Simplify $\sqrt{3+\sqrt{8}}$

$\begingroup$

How can you simplify $\sqrt{3+\sqrt{8}}$?

I tried to change $\sqrt8$ to $2\sqrt2$, but I don't know what to do next. Can you simplify it to one layer (square root)?

$\endgroup$ 2

4 Answers

$\begingroup$

$$\sqrt{3+2\sqrt2}=a+b\sqrt2\iff3+2\sqrt2=a^2+2b^2+2\sqrt2ab.$$So, are there integers $a$ and $b$ such that $a^2+2b^2=3$ and $ab=1$? Yes: $a=b=1$.

$\endgroup$ $\begingroup$

Note that $(1 +\sqrt 2)^2= 1 + 2 + 2\sqrt 2 = 3 + 2\sqrt 2 = 3 + \sqrt 8$.

Hence, $\sqrt{3 + \sqrt 8} = 1 + \sqrt 2$ (the positive square root).

$\endgroup$ $\begingroup$

Just $$\sqrt{3+\sqrt{8}}=\sqrt{1+2\sqrt2+2}=\sqrt{(1+\sqrt2)^2}=1+\sqrt2$$

$\endgroup$ $\begingroup$

HINT

Try to use the following formula

$(a+b)^2=a^2+2ab+b^2$ and $\sqrt{a^2}=\left|a\right|$.

$\endgroup$