# How does $-\sqrt {\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}}$ simplify to $1 - \sqrt 2$?

I've the answer for a question in my textbook to be: $-\sqrt {\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}}$ which i've then simplifed to: $-\sqrt {3 - 2\sqrt 2 }$

However my textbook states $-\sqrt {\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}} = 1 - \sqrt 2$

How is this?

$$\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{2-\sqrt{2}}{2+\sqrt{2}}\frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\left(2-\sqrt{2}\right)^{2}}{2}$$
so $$\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}=\frac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1$$
You can check that both answers are correct by squaring $\pm(1-\sqrt{2})$ and observing that you get $3-2\sqrt{2}$. One way to simplify the answer you got is to assume that it can be written as $a+b\sqrt{2}$ and then equate coefficients in $3-2\sqrt{2} = (a+b\sqrt{2})^2$ to solve for $a$ and $b$.
• Could you show how you would solve for $a$ and $b$ please? Jul 27, 2014 at 19:05
• Square the RHS to get $3-2\sqrt{2} = (a^2+2b^2) + 2ab\sqrt{2}$. Then $a^2+2b^2 = 3$ and $2ab = -2$. Substitute for $b$ in the first equation, using a value from the second, and solve the resulting quadratic in $a^2$. Jul 27, 2014 at 19:07
Alternatively, we can say $x = -\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}$. We can say that $x^2 = -\frac{2-\sqrt{2}}{2+\sqrt{2}} \implies -(2+\sqrt{2})x^2 + (-2 + \sqrt{2}) = 0$. We can now apply the quadratic formula and discard the negative root to obtain that $x = -\frac{2\sqrt{2}}{4+2\sqrt{2}} = -\frac{\sqrt{2}}{2+\sqrt{2}} = -\frac{2}{2+2\sqrt{2}} = -\frac{1}{1+\sqrt{2}} = 1-\sqrt{2}.$