Simplifying √(3+2√(2)): Find the Integer and Irrational

Click For Summary
SUMMARY

The expression √(3 + 2√(2)) simplifies to 1 + √(2). This conclusion is reached by recognizing that 3 can be expressed as 1 + 2, allowing the square root to be rewritten as √(1 + 2√(2) + 2). The discussion emphasizes the legality of manipulating square roots under certain conditions, clarifying that while breaking down square roots can be tricky, the simplification in this case is valid. Squaring both sides confirms the correctness of the simplification.

PREREQUISITES
  • Understanding of square roots and their properties
  • Familiarity with algebraic manipulation of expressions
  • Knowledge of irrational numbers and their characteristics
  • Basic skills in verifying mathematical identities
NEXT STEPS
  • Study the properties of square roots in algebra
  • Learn about simplifying expressions involving irrational numbers
  • Explore techniques for verifying mathematical identities
  • Practice problems involving the manipulation of radical expressions
USEFUL FOR

Students studying algebra, mathematics educators, and anyone interested in understanding the simplification of radical expressions and the properties of irrational numbers.

chillfactor
Messages
12
Reaction score
0

Homework Statement



√(3+2√(2)) can be simplified into a fairly simple sum of two number: one is an integer and one is an irrational number . Find them and show you are correct by squaring both sides of the "equation"

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Here's a hint: note that
\sqrt{3 + 2\sqrt{2}} = \sqrt{1 + 2\sqrt{2} + 2}
 
isnt that illegal. By the same logic I could simplify the square root of nine to the square root of 9 ones and solve and get nine
 
the whole thing is under the square root .. it won't result to 9.
 
Isn't what illegal? All he did was say that 3=2+1. he didn't break up the square root
 
Writing \sqrt{9}= \sqrt{1+ 1+ 1+ 1 +1 + 1+ 1+ 1+ 1} is perfectly legal.

It is thinking that that last is the sum of
\sqrt{1}+ \sqrt{1}+ \sqrt{1}+ \sqrt{1}+ \sqrt{1}+ \sqrt{1}+ \sqrt{1}+ \sqrt{1}+ \sqrt{1}

that is illegal.
 
Last edited by a moderator:
Sorry. I have removed the "answer" to the problem.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K