# Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$

Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?

$$\sqrt{24+8\sqrt{5}}=\sqrt{20+4+2\cdot 2\cdot 2\sqrt{5}}=\sqrt{(2\sqrt5)^2+2^2+2 \cdot 2 \cdot 2\sqrt{5}}=\sqrt{(2+2\sqrt5)^2}=2+2\sqrt5$$

or, to continue on your start, $$\sqrt{6+2\sqrt5}=\sqrt{1^2+(\sqrt5)^2+2\sqrt5}=1+\sqrt5$$

The nested radical: $$\sqrt{6+2\sqrt{5}}=\sqrt{6+\sqrt{20}}$$

can be denested using the identity (that you can easely verify). $$\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}$$

that works if $a^2-b$ is a perfect square. In this case $a^2-b=16$

Hint. One may observe that $$6+2\sqrt{5}=(\sqrt{5})^2+2\sqrt{5}+1=(\sqrt{5}+1)^2.$$

We can generalize this into a formula, but you can ask me later if you want. For now, we denest $\sqrt{24+8\sqrt{5}}$.

To denest, you have to assume that the radical can be rewritten as the sum of two other radicals (surds). So we have $$\sqrt{24+8\sqrt{5}}=\sqrt{x}+\sqrt{y}$$

Squaring both sides gives us $$24+8\sqrt{5}=x+y+2\sqrt{xy}$$

So we have $x+y=24$ and $2\sqrt{xy}=8\sqrt{5}$. So $x\cdot y=80$. This can be easily solved by finding two numbers whose sum is $24$ and their product is $80$.

When $x=20$ and $y=4$, the conditions are met. So $\sqrt{24+8\sqrt{5}}=\sqrt{20}+\sqrt{4}=2+2\sqrt{5}$.

You can start from the identity $$24+8 \sqrt{5} =24+8 \sqrt{5}$$ And write $$24+8 \sqrt{5} =20+4+8 \sqrt{5} = (2+2 \sqrt{5})^{2}$$ Then:

$$\sqrt{24+8 \sqrt{5}}=\sqrt{(2+2 \sqrt{5})^{2}}=2+2 \sqrt{5}$$

To see whether $\sqrt {24+8\sqrt 5 }=A+B\sqrt 5$ with rational $A, B,$ suppose $$24+8\sqrt 5=(A+B\sqrt 5)^2=(A^2+5 B^2)+2 A B\sqrt 5.$$ Since $\sqrt 5$ is not rational, this requires $$24=A^2+5B^2 \;\text {and }\; 2 A B=8.$$ From $2 A B=8$ we have $B=4/A.$ (Note that we cannot have $A=0$ else $24+8\sqrt 5=5 B^2$ is rational, implying $\sqrt 5$ is rational.)

Substituting $4/A$ for $B$ in $24=A^2+5 B^2,$ and multiplying through by $A^2,$ and re-arranging, we get $$(A^2)^2-24 A^2+80=0.$$ This is quadratic in $A^2,$ with solutions $$A^2\in \{4, 20\}.$$ But $A^2=20$ implies $A$ is not rational. This leaves $A^2=4,$ so $A=\pm 2.$ Since we have $B=4/A,$ this implies $(A,B)=(2,2)$ or $(A,B)=(-2,-2).$ Since $A+B\sqrt 5>0,$ we have $A=B=2.$