How do I find the square root of 3 - 2\sqrt 2?

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Homework Help Overview

The problem involves finding the square root of the expression 3 - 2√2, which falls under the subject area of algebra, specifically dealing with square roots and polynomial equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to approach the problem, including setting up equations based on the expression and exploring the implications of different algebraic identities. Some participants question whether the rationality of certain components affects the setup.

Discussion Status

The discussion is active, with participants sharing their attempts and methods. There is no explicit consensus on a single approach, but several lines of reasoning are being explored, including factorization and inspection methods.

Contextual Notes

Some participants express uncertainty about the quickest method to solve the problem, and there are references to the complexity of memorizing formulas for such expressions. Additionally, the discussion includes a mention of trial and error as a viable strategy for identifying solutions.

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Homework Statement


Find the square root of 3 - 2\sqrt 2.


Homework Equations





The Attempt at a Solution


I don't really know how to do this quickly. Could this be done by solving x^2 = 3 - 2\sqrt 2? Or should I solve (a + b)^2 = 3 - 2\sqrt 2? By the way, the answer is 1 - \sqrt 2.
 
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Yeah you got it (a + b)^2 , you know that either a or b is root two, so lay it out like this

a^2 + b^2 = 3
2ab = -2root(2)

It's like factorising, just have a play with it.
 
Ok, here's my working:

(a + b)^2 = 3 - 2 \sqrt 2
a^2 + b^2 + 2ab = 3 - 2 \sqrt 2
a^2 + b^2 = 3 \quad (1) <--- (is this because a^2 + b^2 is a rational?)
2ab = -2 \sqrt 2
ab = - \sqrt 2 \quad (2)

(from 2): a = \frac{-\sqrt 2}{b} \quad (3)
(from 1): a^2 + b^2 = 3
(\frac{-\sqrt 2}{b})^2 + b^2 = 3
\frac{2}{b^2} + b^2 = 3
2 + b^4 = 3b^2
Solving quadratic gives b^2 = 1, 2
b = \pm 1, \pm \sqrt 2
(from 3): a = \frac{-\sqrt 2}{b}
a = \frac{-\sqrt 2}{\pm 1} or a = \frac{-\sqrt 2}{\pm \sqrt 2}
a = \mp \sqrt 2 or a = \mp 1

Hence, roots are \pm(1 - \sqrt 2).

There must be a quick way to do this, or is this the only way?
 
I don't know of any simpler method, however there is a formula for this which I wouldn't recommend trying to memorize if you don't answer questions like these often.

There is another way to approach the same problem, with the same method. Who knows, you might find it interesting:

a^2+b^2=3 (1) , 2ab=-2\sqrt{2} (2)

(a^2+b^2)^2-(2ab)^2=(a^2-b^2)^2

Sub (1) and (2) into equation: 3^2-(-2\sqrt{2})^2=(a^2-b^2)^2

Therefore, (a^2-b^2)^2=1 (3)

but from (1): a^2=3-b^2

Sub (1) into (3): (3-2b^2)^2=1

Solving for b: 3-2b^2=\pm 1 \Rightarrow b^2=\frac{3\pm 1}{2} \Rightarrow b=\pm \sqrt{\frac{3\pm 1}{2}}

Take each case for b, and substitute back into (2) to solve for a.


This other approach doesn't really simplify things, but I only use it just to avoid trying to factorize the quadratic in a2 (yes, I don't enjoy factorizing due to the trail and error nature of it).
 
Just try spotting it out by inspection, trial and error sort of thing, you usually can pick it out after not long at all, took me around 30 seconds or so to see it. Your method is sound however.
 

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