**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.

We have x = 5+6. The objective here is to show that x is an algebraically irrational number. An…

Answered: V5 + v6 Find a monic integral…

Math

Advanced Math

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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Question

**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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