- #1
How can we prove a given identity?
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SUMMARY
The identity \(\sqrt{22+2\sqrt{5}} + \sqrt{5} = \sqrt{11+2\sqrt{29}} + \sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}}\) is proven by rewriting the right-hand side and simplifying both sides. By recognizing that \(16 = 11 + 5\), the expression is transformed into \(\sqrt{22+2\sqrt{5}} = \sqrt{11+2\sqrt{29}} + \sqrt{11-2\sqrt{29}}\). Squaring both sides confirms the equality, resulting in \(22 + 2\sqrt{5} = 22 + 2\sqrt{121 - 4 \cdot 29}\), which holds true.
PREREQUISITES- Understanding of square roots and their properties
- Familiarity with algebraic manipulation and simplification
- Knowledge of the binomial theorem for squaring expressions
- Basic experience with radical expressions and identities
- Study advanced algebraic identities and their proofs
- Learn about the properties of square roots and radicals
- Explore the binomial theorem and its applications in algebra
- Practice solving complex radical equations
Mathematics students, educators, and anyone interested in algebraic proofs and identities will benefit from this discussion.
- #2
Rogerio
- 404
- 1
The original expression :
\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ <br /> \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}\ }\ }
First, note that 16=11+5 .
So, we can rewrite the last part :
\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ <br /> \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{ ( \ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ ) ^2\ }
Or:
\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ <br /> \sqrt{11+2\sqrt{29}\ }\ +\ (\ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ )
Thats the same as:
\sqrt{22+2\sqrt{5}\ }\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{11-2\sqrt{29}\ }
Squaring both sides, we get:
22+2\sqrt{5}\ = 22\ +\ 2\sqrt{11^2\ -\ (2\sqrt{29})^2\ }
Or:
22+2\sqrt{5}\ = 22\ +\ 2\sqrt{121\ -\ 4*29\ }
Which is true.
\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ <br /> \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}\ }\ }
First, note that 16=11+5 .
So, we can rewrite the last part :
\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ <br /> \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{ ( \ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ ) ^2\ }
Or:
\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ <br /> \sqrt{11+2\sqrt{29}\ }\ +\ (\ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ )
Thats the same as:
\sqrt{22+2\sqrt{5}\ }\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{11-2\sqrt{29}\ }
Squaring both sides, we get:
22+2\sqrt{5}\ = 22\ +\ 2\sqrt{11^2\ -\ (2\sqrt{29})^2\ }
Or:
22+2\sqrt{5}\ = 22\ +\ 2\sqrt{121\ -\ 4*29\ }
Which is true.
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