Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $

Question

This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$ having only the left-hand side?

Answer 1

Try to write $2(3+2\sqrt 2)=6+4\sqrt 2$ as a square of $a+b\sqrt 2$ for some rational numbers $a,b$: $$(a+b\sqrt 2)^2=a^2+2b^2+2ab\sqrt2 \;=\; 6+4\sqrt 2$$ By $\Bbb Q$-linear independence of $1$ and $\sqrt 2$, this implies $$ab=2,a^2+2b^2=6$$ Can you take it from here?

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Here are some hints, if needed:

If we try to find integer solutions $a,b$, we easily find $a=2,b=1$ as solution. Therefore: $$\sqrt 2 \cdot \sqrt{3+2\sqrt 2}= \sqrt{6+4\sqrt 2} = \sqrt{(2+\sqrt 2)^2}=2+\sqrt 2$$

Answer 2

You may note that $3+2\sqrt{2}=(1+\sqrt{2})^2$. Hence, we have $$ \sqrt{2}\sqrt{3+2\sqrt{2}}-1=\sqrt{2}(1+\sqrt{2})-1=\sqrt{2}+2-1=\sqrt{2}+1, $$ as desired.

Answer 3

To denest the nested radical $\sqrt{3+2\sqrt{2}}$, we have a useful formula. Namely

Given a radical of the form $\sqrt{X\pm Y}$, we can rewrite it into$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}\tag1$$ With $X>Y$ and $X,Y\in\mathbb{R}$.

In $\sqrt{3+2\sqrt{2}}$, we have $X=3,Y=\sqrt8$. Using $(1)$, we get$$\sqrt{3+2\sqrt{2}}=\left(\frac {3+1}2\right)^{1/2}+\left(\frac {3-1}2\right)^{1/2}=\sqrt2+1\tag2$$

From here, can you continue?