simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$

Question

Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$

To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,

$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$

But even with that I don't get that result.

Answer 1

We can recognize both expressions as squares:

• $3+2\sqrt2=2+2\sqrt2+1=(\sqrt2)^2+2\cdot \sqrt2\cdot 1+1^2=(\sqrt2+1)^2$, and
• $4-2\sqrt3=3-2\sqrt3+1=(\sqrt3-1)^2$.

This means that $\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=2+\sqrt2-\sqrt3$.

Answer 2

Hints: $$3+2\sqrt2=(\sqrt2)^2+2\sqrt2+1\qquad 4-2\sqrt3=(\sqrt3)^2-2\sqrt3+1$$

Answer 3

$$\sqrt{3+2\sqrt2}=\sqrt{3+\sqrt8}=\sqrt\frac{3+\sqrt{9-8}}{2}+\sqrt\frac{3-\sqrt{9-8}}{2}=\sqrt2+1$$ $$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt12}=\sqrt\frac{4+\sqrt{16-12}}{2}-\sqrt\frac{4-\sqrt{16-12}}{2}=\sqrt3-1$$ Now we have: $$\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=\sqrt2+2-\sqrt3$$