Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$)

Start by trying to simplify $\sqrt{6-4\sqrt{2}}$. Let's assume there is some number $p+q\sqrt{2}$ for which $$\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$$ Squaring both sides gives $$6-4\sqrt{2} = (p + q\sqrt{2})^2 = p^2+2q^2 + 2pq\sqrt{2}$$ Comparing coefficients gives $6=p^2+2q^2$ and $-4=2pq$, i.e. $-2=pq$.

We need to solve $p^2+2q^2=6$ and $pq = -2$ simultaneously.

If $pq=-2$ then $q=-\frac{2}{p}$ and we can substitute this into $p^2+2q^2=6$. We get \begin{eqnarray*} p^2+2q^2 &=& 6 \\ \\ p^2 + 2\left(-\frac{2}{p}\right)^2 &=& 6 \\ \\ p^2 + \frac{8}{p^2} &=& 6 \\ \\ p^4+8 &=& 6p^2 \\ \\ p^4-6p^2+8 &=& 0 \\ \\ (p^2-2)(p^2-4) &=& 0 \end{eqnarray*}

Either $p^2-2=0$ or $p^2-4=0$, i.e. $p=\pm\sqrt{2}$ or $p=\pm 2$. The only valid solutions are $p = \pm2$ because we usually assume that $p$ and $q$ are rational numbers, i.e. fractions.

If $p=\pm 2$ then $pq=2$ gives $\pm 2q=-2$, and so $q=\mp 1$. Hence $$p+q\sqrt{2} = \pm(2-\sqrt{2})$$

Recall that $\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$ and since, by definition, $\sqrt{6-4\sqrt{2}} \ge 0$ we conclude $$\sqrt{6-4\sqrt{2}} = 2-\sqrt{2}$$

Finally: $$\sqrt{6-4\sqrt{2}} \ \ {\color{red}{+\sqrt{2}}}= 2-\sqrt{2} \ \ {\color{red}{+\sqrt{2}}} = 2$$

Since \begin{align} 6-4\sqrt{2}&=4-2\cdot 2\cdot\sqrt{2}+2\\ &=(2-\sqrt{2})^2 \end{align} Where $2-\sqrt{2}>0$, it follows $\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$.

Then $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\color{blue}{2}$$

In general, $\sqrt{\left(a+b\right)\pm 2\sqrt{a\cdot b}} = \sqrt{a}\pm\sqrt{b}$. So now you are left to find $a$ and $b$ which satisfies $a+b = 6$ and also $a\cdot b = 8$. The answer should be $2$. edited