# Homework Help: Finding the square root

1. Jun 20, 2009

### deancodemo

1. The problem statement, all variables and given/known data
Find the square root of $$3 - 2\sqrt 2$$.

2. Relevant equations

3. 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 $$1 - \sqrt 2$$.

2. Jun 20, 2009

### Chewy0087

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.

3. Jun 20, 2009

### deancodemo

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?

4. Jun 20, 2009

### Mentallic

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).

5. Jun 21, 2009

### Chewy0087

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.