# How do you simplify sqrt(3-2*sqrt(2))?

We have that

$\sqrt{3 - 2 \cdot \sqrt{2}} = \sqrt{{\left(\sqrt{2}\right)}^{2} - 2 \sqrt{2} \cdot 1 + {1}^{2}} = \left({\sqrt{\left(\sqrt{2} - 1\right)}}^{2}\right) = \sqrt{2} - 1$

Jun 27, 2018

Take the needed answer form with variables and solve for them

#### Explanation:

The answer above is completely correct, but it may help the student to see how to reach it:

We seek a square root of $3 - 2 \sqrt{2}$. This must have the form $a + b \sqrt{2}$.

${\left(a + b \sqrt{2}\right)}^{2} = {a}^{2} + 2 a b \sqrt{2} + 2 {b}^{2}$

So
${a}^{2} + 2 {b}^{2} = 3$
and
$2 a b = - 2 \Rightarrow a b = - 1$
These are two simultaneous equations for $a$ and $b$.

Rearrange the second:
$b = - \frac{1}{a}$
Substitute into the first:
${a}^{2} + \frac{2}{a} ^ 2 = 3$
${a}^{4} + 2 = 3 {a}^{2}$
${a}^{4} - 3 {a}^{2} + 2 = 0$

Note that this is a quadratic in ${a}^{2}$:
${\left({a}^{2}\right)}^{2} - 3 \left({a}^{2}\right) + 2 = 0$
Factorise:
$\left({a}^{2} - 2\right) \left({a}^{2} - 1\right) = 0$

This gives us two possible solutions for ${a}^{2}$: $2$ and $1$, and so the four solutions for $a$: $\pm \sqrt{2}$ and $\pm 1$.

We are looking for integer solutions for $a$, and so $\pm 1$ are possible solutions. But the other two are possible too - they can simply be folded in to the $\sqrt{2}$ term. This wouldn't have been possible if we'd had the root of some other number in the solution for $a$, but this solution is a special case.

Now use the second equation to deduce the four equivalent solutions for $b$:
$b = - \frac{1}{a}$
$b = \overline{+} \frac{1}{\sqrt{2}} = \overline{+} \frac{1}{2} \sqrt{2}$ and $\overline{+} 1$.

So we have the four solution pairs $\left(a , b\right)$:
$\left(\sqrt{2} , - \frac{1}{2} \sqrt{2}\right)$
$\left(- \sqrt{2} , \frac{1}{2} \sqrt{2}\right)$
$\left(1 , - 1\right)$
$\left(- 1 , 1\right)$

This is a bit suspicious - we expect only two solutions, positive and negative square roots, so we wonder if some of these are identical to each other: When we substitute them in to our desired expression $a + b \sqrt{2}$, we get:

$\sqrt{2} - \frac{1}{2} \sqrt{2} \sqrt{2} = - 1 + \sqrt{2}$
$- \sqrt{2} + \frac{1}{2} \sqrt{2} \sqrt{2} = 1 - \sqrt{2}$
$1 - \sqrt{2}$
$- 1 + \sqrt{2}$

So the two solutions with $\sqrt{2}$ are identical to the two simpler solutions, so we can get rid of them. We now have two solutions, positive and negative square roots:
$1 - \sqrt{2}$
$- 1 + \sqrt{2}$

When we take the written square root of a quantity, it is implied that the desired root is the positive root, the "principal value" of the square root function. So we take the single solution that comes from $a = + 1$:
$1 - \sqrt{2}$

Double check: Make sure that this produces the desired answer:
${\left(1 - \sqrt{2}\right)}^{2} = 1 - 2 \sqrt{2} + 2 = 3 - 2 \sqrt{2}$