Find the 4th Root of (𝟕+√𝟒𝟖) | Olympiad Math

This is a challenging question. The method we use is not common. We define a as the answer. Here are the steps to reach the final answer:
Step 1: Find 1/a. Step 2: Find a^4 + a^(-4) and a^4 - a^(-4). Step 3: Find a^2 + a^(-2) and a^2 - a^(-2).
Step 4: Find a +1/a and a -1/a.

[佐藤広-cp]

This problem is rooted to the fourth power, so it looks a bit difficult. However, I didn't realize that it was well made and could be solved as easily as in the video. When I tried to solve the same problem as the analogy, it became as follows.
₄√(7+√48)=₄√(7+2√12)=₄√{√4)²+(√3)²+2*√4*√3}=₄√(√4+√3))²=₄√(2+√3)²=√(2+√3), let α=√(2+√3), β=√(2-√3), then α²+β²=4, and α²β²=1, α²β²=(αβ)²=1⇒αβ=1 (α>β>0), (α+β)²=α² +β²+2αβ=4+2*1=6⇒α+β=√6 (α>β>0)
The quadratic equation that solves α and β is (t-α)(t-β)=t²-(α+β)t+αβ=0, so t²-√6t+1=0. Using the formula for the solution of the quadratic equation, t=[√6±√{(-√6) ²-4 * 1 * 1}]/ 2=(√6±√2)/2, 2 solutions are obtained.
Since the relationship between these two solutions is α>β>0, α=₄√(7 +√48)=√(2+√3)=(√6+√2)/2