algebra precalculus - Simplifying $sqrt[4]{161-72 sqrt{5}}$

$$\sqrt[4]{161-72 \sqrt{5}}$$

I tried to solve this as follows:

the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:

$$a^4+30 a^2 b^2+25 b^4=161$$ $$4 a^3 b+20 a b^3=-72$$

In an attempt to solve this, I first tried to factor stuff and rewrite it as:

$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$ $$4 a b \left(a^2+5 b^2\right)=-72$$

Then letting $p = a^2 + 5b^2$ and $q = ab$ you get

$$4 p q=-72$$ $$p^2+10 q^2=161$$

However, solving this yields messy roots. Am I going on the right path?

Answer

$$\sqrt[4]{161-72\sqrt5}=\sqrt[4]{81-72\sqrt5+80}=\sqrt[4]{(9-4\sqrt{5})^2}=\sqrt{9-4\sqrt{5}}=\sqrt{4-4\sqrt{5}+5}=\sqrt{(2-\sqrt{5})^2}=\sqrt5-2$$ The trick is to notice that $72$ factors into $2*9*4$ and since $9^2+(4\sqrt5)^2=161$ you get this