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!
 A: 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.
A: $$(1+\sqrt3)\sqrt{2-\sqrt3}=\sqrt{(4+2\sqrt3)(2-\sqrt3)}=\sqrt{2(4-3)}=\sqrt2.$$
A: You have $x\sqrt y$.  Bring $x$ inside the square root to get $\sqrt{x^2y}$
A: This is what the OP means by the 'setting equal to X' method.
Let $x = (1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$, and thus:
$$x^2 = (1 + \sqrt{3})^2 \cdot (2 - \sqrt{3}) =(4+2\sqrt{3})(2-\sqrt{3}) =8-2\sqrt{3}^2 = 2 \require{cancel}$$
$$\Rightarrow x = \cancel{-\sqrt{2}}, \sqrt{2}$$
The reason why we have crossed out $-\sqrt{2}$ is because $1 + \sqrt{3} > 1 > 0$, and $\sqrt{2 - \sqrt{3}} > 0$, thus $x = (1 + \sqrt{3})(\sqrt{2 - \sqrt{3}}) > 0$. Hence $x = \sqrt{2}$ is the only possible value.
