<s>[INST] Simplify $(1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$ Can someone help me simplify $(1+\sqrt{3})\times\sqrt{2-\sqrt{3}}$?
The end result is $\sqrt{2}$, however, I honestly do not know how to get there using my current skills.
I asked a teacher/tutor and he proposed setting the expression equal to X and working backwards, squaring both sides so that:
$$X^2 = (1 + \sqrt{3})^2 \cdot (2 - \sqrt{3}) =(4+2\sqrt{3})(2-\sqrt{3}) =8-2\sqrt{3}^2 = 2 \require{cancel}$$
$$\Rightarrow X = \sqrt{2}$$
My main question is:

*

*What are the steps to simplify this without setting equal to X?

*I tried watching a youtube video, but had no success - is difference of squares applicable here?

Thanks!
 [/INST] In general, $\sqrt{2 - \sqrt3}$ can be denested using the equation $$\sqrt{X\pm Y} = \sqrt{\frac{X + \sqrt{X^2-Y^2}}{2}} \pm \sqrt{\frac{X - \sqrt{X^2-Y^2}}{2}}.$$ (source)
Setting $X = 2$ and $Y = \sqrt3$ gives $X^2-Y^2=1$ and therefore $$\sqrt{2-\sqrt3} = \sqrt{\frac{2+1}{2}} - \sqrt{\frac{2-1}{2}} = \frac{\sqrt3 - 1}{\sqrt 2}.$$
Multiplying by $\sqrt3 + 1$ turns the numerator into $3-1 = 2$, and $\frac{2}{\sqrt2} = \sqrt2$.
This is longer than bringing $\sqrt3+1$ into the radical in this particular case, but works more generally.
</s>