
\]
\[
x^4 - 22x^2 + 121 = 120
\]
5. **Simplify:**
\[
x^4 - 22x^2 + 1 = 0
\]
This polynomial is monic (leading coefficient is 1) and integral, and \( x \) is its root.
**Proof of Irrationality:**
Since \( x = \sqrt{5} + \sqrt{6} \) represents a sum of two irrational numbers, it cannot be expressed as a ratio of two integers, and therefore it is algebraically irrational. The polynomial \( x^4 - 22x^2 + 1 = 0 \) further confirms that \( x \) is a root that satisfies this condition.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F004a8e3c-a278-4414-bbaf-77c8d726d7bd%2Fa033fce5-ae23-4170-b113-f5212bf264fa%2F3c7z0m6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find a monic integral polynomial for which \( x \) is a root, where \( x = \sqrt{5} + \sqrt{6} \). Prove that \( x \) is algebraically irrational.
To solve this problem:
1. **Set Up the Expression:**
Let \( x = \sqrt{5} + \sqrt{6} \).
2. **Square Both Sides:**
\[
x^2 = (\sqrt{5} + \sqrt{6})^2 = 5 + 2\sqrt{30} + 6 = 11 + 2\sqrt{30}
\]
3. **Isolate the Square Root:**
\[
x^2 - 11 = 2\sqrt{30}
\]
4. **Square Again to Eliminate the Square Root:**
\[
(x^2 - 11)^2 = (2\sqrt{30})^2 = 120
\]
\[
x^4 - 22x^2 + 121 = 120
\]
5. **Simplify:**
\[
x^4 - 22x^2 + 1 = 0
\]
This polynomial is monic (leading coefficient is 1) and integral, and \( x \) is its root.
**Proof of Irrationality:**
Since \( x = \sqrt{5} + \sqrt{6} \) represents a sum of two irrational numbers, it cannot be expressed as a ratio of two integers, and therefore it is algebraically irrational. The polynomial \( x^4 - 22x^2 + 1 = 0 \) further confirms that \( x \) is a root that satisfies this condition.
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